Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.4

Choose the correct answer

Question 1.
Normal distribution was invented by
(a) Laplace
(b) De-Moivre
(c) Gauss
(d) all the above
Solution:
(b) Demoivre

Question 2.
If X ~ N(9, 81) the standard normal variate Z will be
(a) Z = \(\frac { X-81 }{9}\)
(b) Z = \(\frac { X-9 }{81}\)
(c) Z = \(\frac { X-9 }{9}\)
(d) Z = \(\frac { 9-X }{9}\)
Solution:
(b) Z = \(\frac { X-9 }{81}\)
Hint:
Here µ = 9
σ² = 81
∴ σ = 9
Z = \(\frac { X-µ }{σ}\) = \(\frac { X-9 }{9}\)

Question 3.
If Z is a standard normal variate, the proportion of items lying between Z = -0.5 and Z = -3.0 is
(a) 0.4987
(b) 0.1915
(c) 0.3072
(d) 0.3098
Solution:
(c) 0.3072
Hint:

p(-3.0 < z < -0.5)
= p(0.5 < z < 3.0) – p(0 < z < o.5)
= 0.4987 – 0.1915 = 0.3072

Question 4.
If X ~ N(µ, σ2), the maximum probability at the point of inflexion of normal distribution
(a) (\(\frac { 1 }{\sqrt{2π}}\))e^{\(\frac { -1 }{2}\)}
(b) (\(\frac { 1 }{\sqrt{2π}}\))e^{\(\frac { 1 }{2}\)}
(c) (\(\frac { 1 }{σ\sqrt{2π}}\))e^{\(\frac { -1 }{2}\)}
(d) (\(\frac { 1 }{\sqrt{2π}}\))
Solution:
(c) (\(\frac { 1 }{σ\sqrt{2π}}\))e^{\(\frac { -1 }{2}\)}

Question 5.
In a parametric distribution the mean is equal to variance is
(a) binomial
(b) normal
(c) poisson
(d) all the above
Solution:
(c) poisson

Question 6.
In turning out certain toys in a manufacturing company, the average number of defectives is 1%. The probability that the sample of 100 toys there will be 3 defectives is
(a) 0.0613
(b) 0.613
(c) 0.00613
(d) 0.3913
Solution:
(a) 0.0613
Hint:
Given
p = 0.01 and n = 100
λ = np = 0.01 × 100 = 1
p(x = x) = \(\frac { e^{-λ}λ^x }{x!}\)
p(x = x) = \(\frac { e^{-1}(1)^3 }{3!}\) = \(\frac { 0.3678 }{6}\) = 0.0613

Question 7.
The parameters of the normal distribution
f(x) (\(\frac { 1 }{\sqrt{72π}}\)) \(\frac { e^{-(x-10)^2} }{72}\) – ∞ < x < ∞
(a) (10, 6)
(b) (10, 36)
(c) (6, 10)
(d) (36, 10)
Solution:
(b) (10, 36)
Hint:

Here σ = 6 and µ = 10
∴ σ = (6)² = 36

Question 8.
A manufacturer produces switches and experiences that 2 per cent switches are defective. The probability that in a box of 50 switches, there are at most two defective is:
(a) 2.5 e^-1
(b) e^-1
(c) 2 e^-1
(d) none of the above
Solution:
(a) 2.5 e^-1
Hint:

Question 9.
An experiment succeeds twice as often as it fails. Die chance that in the next six trials, there shall be at least four successes is
(a) 240/729
(b) 489/729
(c) 496/729
(d) 251/729
Solution:
(c) 496/729
Hint:
p = 2 q ⇒ p = 2(1 – p)
p (2 – 2p) ⇒ 3p = 2
p = 2/3 and q = 1 – p ⇒ q = 1 – 2/3
q = 1/3 and n = 6
p(X = x) = nc_xp^xq^n-x
p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)

Question 10.
If for a binomial distribution b(n, p) mean = 4 and variance = 4/3, the probability, P(X ≥ 5) is equal to:
(a) (2/3)⁶
(b) (2/3)⁵( 1/3)
(c) (1/3)⁶
(d) 4(2/3)⁶
Solution:
(d) 4(2/3)⁶
Hint:
In a binomial distribution
mean np = 4 → (1)
variance npq = 4/3 → (2)
(2) ÷ (1) ⇒ npq = 4/3 ⇒ q = 1/3
p = (1 – q) ⇒ p = 1 – 1/3
∴ p = 2/3
p(X = x) = nc_rp^xq^n-x
p(x ≥ 5) = p(x = 5) + p(x = 6)

Question 11.
The average percentage of failure in a certain examination is 40. The probability that out of a group of 6 candidates atleast 4 passed in the examination are:
(a) 0.5443
(b) 0.4543
(c) 0.5543
(d) 0.4573
Solution:
(b) 0.4543
Hint:
given:
n = 6
q = 40/100 = 2/5
p = 1 – q = 1 – 2/5 = 3/5
p(X = x) = nc_xp^xq^n-x
p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)

= \(\frac { 8505 }{5^6}\) = \(\frac { 1701 }{3125}\)
= 0.54432

Question 12.
Forty percent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. For a full flight, what is the mean of the number of passengers who do not check in any luggage?
(a) 6.00
(b) 6.45
(c) 7.20
(d) 7.50
Solution:
(a) 6.00
Hint:
Given
P = \(\frac { 40 }{100}\) and n = 15
Mean np = \(\frac { 40 }{100}\) × 15
= 0.4 × 15
= 6.00

Question 13.
Which of the following statements is/are true regarding the normal distribution curve?
(a) it is symmetrical and bell shaped curve
(b) it is asymptotic in that each end approaches the horizontal axis but never reaches it
(c) its mean, median and mode are located at the same point
(d) all of the above statements are true.
Solution:
(d) all of these above statements are true

Question 14.
Which of the following cannot generate a Poisson distribution?
(a) The number of telephone calls received in a ten-minute interval
(b) The number of customers arriving at a petrol station
(c) The number of bacteria found in a cubic feet of soil
(d) The number of misprints per page
Solution:
(b) The number of customers arriving a petrol station

Question 15.
The random variable X is normally distributed with a mean of 70 and a standard deviation of 10. What is the probability that X is between 72 and 84?
(a) 0.683
(b) 0.954
(c) 0.271
(d) 0.340
Solution:
(d) 0.340
Hint:
In a normal distribution
S.D(σ) = 10
mean(µ) = 70 and

= p(0.2 < z < 1.4) – p(0 < z < 0.2)
= 0.4192 – 0.0793
= 0.3399

Question 16.
The starting annual salaries of newly qualified chartered accountants (CA’s) in South Africa follow a normal distribution with a mean of Rs 180,000 and a standard deviation of Rs 10,000. What is the probability that a randomly selected newly qualified CA will earn between Rs 165,000 and Rs 175,000 per annum?
(a) 0.819
(b) 0.242
(c) 0.286
(d) 0.533
Solution:
(b) 0.242
Hint:
In a normal distribution
µ = 180,000 and σ = 10,000
z = \(\frac { X-µ }{σ}\) = \(\frac { X-180000 }{10000}\)
p(165,000 < x < 1,75,000) = ?
when x = 165,000
z = \(\frac { 165000-180000 }{10000}\) = \(\frac { 15000 }{10000}\)
z = -1.5

when x = 175,000
z = \(\frac { 175000-180000 }{10000}\) = \(\frac { -5000 }{10000}\) = \(\frac { -1 }{2}\)
z = -0.5
p(165,000 < x < 175,000)
= p(-1.5 < z < -0.5)
= p(0.5 < z < 1.5)
= p(0 < z < 1.5) – p(0 < z < 0.5)
= 0.4332 – 0.1915 = 0.2417

Question 17.
In a large statistics class the heights of the students are normally distributed with a mean of 172 cm and a variance of 25 cm. What proportion of students are between 165 cm and 181 cm in height?
(a) 0.954
(b) 0.601
(c) 0.718
(d) 0.883
Solution:
(d) 0.883
Hint:

In a normal distribution
µ = 172; σ² = 25 then σ = 5
z = \(\frac { x-µ }{σ}\) = \(\frac { x-172 }{5}\)
p(165 < x < 181) = ?
when x = 165 z = \(\frac { 165-172 }{5}\) = \(\frac { -7 }{5}\) = -1.4
when x = 181 z = \(\frac { 181-172 }{5}\) = \(\frac { 9}{5}\) = 1.8
p(165 < x < 181) = p(-1.4 < z < 1.8)
= p(-1.4 < z < 0) + p(0 < z < 1.8)
= p(0 < z < 1.4) + p(0 < z < 1.8)
= 0.4192 + 0.4641
= 0.8833

Question 18.
A statistical analysis of long-distance” telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What proportion of calls lasts less than 180 seconds?
(a) 0.214
(b) 0.094
(c) 0933
(d) 0.067
Solution:
(d) 0.067
Hint:
In a normal distribution

µ = 240 and σ = 40

Question 19.
Cape town is estimated to have 21% of homes whose owners subscribe to the satelite service, DSTV. If a random sample of your home in taken, what is the probability that all four home subscribe to DSTV?
(a) 0.2100
(b) 0.5000
(c) 0.8791
(d) 0.0019
Solution:
(d) 0.0019
Hint:
p = \(\frac { 21 }{100}\) = 0.21
p(x = 4) = (0.21) × (0.21) × (0.21) × (0.21)
= 0.00194481

Question 20.
Using the standard normal table, the sum of the probabilities to the right of z = 2.18 and to the left of z = -1.75 is:
(a) 0.4854
(b) 0.4599
(c) 0.0146
(d) 0.0547
Solution:
(d) 0.0547
Hint:

p(z < – 1.75) + p(z > 2.18)
= p(-∞ < z < 0) – p(-1.75 < z < 0) + p(0 < z < ∞) – p(0 < z < 2.19)
= 0.5 – p(0 < z < 1.75) + 0.5 – p(0 < z < 2.18)
= (0.5 – 0.4599) + (0.5 – 0.4854)
= 0.0401 + 0.0146
= 0.0547

Question 21.
The time until first failure of a brand of inkjet printers is normally distributed with a mean of 1,500 hours and a standard deviation of 200 hours. What proportion of printers fails before 1000 hours?
(a) 0.0062
(b) 0.0668
(c) 0.8413
(d) 0.0228
Solution:
(a) 0.0062
Hint:

Question 22.
The weights of newborn human babies are normally distributed with a mean of 3.2 kg and a standard deviation of 1.1 kg. What is the probability that a randomly selected newborn baby weighs less than 2.0 kg?
(a) 0.138
(b) 0.428
(c) 0.766
(d) 0.262
Solution:
(a) 0.138
Hint:

Question 23.
Monthly expenditure on their credit cards, by credit card holders from a certain bank, follows a normal distribution with a mean of Rs 1,295.00 and a standard deviation of Rs 750.00. What proportion of credit card holders spend more than Rs 1,500.00 on their credit cards per month?
(a) 0.487
(b) 0.392
(c) 0.500
(d) 0.791
Solution:
(b) 0.392
Hint:
In a normal distribution

mean m = 1295 and S.D σ = 7.50
p(x > 1500) = ?
when x = 1500
z = \(\frac { 1500-1295 }{750}\) = \(\frac { 205 }{750}\) = 0.273
(ie) p(x > 1500) = p(z > 0.273)
= 0.5 – p(0 < z < 0.273)
= 0.5 – 0.1064
= 0.3936

Question 24.
Let z be a standard normal variable. If the area to the right of z is 0.8413, then the value of z must be:
(a) 1.00
(b) -1.00
(c) 0.00
(d) -0.41
Solution:
(b) -1.00
Hint:

p(-c < z < ∞) = 0.8413
p(-c < z < 0) + 0.5 = 0.8413
p(-c < z < 0) = 0.813 – 0.5 = 0.3413
p(0 < z < c) = 0.3413 ⇒ c = 1.00
∴ -c = -1.00

Question 25.
If the area to the left of a value of z (z has a standard normal distribution) is 0.0793, what is the value of z?
(a) -1.41
(b) 1.41
(c) -2.25
(d) 2.25
Solution:
(a) -1.41
Hint:

p(z < -c) = 0.0793 ie p(z > c) = 0.0793
p(0 < z < ∞) – p(0 < z < c) = 0.0793
0. 5 – p(0 < z < c) = 0.0793
p(0 < z < c) = 0.5 – 0.0793
p(0 < z < c) = 0.4207
c = 1.41
then -c = -1.41

Question 26.
If P(Z > z) = 0.8508 what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) -1.04
(d) 1.04
Solution:
(c) -1.04
Hint:
since the area is greater than 0.5 then the z must be negative

p(z < Z) = 0.8508
p(-z < Z < 0) + p(0 < z < ∞) = 0.8508
p(-z < Z < 0) + 0.5 = 0.8508
p(-z < Z < 0) = 0.8508 – 0.5 = 0.3508
p(0 < Z < z) = 0.3508
z = 1.04 then -z = -1.04

Question 27.
If P(Z > z) = 0.5832 what is the value of z (z has a standard normal distribution)?
(a) -0.48
(b) 0.48
(c) 1.04
(d) -0.21
Solution:
(d) -0.21
Hint:
p(z > Z) = 0.5832
since area >0.5 then then z must be negative

p(- z < Z < ∞) = 0.5832
p(-z < Z < 0) + p(0 < Z < ∞) = 0.5832
p(-z < Z < 0) + 0.5 = 0.5832
p(-z < Z < 0) = 0.0832
z = 0.21
then -z = -0.21

Question 28.
In a binomial distribution, the probability of success is twice as that of failure. Then out of 4 trials, the probability of no success is
(a) 16/81
(b) 1/16
(c) 2/27
(d) 1/81
Solution:
(d) 1/81
Hint:
In a binomial distribution
p = 2q ⇒ p = 2(1 – p)
p = 2 – 2p ⇒ 3p = 2 ⇒ p = 2/ 3
q = 1 – p = 1 – 2/3
∴ q = 1/3
p(X = x) = nc_xp^xq^n-x = 4c_x (\(\frac { 2 }{3}\))^x (\(\frac { 1 }{3}\))^4-x
p(X = 0) = 4c₀ (\(\frac { 2 }{3}\))⁰ (\(\frac { 1 }{3}\))^4-0
= (1)(1)(\(\frac { 1 }{3}\))⁴ = \(\frac { 1 }{81}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Miscellaneous Problems

Question 1.
A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain
(a) no more than 2 rejects?
(b) at least 2 rejects?
Solution:
In a binomial distribution
n = 10; p = \(\frac { 12 }{100}\) = \(\frac { 3 }{25}\); q = 1 – p = 1 – \(\frac { 3 }{25}\); q = \(\frac { 22 }{25}\)
p(X = x) = nc_xp^xq^n-x
(a)p(no more than 2 rejects)
p(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

(b) p (at least 2 rejects) = p (x ≥ 2)

Question 2.
Hospital records show that of patients suffering from a certain disease 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?
Solution:
Let x be the random variable of patience suffering from a certain disease
In a binomial distribution
p (X = x) = nc_xp^xq^n-x

Question 3.
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.
Solution:
In a poisson distribution
mean(λ) = \(\frac { 3 }{20}\) = 0.15
p(X = x) = \(\frac { e^{-λ}λ^x }{x!}\)
p(not be more than one failure) = p(x ≤ 1)
p(x = 0) + p(x = 1)

= 0.86074 × (1.15)
= 0.98981

Question 4.
Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
1. Find the probability that none passes in a given minute.
2. What is the expected number passing in two minutes?
Solution:
In a poisson distribution
Average per hour = 300 vehicles
mean per minute = \(\frac { 300 }{60}\) = 5
∴ λ = 5

= e^-5(12.5)
= 0.0067379 × 12.5
= 0.08422375
= 0.08422375 × 10²

Question 5.
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Raghul wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Raghul takes the test and scores 585. Will he be admitted to this university?
Solution:
Let x denotes the scores of a national test mean
µ = 500 and standard deviation σ = 100
standard normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-5000 }{100}\)
when x = 585
z = \(\frac { 585-500 }{100}\) = \(\frac { 85 }{100}\) = 0.85
p(x ≤ 585) = p(z ≤ 0.85)
p(z ≤ 0.85) = p(-∞ < z < 0) + p(0 < z < 0.85)
= 0.5 + 0.3023
= 0.8023
for n = 100;
p(z ≤ 0.85) = 100 × 0.8023
= 80.23
∴ Raehul scores 80.23%

We can determine the scores of 70% of the students as follows:
from the table for the area 0.35
We get z₁ = -1.4(as z₁ lies to left of z = 0)
similarly z₂ = 1.4

Now z₁ = \(\frac { x_1-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)
-1.4 × 100 = x₁ – 500 ⇒ x₁ 500 – 140
x₁ = 360
Again z₂ = \(\frac { x_2-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)
1.4 × 100 = x₂ – 500 ⇒ x₂ = 140 + 500
= x₂ = 640
Hence 70% of students score between 360 and 640
But Raghul scored 585. His score is not better than the score of 70% of the students.
∴ He will not be admitted to the university.

Question 6.
The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time.
(i) less than 19.5 hours?
(ii) between 20 and 22 hours?
Solution:
Let x denotes the time taken to assemable cars mean µ = 20 hours and S.D σ = 2 hours
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-20 }{2}\)
(i) p(less than 19.5 hours) = p(x < 19.5)
when x = 19.5
z = \(\frac { 19.5 }{2}\) = \(\frac { -0.5 }{2}\) = 0.25
p(x < 19.5) = p(z < – 0.25)
= p(-∞ < z < 0) – p(-0.25 < z < 0)
= 0.5 – p(0 < z < 0.25)
= 0.5 – 0.0987
= 0.4013

(ii) p(between 20 and 22 hours) = p(20 < x < 22)

when x = 20
z = \(\frac { 20-20 }{2}\) = \(\frac { 0 }{2}\) = 0
when x = 22;
z = \(\frac { 22-20 }{2}\) = \(\frac { 2 }{2}\) = 1
p(20 < x < 22) = p(0 < z < 1)
= 0.3413

Question 7.
The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000.
(a) What percent of people earn less than $40,000?
(b) What percent of people earn between $45,000 and $65,000?
(c) What percent of people earn more than $75,000
Solution:
Let x denotes the annual salaries of employees in a large company
mean µ = 50,000 and S.D σ = 20,000
Standard normal variate z = \(\frac { x-µ }{σ}\)
(a) p(people earn less than $40,000) = p(x < 40,000)
when x = 40,000
z = \(\frac { 40,000-50,000 }{20,000}\) = \(\frac { 10,000 }{20,000}\)
z = -0.5
p(x < 40,000) = p(z < -0.5)
= p(-∞ < z < 0) – p(-0.5 < z < 0)
= 0.5 -p(-0.5 < z <0)
= 0.5 – p(0 < z < 0.5) (due to symmetry)
= 0.5 – 0.01915
= 0.3085
= p(x < 40,000) in percentage = 0.3085 × 100 = 30.85

(b) p(people ear between $45,000 and $65,000)
p(45000 < x < 65000)
When x = 45,000;
z = \(\frac { 45,000-50,000 }{20,000}\) = \(\frac { -5000 }{20,000}\) = \(\frac { -1 }{4}\)
z = -0.25
when x = 65,000;
z = \(\frac { 65,000-50,000 }{20,000}\) = \(\frac { 15000 }{20,000}\) = \(\frac { 3 }{4}\)
z = 0.75
p(45000 < x < 65000) = p(-0.25 < z < 0.75)
= p(-0.25 < z < 0) + p(0 < z < 0.75)
= p(0 < z < 0.25) + p(0 < z < 0.75)
= p(0 < z < 0.25) + p(0 < z < 0.75)
= 0.0987 + 0.2734 = 0.3721
p(45000 < x < 65000) in percentage = 0.3721 × 100
= 37.21

p(people earn more than$75,000) = p(x > 70000)
when x = 75,000;
z = \(\frac { 75,000-50,000 }{20,000}\) = \(\frac { 25000 }{20,000}\) = \(\frac { 5 }{4}\)
z = 1.25
p(x > 75,000) = p(x > 1.25)
= p(0 < z < ∞) – p(0 < z < 1.25) = 0.5 – 0.3944 = 0.1056 p(x > 750,000)in percent = 01056 × 100
= 10.56

Question 8.
X is a normally normally distributed variable with mean µ = 30 and standard deviation σ = 4. Find
(a) P(x < 40) (b) P(x > 21)
(c) P(30 < x < 35)
Solution:
x is a normally distributed variable with mean µ = 30 and standard deviation σ = 4

Then the normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-30 }{4}\)

(a) p(x < 40) = ?
when x = 40;
z = \(\frac { 40-30 }{4}\) = \(\frac { 10 }{4}\) = 2.5
p(x < 40) = p(z < 2.5)
= p(-∞ < z < 0) + p(0 < z < 2.5) = 0.5 + 0.4938 = 0.9938 (b) p(x > 21) = ?
when x = 21;
z = \(\frac { 40-30 }{4}\) = \(\frac { 10 }{4}\) = -2.25
p(x > 21) = p(z > -2.25)
= p(-2.25 < z < 0) + p(0 < z < ∞)
= p(0 < z < 2.25) + 0.5
= 0.4878 + 0.5
= 0.9878

(c) p(30 < x < 35) = ?
when x = 30;
z = \(\frac { 30-30 }{4}\) = \(\frac { 0 }{4}\) = 0
when x = 35;
z = \(\frac { 35-30 }{4}\) = \(\frac { 5 }{4}\) = 1.25
p(30 < x < 35) = p(0 < z < 1.25)
= 0.3944

Question 9.
The birth weight of babies is Normally distributed with mean 3,500 g and standard deviation 500 g. What is the probability that a baby is born that weighs less than 3,100 g?
Solution:
Let x be a normally distributed variable with mean 3,500 g and standard deviation 500 g

Here µ = 3500 and σ = 500
The standard normal variate z = \(\frac { x-µ }{σ}\)
p(weight less than variate 3100 g) = p(x < 3100)
when x = 3100;
z = \(\frac { 3100-3500 }{500}\) = \(\frac { -400 }{500}\) = \(\frac { -4 }{5}\)
z = -0.8
∴ p(z < 3100) = p(z < -0.8)
= p(-∞ < z < 0) – p(-0.8 < z < 0)
= 0.5 – p(0 < z < 0.8)
= 0.5 – 0.2881
= 0.2119

Question 10.
People’s monthly electric bills in chennai are normally distributed with a mean of Rs 225 and a standard deviation of Rs 55. Those people spend a lot of time online. In a group of 500 customers, how many would we expect to have a bill that is Rs 100 or less?
Solution:
Let X be a normally distributed variable with a mean of Rs 225 and a standard deviation of Rs 55

Here µ = 225 and σ = 55
The standard normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-225 }{55}\)
p(a bill have Rs 100 or less) = p(x ≤ 100)
when x = 100;
z = \(\frac { 100-225 }{55}\) = \(\frac { -125 }{55}\) = -2.27
p(x ≤ 100) = p(z < -2.27)
p(z < -2.27) = p(-∞ < z < 0) – p(-2.27 < z < 0)
= 0.5 – p(0 < z < 2.27)
= 0.5 – 04884
= 0.0116

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3

Question 1.
Define normal distribution.
Solution:
A random variable X is said to follow a normal distribution with parameters mean µ and varaince σ², if its probability density function is given by

Question 2.
Define standard normal variate.
Solution:
A random variable Z = (X – µ)/σ follows the standard normal distribution. Z is called the standard normal variate with mean 0 and standard deviation 1 i.e Z – N (0, 1). Its Probability density function is given by:
φ(z) = \(\frac { 1 }{\sqrt {2π}}\) e^-x²/2 -∞ < z < ∞

Question 3.
Write down the conditions in which the normal distribution is a limiting case of binomial distribution.
Solution:
The normal distribution of a variable when represented graphically, takes the shape of a symmetrical curve, known as the Normal Curve. The curve is asymptotic to x-axis on its either side.

Question 4. m
Write down any five characteristics of normal probability curve.
Solution:
Chief Characteristics or Properties of Normal Probability distribution and Normal probability Curve.
The normal probability curve with mean µ and standard deviation σ has the following properties:
(i) the curve is bell-shaped and symmetrical about the line x = u.
(ii) Mean, median and mode of the distribution coincide.
(iii) x-axis is an asymptote to the curve, (tails of the curve never touches the horizontal (x) axis)
(iv) No portion of the curve lies below the x-axis as f(x) being the probability function can never be negative.
(v) The points of inflexion of the curve are x = µ ± σ

Question 5.
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and standard deviation of 60 hours. Estimate the number of bulbs likely to burn for
(i) more than 2,150 hours
(ii) less than 1,950 hours
(iii) more 1,920 hours but less than 2,100 hours.
Solution:
Let x denote the burning of the bulb follows normal distribution with mean 2,040 and standard deviation 60 hours.
Here m = 2040; σ = 60 and N = 2000
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-2040 }{60}\)
(i) p(morethan 2,150 hours)
p(x > 2150)
when x = 2150
z = \(\frac { 2150-2040 }{60}\) = \(\frac { 110 }{60}\)= 1.833
p(x > 2150) = p(z > 1.833)
= p(0 < z < ∞) – p(0 < z < 1.833)
= 0.5 – 0.4664
= 0.0336
∴ Number of bulbs whose burning time is more than 2150 hours=0.0336 × 2000
= 67.2 = 67(approximately)

(ii) p(less than 1950 hours)
p(x < 1950)
when x = 1950
z = \(\frac { 1950-3040 }{60}\) = \(\frac { -90 }{60}\)= -1.5
p(x < 1950) = p(z < -1.5) = p(z > 1.5)
= 0.5 – 0.4332
= 0.068

Numbers of bulbs whose burning time is less than
1950 = 0.0668 × 2000 = 133.6
= 134 (approximately)

(iii) p(more 1,920 hours but less than 2,100 hours)
= p(1920 < x < 2100)
when x = 1950
z = \(\frac { 1920-2040 }{60}\) = \(\frac { -120 }{60}\)= -2
when x = 2100
z = \(\frac { 2100-2040 }{60}\) = \(\frac { -60 }{60}\)= -2
∴ p(1920 < x < 2040) = p(-2 < z < 1)
= p(0 < z < 2) + p(0 < z < 1)
= 0.4772 + 0.3413
= 0.8185

∴ Number of bulbs whose burning time more than 1920 hours but less than 2100 hours) = 0.8185 × 2000
= 1637

Question 6.
In a distribution 30% of the items are under 50 and 10% are over 86. Find the mean and standard deviation of the distribution.
Solution:

z = \(\frac { x-µ }{σ}\)
Given that
p(x < 50) = 0.3 p(x > 86) = 0.1
p(z < -c) = 0.3
p(-c < z < 0) = 0.5 – 0.3
p(-c < z < 0) = 0.2 {from the table}
p(0 < z < c) = 0.2
c = 0.53
then -c = -0.53
∴ \(\frac { 50-µ }{σ}\) = -0.53
50 – µ = -0.53σ
µ – 0.53σ = 50 → 1
p(x < 50) = 0.1
p(0 < z < ∞) = -p(0 < z < c₁) = 0.1
p(0 < z < ∞) = p(0 < z < c₁) + 0.1
0.5 = p(0 < z < c₁) + 0.1
p(0 < z < c₁) = 0.5 – 0.1
p = (0 < z < c₁) = 0.4
c₁ = 1.29
∴ \(\frac { 86-µ }{σ}\) = 1.29
86 – µ = 1.29 σ
µ + 1.29σ = 86 → 2
solving eqn 1 & 2
eqn 2 ⇒ m + 1.29σ = 86
eqn 1 ⇒ m + 0.53σ = 50
– + – ………….
………… 1.82 ………… σ = 36 ………….
σ = \(\frac { 36 }{1.82}\) ∴ = 19.78
Substitute σ = 19.78 in eqn 1
µ – 0.53(19.78) = 50
µ -10.48 = 50
µ = 50 + 10.48
µ = 60.48
Mean = 60.48 and standard deviation = 19.78

Question 7.
X is normally distributed with mean 12 and sd 4. Find P (x ≤) 20 and P (0 ≤ x ≤ 12)
Solution:
x is normally distribution with mean 12 and sd 4
∴ µ = 12 and σ = 4
Standard normal variable
z = \(\frac { x-µ }{σ}\) = \(\frac { x-12 }{4}\)
(i) p(x ≤ 20)

when x = 20
z = \(\frac { 20-12 }{4}\) = \(\frac { 8 }{4}\) = 2
v(x ≤ 20) = \(\frac { 8 }{4}\) = 2
p(x ≤ 20) = p(z ≤ 2)
= 0.5 + p(0 < z < 2)
= 0.5 + 0.4772
= 0.9772

(ii) p(0 < x < 12 )

when x = 0
z = \(\frac { 0-12 }{4}\) = \(\frac { -12 }{4}\) = -3
when x = 12
z = \(\frac { 12-12 }{4}\) = \(\frac { 0 }{4}\) = 0
p(0 ≤ x ≤ 12) = p(-3 ≤ z ≤ 0)
= p(0 ≤ z ≤ 3)
= 0.4987

Question 8.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height
(a) greater than 2 inches
(b) less than or equal to 64 inches
(c) between 65 and 71 inches.
Solution:
let x denote the height of a student N = 500; m = 68.0 inches and σ = 3.0 inches the standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-68 }{3}\)
a(greater than 72 inches)
p = p(x > 72)
when x = 72
z = \(\frac { 72-68 }{3}\) = \(\frac { 4 }{3}\) = 1.33
p(x > 72) = p(z > 1.33)
= 0.5 – 0.4082
= 0.0918

Number of students whose height are greater than 72 inches
= 0.0918 × 500
= 45.9
= 46 (approximately)

(b) p(less than or equal to 64 inches)
p(x ≤ 64)
when x = 64
z = \(\frac { 64-68 }{3}\) = \(\frac { -4 }{3}\) = -1.33
p(x ≤ 64) = p(z ≤ -1.33)
p(z ≥ -1.33)
= 0.5 – 0.4082
= 0.0918

∴ Number of heights whose ate less than or equal to 64 inches =0.0918 × 500
= 45.9
= 46 (approximately)

(c) p(between 65 and 71 inches)
p(65 ≤ x ≤ 71)
when x = 65

z = \(\frac { 65-68 }{3}\) = \(\frac { -3 }{3}\) = -1
when x = 71
z = \(\frac { 71-68 }{3}\) = \(\frac { 3 }{3}\) = 1
p(65 ≤ x ≤ 71) = p(-1 < z < 1)
= p(-1 < z < 0) + p(0 < z < 1)
= p(0 < z < 1) + p(0 < z < 1)
= 2 × [p(0 < z < 1)]
= 2 × 0.3413
= 0.6826
∴ Number of students whose height between 65 and 7 inches = 0.6826 × 500
= 341.3
= 342 (approximately)

Question 9.
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second. Find the probability that it will take less than 16.35 seconds to develop prints.
Solution:
let x be the random variable have long the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second

µ = 16,28 and σ = 0.12
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-16.28 }{0.12}\) = 1
p(less than 16.35 seconds) = p(x < 16.35)
when x = 16.35
z = \(\frac { 16.35-16.28 }{0.12}\) = \(\frac { 0.07 }{0.12}\) = 0.583
p(x< 16.35) = p(z < 0.583)
= p(—∞ < z < 0) + p(0 < z < 0.583)
= 0.5 + 0.2190
= 0.7190

Question 10.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height (a) greater than 2 inches (b) less than or equal to 64 inches (c) between 65 and 71 inches.
Solution:
Let x be a normal variate with mean 400 labour days and standard deviation of 100 labour days
m = 400 and σ = 100
The construction work should be completed within 450 days.
The standard normal variate
\(\frac { x-µ }{6}\) = \(\frac { x-400 }{100}\)
personality for 1 labour day = Rs 10,000
If personality amount is = 2,00,000 than No of excess

days = \(\frac { 200000 }{10000}\) = 20
∴ x = 450 + 20 = 470
when x = 470
z = \(\frac { 470-400 }{100}\) = \(\frac { 70 }{100}\) = 0.7
= p(x ≥ 470) = p(z ≥ 0.7)
= 0.5 – 0.2580
= 0.2420

(ii) p(at most 500 days) = p(x ≤ 500 )
when x = 500

z = \(\frac { 500-400 }{100}\) = \(\frac { 100 }{100}\) = 1
p(x ≤ 500) = p(z ≤ 1)
= p(∞ < z < 0) -r- p(0 < z < 1)
= 0.5 + 0.3415
= 0.8413

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Define possion distribution.
Solution:
Poisson distribution was derived in 1837 by a French Mathematician Simeon D. Poisson.
A random variable X is said to follow a Poisson distribution with parameter X if it assumes only non-negative values and its probability mass function is given by

Question 2.
Write any 2 examples for possion distribution
Solution:
1. The number of alpha particles emitted by a radioactive substance in a fraction of a second.
2. Number of road accidents occurring at a particular interval of time per day.

Question 3.
Write the condition for which the possion distribution is limiting case of binomial distribution
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:
(i) n, the number of trials is indefinitely large i.e n → ∞
(ii) p, the constant probability of success in each trial is very small, i.e. p → 0.
(iii) np = λ is finite. Thus p = \(\frac { λ }{n}\) and q = 1 – (\(\frac { λ }{n}\))
where λ, is a positive real number.

Question 4.
Derive the mean and variance of possion distribution.
Solution:
Derivation of Mean and variance of Poisson distribution

Variance (X) = E(X²) – E(X)²

Variance (X) = E(X²) – E(X)²
= λ² + λ – (λ)²
= λ

Question 5.
Mention the properties of possion distribution.
Solution:
Poisson distribution is the only distribution in which the mean and variance are equal.

Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? Give e^(-2.8) = 0.06
Solution:
Since the mortality rate for a certain disease in 7 in loop
∴ p = \(\frac { 7 }{1000}\) and n = 400
The value of mean A = λp = 400 × \(\frac { 7 }{1000}\)
∴ λ = 2.8
Let x be a random variable following
distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
∴ the distribution is p(x = 2) = \(\frac { e^{-2-8}(2.8)^2 }{2!}\)
\(\frac { 0.06×7.84 }{2}\)
= 0.2352

Question 7.
Mention the properties of possion distribution.
Solution:
p(defective bulbs) = \(\frac { 5 }{100}\)
n = 120
The value of mean λ = np = 120 × \(\frac { 5 }{100}\)
λ = 6
Hence, x follows possion distribution with
P(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(x = 0) = \(\frac { e^{-6}(6)^0 }{0!}\) = e^-6
= 0.0025

Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a possion variate, with mean 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused
Solution:
In a possion distribution n=2
mean λ = 1.5
x follows poison distribution
With in p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(neither car is used) = p(x = 0)
\(\frac { e^{-1.5}(1.5)^0 }{0!}\) = e^-1.5
= 0.2231

(ii) p(some demand is refused) = p(x > 2)

= 1 – 0.2231 [1 + 1.5 + 1.125]
= 1 – 0.2231 [3.625]
= 1-0.8087
= 0.1913

Question 9.
The average number of phone calls per minute into the switch board of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be (i) no phone at all (ii) exactly 3 calls (iii) atleast 5 calls
Solution:
The average number of phone cells per minute into the switch board of a company is λ = 2.5
x follows poisson distribution with

(iii) p(atleast 5 calls) = p(x ≥ 5)
= p(x = 5) + p(x = 6) + …………..
= 1 – p(x < 5)

Question 10.
The distribution of the number of road accidents pre day in a city is possion with mean 4. Find the number of days out of 100 days when there will be (i) no accident (ii) atleast 2 accidents and (iii) at most 3 accidents.
Solution:
In a possion distribution
mean λ = 4
n = 100
x follows possion distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) = \(\frac { e^{-4}(4)^x }{x!}\)
(i) p(no accident) = p(x = 0)
= \(\frac { e^{-4}(4)^0 }{0!}\) = e^-4 = 0.0183
out of 100 days there will be no accident
= n × p(x = 0)
= 100 × 0.0183 = 1.83
= 2 days (approximately)

(ii) p(atleast 2 accidents)
= p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 4) + …………
= 1 – p(x < 2)
= 1 – [p(x = 0) + p(x = 1)]
= 1 – [\(\frac { e^{-4}(4)^0 }{0!}\) + \(\frac { e^{-4}(4)^1 }{1!}\)]
= 1 – e^-4 [l + 4]
= 1 – 0.0183(5) = 1 – 0.0915
= 0.9085
= Out of 100 days there will be atleast 2 accidents = n × p(x ≥ 2)
= 100 × 0.9085
= 90.85
= 91 days (approximately)

(iii) p(atmost 3 accident) = p(x ≤ 3)
= p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

out of 100 days there will be at most 3 acccident = n × p(x ≤ 3)
= 100 × 0.4331
= 43.31
= 43 days(approximately)

Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200/ calculate the probability that in a factory employing 300 workers there will be atleast two fatal accidents in year, (given e^-0.25 = 0.7788 Solution:
Let p be the probability of a fatal accident in a factory during the yeart
p = \(\frac { 1 }{1200}\) and n = 300 1200
λ = np = 300 × \(\frac { 1 }{1200}\) = \(\frac { 1 }{4}\)
λ = 0.25
x follows poison distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) + \(\frac { e^{-0.25}(0.25) }{x!}\)
p(atleasttwo fatal accidents) = p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 3) + p(x = 4) + ……….
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}
= 1 – 0.7788 [1 + 0.25]
= 1 – 0.7788(1.25)
= 1 – 0.9735 = 0.0265
∴ p(x ≥ 2) = 0.0265

Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute (i) No customer appears (ii) three or more customers appear.
Solution:
The average number of customers ,who appear in a counter of a certain bank per minute = 2
∴ λ = 2
x follows poisson distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
x follows poisson distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(no customber appears) = p(x = 0)
= \(\frac { e^{-2}(2)^0 }{0!}\) = e^-2
= 0.1353

(ii) p(three or more customers appears) = p(x ≥ 3)
= p(x = 3) + p{x = 4) + p(x = 5) + ……
= 1 – p(x < 3)
= 1 – {p(x = 0) + p(x = 1) + p(x = 2)}

= 1 – 2^-2 [1 + 2 + 2]
= 1 – 0.1353(5)
= 1 – 0.6765
= 0.3235

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1.
Define Biomial distribution.
Solution:
Binomial distribution was discovered by James Bernoulli (1654-1705) in the year 1700.
A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non-negative value and its probability mass function in given by

Question 2.
Define Bernouli trials.
Solution:
A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q respectively, is called a Bernoulli trial.
Some examples of Bernoulli trials are:
(i) Tossing of a coin (Head or tail)
(ii) Throwing of a die (getting even or odd number)

Question 3.
Derive the mean and variance of bionomial distribution
Solution:
Derivation of the Mean and Variance of Binomial distribution:
The mean of the binomial distribution

= np(q + p)n – 1 [since p + q = 1]
= np
E(X) = np
The mean of the binomial distribution is np.
Var(X) = E(X²) – E(X²)

= n(n – 1)p²(q + p)(n – 2) + np
n(n – 1 )p² + np
Variance = E(X²) – [E(X)]²
= n²p² – np² + np – n²p²
= np(1 – p) = npq
Hence, mean of the BD is np and the Variance is npq.

Question 4.
Write down the condition for which the bionomial distribution can be used.
Solution:
The Binomial distribution can be used under the following conditions:
1. The number of trials ‘n finite
2. The trials are independent of each other.
3. The probability of success ‘p’ is constant for each trial.
4. In every trial there are only two possible outcomes – success or failure.

Question 5.
Mention the properties of bionomial distribution.
Solution:
Properties of Binomial distribution
1. Binomial distribution is symmetrical if p = q = 0.5, It is skew symmetric if p ≠ q. It is positively skewed if p < 0.5 and it is negatively skewed it p > 0.5.
2. For Binomial distribution, variance is less than mean
Variance npq = (np) q < np

Question 6.
If 5% of the items produced turn out to be defective, then find out the probability that out of 10 items selected at random there are
(i) exactly three defectives
(ii) atleast two defectives
(iii) exactly 4 defectives
(iv) find the mean and variance.
Solution:
Probability of getting a defective item

In binomial distribution
p(X = x) = nC_xp^xq^n-x

Let x = (19)²
log x = 7 log(19)⁷
= 7 log 19
= 7 × 1.2788
= 8.9516
x = Antilog 8.956
= 8.945 × 10⁸
ut y = (20)¹⁰
log y = (10) log 20
= 10 × 1.3010
= 13.010
y = Anti log 13.010
= 1.023 × 10³

(ii) p(atleast two defectives)
= p(x ≥ 2)
= p(x = 2) = p(x = 3) + ………….. + p(x = 10)
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}

Let x = (19)⁹
log x = 9 log 19
= 9 × 1.2788
= 11.5092
x = Antilog 11.5092
= 3.229 × 1011

(iii) p(extactly 4 defectives) = p(X = 4)

Let x = (19)⁶
log x = 6 log 19
= 6 × 1.2788
log x = 7.66728
Antilog (7.66728)
x = 4.648 × 10⁷

(iv) mean E(x) = np
= 10 × \(\frac { 1 }{20}\) = \(\frac { 1 }{2}\) = 0.5
Varaince = npq
= 10 × \(\frac { 1 }{20}\) = \(\frac { 19 }{20}\) = \(\frac { 19 }{40}\) = 0.475

Question 7.
In a particular university 40% of the students are having news paper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that
(i) none of those selected have news paper reading habit
(ii) all those selected have news paper reading habit
(iii) atleast two third have news paper reading habit.
Solution:
let p to the probability of having newspaper reading habit
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1\(\frac { 2 }{5}\) = \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\) and n = 9
In the binomial distribution p(x = 4) = nc_x p^xq^n-r
The binomial distribution P(x) = 9c_x (\(\frac { 2 }{5}\))^x (\(\frac { 3 }{5}\))^9-x

(i) p(none of those selected have newspaper reading

(ii) P(all those selected have newspaper reading habit)

(iii) p(at least two third have newspaper reading habit)

Question 8.
In a family of 3 children, what is the probability that there will be exactly 2 girls?
Solution:
let p be the probability of getting a girl child= 1/2 q = 1 – p ⇒ = 1 – 1/2
∴ q = \(\frac { 1 }{2}\) and n = 3
In a binomial distibution
p(x = x) nc_x p^xq^n-x
p(exactly 2 girls) p(x = 2)

Question 9.
Defects in yarn manufactured by a local mill can be approximated by a distribution with a mean of 1.2 defects for every 6 metres of length. If lengths of 6 metres are to be inspected, find the probability of less than 2 defects.
Solution:
Given n = 6
mean np = 1.2
⇒ 6p = 1.2
p = \(\frac { 1.2 }{6}\) p = 0.2 (or) p = \(\frac { 1 }{5}\)
q = 1 – p = 1 – \(\frac { 1 }{5}\)
∴ q = \(\frac { 4 }{5}\)
The binomial distribution
p(x) = 6C_x (0.2)^x(0.8)^6-x
p(x < 2) = p(x = 0) + p(x = 1)

Question 10.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
Solution:
n = 4
probability of defective bolts p = 18/100
9 = 1 – p = 1.018 = 0.82
The binoial distribution p(x) = 4C_x(0.18)^x(0.82)^4-x
= 4C₁(0.18)¹(0.82)^4-1
= 4 × 0.18 × (0.82)³
= 0.72 × 0.551368
= 0.39698496
p(X = 1)= 0.3969 approximately

(ii) p(none will be defective)p(x = 0)
= 4C₀(0.18)°(0.82)^4-0
= (1)(1)(0.45212176)
p(x = 0) = 0.45212

(iii) p(almost 2 will be defective) = p(x ≤ 2)
= p(x = 0) + (p(x = 1) + p(x = 2)
= 4C₀(0.18)°(0.82)^4-0 + 4C₁(0.18)₁(0.82)^4-1 + 4C₂ (0.18)² (0.82)^4-2
= (0.82)⁴ + 4 × (0.18) × (0.82)³ + \(\frac { 4×3 }{1×2}\) × (0.18)² (0.82)²
= 0.45212176 + (0.72 × 0.551368) + (6 × 0.0324 × 0.6724)
= 0.45212176 + 0.39698496 + 013071456
= 0.97982128
= 0.9798

Question 11.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
(i) exactly one will be defective
(ii) none will be defective
(iii) atmost 2 will be defective
Solution:
p = 0.09 = \(\frac { 9 }{100}\)
q = 1 – p ⇒ q – 1 – \(\frac { 9 }{100}\)
q = \(\frac { 100-9 }{100}\)
q = \(\frac { 91 }{100}\)
In a binomial distribution p(x = x) = nc_xp^xq^n-x
p(atleast one success) = p(x ≥ 1)

Taking log on both sides log ≤ (0.66) ≤ n log(0.91)
∴ n ≥ log \(\frac { log(0.66) }{log(0.91)}\)
n ≥ 5
∴ 5 or more trials are needed

Question 12.
Among 28 professors of a certain department, 18 drive foreign cars and 10 drive local made cars. If 5 of these professors are selected at random, what is the probability that atleast 3 of them drive foreign cars?
Solution:
Let p be the probability of foreign cars driven by the professors
p = \(\frac { 18 }{28}\) = \(\frac { 9 }{14}\)
Let q be the probability of local made cars driv¬en by the professors
q = \(\frac { 10 }{28}\) = \(\frac { 5 }{14}\)
and n = 5
The binomial distribution p(x = x) = nc_xp^xq^n-x

Question 13.
Out of 750 families with 4 children each, how many families would be expected to have (i) atleast one boy (ii) atmost 2 girls (iii) and children of both sexes? Assume equal probabilities for boys and girls?
Solution:
Assume equal probabilities for boys girls let p be the probability of having a boy Let x be the random variable for getting either a boy or a girl
∴ p = \(\frac { 1 }{2}\) and q \(\frac { 1 }{2}\) and n = 4
In binomial distribution p(x = 4) = nc_xp^xq^n-x
Here the binomial distribution is p(X= x)

= 1 – 0.0625
= 0.9375
for 750 families (p ≥ 1) = 750 × 0.9375
= 703.125
= 703(approximately)

(ii) p(almost 2 girls) = p(x ≤ 2)
= p(x = 0) = p(x = 1) + p(x = 2)

For 750 families p(x ≤ 2) = 0.6875 × 750
= 515.625
= 516 (approximately)

(iii) p(children of both sexes) = p(x = 1) + p(x = 2)p(x = 3)

= 0.875 x 750
For 750 families p(x = 2) = 0.875 × 750
= 656.25
= 656 (approximately)

Question 14.
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers,
Solution:
Given n = 5
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1 – \(\frac { 2 }{5}\) \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\)
The binomial distributionp (X = x) = 15c_x (\(\frac { 2 }{5}\))^x (\(\frac { 3 }{5}\))^15-x
(i) p(probability that 3 will have a laptop) = p(x = 3)

(ii) p(12 of the traels will not have a laptop)
= 1 – p(x = 12)
= 1 – 15c₁₂ (\(\frac { 2 }{5}\))¹² (\(\frac { 3 }{5}\))^15-12

(iii) p(at least three of the travelers have a laptop)
= p(x ≥ 3)
p(x ≥ 3) = 1 – p (x < 3)
= 1 – [p(x = 0) + p(x = 1) + p(x = 2)]
p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)

Question 15.
A pair 5f dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes.
Solution:
In a throw of a pair dice the doublets are (1, 1),(1, 2) (3, 3),(4, 4),(5, 5),(6, 6)
probability of getting a doublet p = \(\frac { 6 }{36}\) = \(\frac { 1 }{6}\)
⇒ q = 1 – q = \(\frac { 5 }{6}\) and n = 4 is given
The probability of success = \(\left[\begin{array}{l}
4 \\
x
\end{array}\right]\) [latex]\frac { 1 }{6}[/latex]^x [latex]\frac { 5 }{6}[/latex]^4-x
Therefore the probability of 2 success are

Question 16.
The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution.
Solution:
In a binomial distribution
mean np = 5 → (1)
Standard deviation \(\sqrt { npq}\) = 2
squaring on body sides
npq = 4 → (2)

n = 5 × 5 ⇒ n = 25
∴ the binomial distribution is
P(X = x) = nc_xp^xq^n-x
(i.e) p(X = x) = \(\left[\begin{array}{l}
25 \\
x
\end{array}\right]\) (\(\frac { 1 }{5}\))^x (\(\frac { 4 }{5}\))^(25-x)

Question 17.
Determine the binomial distribution for which the mean is 4 and variance 3. Also find P(X = 15)
Solution:
In a binomial distribution
mean np = 4 → (1)
variance npq = 3 → (2)

n = 4 × 4 ⇒ n = 16
The binomial distribution is p(X = x)nc_xp^xq^n-x

Question 18.
Assume that a drug causes a serious side effect at a rate of three patients per one hundred. What is the probability that atleast one person will have side effects in a random sample of ten patients taking the drug?
Solution:
Here n = 10

= 0.5344

Question 19.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease.
Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
n = 5
Let probability of recovery p = 0.73
q = 1 – p = 1 – 0.73
∴ q = 0.27
The binomial distribution is
p (x = x) = nC_xp^xq^n-x
p(x = x) = 5C_x(0.73)^x(0.27)^5-x
p(atleast 3 of the 5 mice recover) = p(x ≥ 3)
= p(x = 3) + p(x = 4) + p(x = 5)
= 5C₃ (0.73)³(0.27)^5-3 + 5C₄ (0.73)⁴ (0.27)^5-4 + 5C₅ (0.73)⁵ (0.27)^5-5
5C₂ (0.73)³ (0.27)² + 5c₁ (0.73)⁴ (0.27)¹ + 5c₀(0.73)⁵(0.27)°
[\(\frac { 5×4 }{1×2}\) × 0.389017 × 0.0729] + [5 × 0.28398241 × 0.27] + (1 × 0.2073071593 × 1)
= 0.283593393 + 0.3833762535 + 0.2073071593
= 0.2836 + 0.3834 + 0.2073
= 0.8743

Question 20.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease. Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
Success = 2 × fails
p = 2q ⇒ p = 2(1 – p)
p = 2 – 2p ⇒ p + 2p = 2
3p = 2 and p = 2/3
q = 1 – p = 1 – 2/3
q = 1/3 and n = 5
The binomial destribution is
p (X = x) = nC_xp^xq^n-x
= 5C(2/3)^x (1/3)
(i) p(three successes) = p(x = 3)

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 12th Business Maths and Statistics Book Answers Solutions Guide Pdf Free Download of Volume 1 and Volume 2 in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.

Let us look at these TN Board Samacheer Kalvi 12th Std Business Maths Guide Pdf Free Download of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas and revise our understanding of the subject.

Students can also read Tamil Nadu 12th Business Maths Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 12th Business Maths Book Solutions Answers Guide

Samacheer Kalvi 12th Business Maths Book Back Answers

Tamilnadu State Board Samacheer Kalvi 12th Business Maths Book Volume 1 Solutions

Samacheer Kalvi 12th Business Maths Guide Chapter 1 Applications of Matrices and Determinants

• Chapter 1 Applications of Matrices and Determinants Ex 1.1
• Chapter 1 Applications of Matrices and Determinants Ex 1.2
• Chapter 1 Applications of Matrices and Determinants Ex 1.3
• Chapter 1 Applications of Matrices and Determinants Ex 1.4
• Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

12th Business Maths Book Back Answers Chapter 2 Integral Calculus I

• Chapter 2 Integral Calculus I Ex 2.1
• Chapter 2 Integral Calculus I Ex 2.2
• Chapter 2 Integral Calculus I Ex 2.3
• Chapter 2 Integral Calculus I Ex 2.4
• Chapter 2 Integral Calculus I Ex 2.5
• Chapter 2 Integral Calculus I Ex 2.6
• Chapter 2 Integral Calculus I Ex 2.7
• Chapter 2 Integral Calculus I Ex 2.8
• Chapter 2 Integral Calculus I Ex 2.9
• Chapter 2 Integral Calculus I Ex 2.10
• Chapter 2 Integral Calculus I Ex 2.11
• Chapter 2 Integral Calculus I Ex 2.12
• Chapter 2 Integral Calculus I Miscellaneous Problems

12th Business Maths Solution Book Chapter 3 Integral Calculus II

• Chapter 3 Integral Calculus II Ex 3.1
• Chapter 3 Integral Calculus II Ex 3.2
• Chapter 3 Integral Calculus II Ex 3.3
• Chapter 3 Integral Calculus II Ex 3.4
• Chapter 3 Integral Calculus II Miscellaneous Problems

12th Samacheer Business Maths Book Chapter 4 Differential Equations

• Chapter 4 Differential Equations Ex 4.1
• Chapter 4 Differential Equations Ex 4.2
• Chapter 4 Differential Equations Ex 4.3
• Chapter 4 Differential Equations Ex 4.4
• Chapter 4 Differential Equations Ex 4.5
• Chapter 4 Differential Equations Ex 4.6
• Chapter 4 Differential Equations Miscellaneous Problems

12th Business Maths Guide Volume 1 Chapter 5 Numerical Methods

• Chapter 5 Numerical Methods Ex 5.1
• Chapter 5 Numerical Methods Ex 5.2
• Chapter 5 Numerical Methods Ex 5.3
• Chapter 5 Numerical Methods Miscellaneous Problems

Tamilnadu State Board Samacheer Kalvi 12th Business Maths Book Volume 2 Solutions

12th Business Maths Guide Volume 2 Chapter 6 Random Variable and Mathematical Expectation

• Chapter 6 Random Variable and Mathematical Expectation Ex 6.1
• Chapter 6 Random Variable and Mathematical Expectation Ex 6.2
• Chapter 6 Random Variable and Mathematical Expectation Ex 6.3
• Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

TN 12th Business Maths Solution Book Chapter 7 Probability Distributions

• Chapter 7 Probability Distributions Ex 7.1
• Chapter 7 Probability Distributions Ex 7.2
• Chapter 7 Probability Distributions Ex 7.3
• Chapter 7 Probability Distributions Ex 7.4
• Chapter 7 Probability Distributions Miscellaneous Problems

12th Std Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference

• Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1
• Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2
• Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3
• Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

12th Business Maths Book Answer Key Chapter 9 Applied Statistics

• Chapter 9 Applied Statistics Ex 9.1
• Chapter 9 Applied Statistics Ex 9.2
• Chapter 9 Applied Statistics Ex 9.3
• Chapter 9 Applied Statistics Ex 9.4
• Chapter 9 Applied Statistics Miscellaneous Problems

12th Business Maths Book Pdf Chapter 10 Operations Research

• Chapter 10 Operations Research Ex 10.1
• Chapter 10 Operations Research Ex 10.2
• Chapter 10 Operations Research Ex 10.3
• Chapter 10 Operations Research Ex 10.4
• Chapter 10 Operations Research Miscellaneous Problems

We hope these Tamilnadu State Board Class 12th Business Maths Book Solutions Answers Guide Volume 1 and Volume 2 Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 12th Standard Business Maths Guide Pdf Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas, drop a comment below and we will get back to you as soon as possible.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 1.
The probability function of a random variable X is given by

Evaluate the following probabilities.
(i) P(X ≤ 0)
(ii) P(X ≤ 0)
(iii) P( X ≤ 2) and
(iv) P(0 ≤ X ≤ 10)
Solution:
(i) From the data

x	-2	0	10
P(x = x)	1/4	1/4	1/2

(i) P(x ≤ 0) = P( x = -2) + P(x = 0)
= \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\) = \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(ii) P(x < 0) = P(x = -2) = \(\frac { 1 }{4}\)

(iii) P(|x| ≤ 2) = P(-2 ≤ X ≤ 2)
= P(X = -2) + P(X = 0) = \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\)
= \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(iv) p(0 ≤ X ≤ 10) = P (X = 0) + P(X = 10)
= \(\frac { 1 }{4}\) + \(\frac { 1 }{2}\) = \(\frac { 1+2 }{4}\) = \(\frac { 3 }{4}\)

Question 2.
The probability function of a random variable X is given by

(a) Compute: (i) P(1 ≤ X ≤ 2) and (ii) P(X = 3).
(b) Is X a discrete random variable? Justify your answer.
Solution:
W.K.T Probability density Function
f(x) = \(\frac { d[F(x)] }{dx}\)

(ii) P(X = 3) = 0
(b) X is not discrete since f is not a step function.

Question 3.
The p.d.f. of X is defined as
f(x) = \(\left\{\begin{array}{l}
\mathrm{k}, \text { for } 0<x \leq 4 \\
0, \text { otherwise }
\end{array}\right.\)
Solution:
Let X and a random variable if a Probability density function

Question 4.
The probability distribution function of a discrete random variable X is

where k is some constant. Find (a) k and (b) P(X > 2).
Solution:
(a) Let X be the random variable of a probability distribution function
W.K.T Σpi = 1
P(x = 1) + P(x = 3) + P(x = 5) = 1
2k + 3k + 4k = 1
9k – 1 ⇒ k = 1/9

(b) P(x > 1) = P(x = 3) + P(x = 5)
= 3k + 4k = 7k
= 7(1/9) = 7/9

Question 5.
The probability distribution function of a discrete random variable X is
f(x) = \(\left\{\begin{array}{l}
\mathrm{a}+\mathrm{b} x^{2}, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
where a and b are some constants. Find (i) a and b if E(X) = 3/5 (ii) Var(X).
Solution:
Let X be due continuous variable of density function

Question 6.
Prove that if E(X) = 0, then V(X) = E(X²)
Solution:
V(X) = E(X²) – [E(X)]²
= E(X²) – 0 {Given that E(X) = 0}
Var(X) = E(X²)
Hence proved.

Question 7.
What is the expected value of a game that works as follows: I flip a coin and, if tails pay you ₹ 2; if heads pay you ₹ 1. In either case I also pay you ₹ 50.
Solution:
Let x be the remain variable denoting the amount paying for a game of flip coin then x and takes 2 and 1
P(X = 2) = \(\frac { 1 }{2}\) (getting a head)
p(X = 1) = \(\frac { 1 }{2}\) (getting a tail)
Hence the probability of X is

x	2	1
P(x = x)	1/2	1/2

Expected value E(X) = \(\sum_{ x }\)x P(x)
= (2)(\(\frac { 1 }{2}\)) + 1 (\(\frac { 1 }{2}\))
= 1 + \(\frac { 1 }{2}\) = \(\frac { 3 }{2}\)
Since I pay you ₹ 50 in either case
E(X) = 50 × 3/2 = ₹ 75

Question 8.
Prove that, (i) V(aX) = a²V(X) , and (ii) V(X + b) = V(X).
Solution:
LHS = V(ax)
= E(ax)² – [E(ax)]²
= a² E(x²) – [aE(x)]²
= a²E(x²) – a²E(x)]²
= a²E(x²) – E(x)²]
= a²v(x)
= RHS
Hence proved

(ii) LHS = V(x + b)
= E (x + b)² – [E(x + b)]²
= E(x² + 2bx + b²) – [E(x) + b]² –
= E(x²) + 2bE(x) + b² – [E(x)]² + b² + 2bE(x)]
= E(x²) + 2bE(x) + b² -[E(x)]² + b² – 2bE(x)]
= E(x²) – [E(x)]²
= V(x)
= RHS
LHS = RHS Hence proved.

Question 9.
Consider a random variable X with p.d.f.
f(x) = \(\left\{\begin{array}{l}
3 x^{2}, \text { if } 00 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable

Question 10.
The time to failure in thousands of hours of an important piece of electronic equipment used in a manufactured DVD player has the density function
f(x) = \(\left\{\begin{array}{l}
2 e^{-2 x}, x>0 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.
Find the expected value for the random variable of an unbiased die
Solution:
When a un based die is thrown , any one of the number 1, 2, 3, 4, 5, 6, can turn up that x denote the random variable taking the values from 1 to 6

Question 2.
Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table

Solution:
Let X be a random variable taking values 0, 1, 2, 3,
Expected value E(x) = Σp₁x₁
= (0.2 × 0) + (0.1 × 1) + (0.4 × 2) + (0.3 × 3)
= 0 + 0.1 + 0.8 + 0.9
∴ E(x) = 0.18

Question 3.
The following table is describing about the probability mass function of the random variable X

Find the standard deviation of x.
Solution:
Let x be the random variable taking the values 3, 4, 5
E(x) = Σpixi
= (0.1 × 3) + (0.1 × 4) + (0.2 × 5)
0.3 + 0.4 + 1.0
E(x) = 1.7
E(x²) = Σpixi²
= (0.1 × 3²)+ (0.1 × 4²) + (0.2 × 5²)
= (0.1 × 9) + (0.1 × 16) + (0.2 × 25)
E(x²) = 7.5
Var(x) = E(x²) – (E(x)]²
= 7.5 – (1.7)²
= 7.5 – 2.89
Var(x) = 4.61
Standard deviation(S.D) = \(\sqrt { var (x)}\)
= \(\sqrt { 4.61}\)
σ = 2.15

Question 4.
Let X be a continuous random variable with probability density function
f_x (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected value of X.
Solution:
Let x be a continuous random variable. In the probability density function, Expected

Question 5.
Let X be a continuous random variable with probability density function
f_x (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the mean and variance of X.
Solution:
Let x be a continuous random variable. In the probability density function,

Question 6.
In an investment, a man can make a profit I of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38 . Find the expected gain.
Solution:

X	5000	-8000
P(x = x)	0.62	0.38

Let x be the random variable of getting gain in an Investment
E(x) be the random variable of getting gain in an Investment
E(x) = ΣPixi
= (0.62 × 5000) + [0.38 × (-8000)]
= 3100 – 3040
E(x) = 60
∴ Expected gain = ₹ 60

Question 7.
What are the properties of mathematical expectation?
Properties of Mathematical expectation
Solution:
i. E(a) = a, where ‘a’ is a constant.
ii. E(aX) = aE(X)
iii. E(aX + b) = aE(X) + b, where ‘a’ and ‘b’ are constants.
iv. If X ≥ 0, then E(X) ≥ 0
v. V(a) = 0
vi. If X is random variable, then V(aX + b) = a²V(X)

Question 8.
What do you understand by mathematical expectation?
Solution:
The average value of a random Phenomenon is termed as mathematical expectation or expected value.
The expected value is weighted average of the values of a random variable may assume

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
The variance of X is defined by
Var(X) = Σ[x – E(X)]² p(x)
If X is discrete random variable with probability mass function p(x).
Var(X) = \(\int_{ -∞ }^{∞}\) [x- E(X)]² f_x (x) dx
If X is continuous random variable with probability density function f_x (x).

Question 10.
Define mathematical expectation in tears of discrete random variable?
Solution:
Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by
E(X) = \(\sum_{ x }\) x p(x) ……(1)

Question 11.
State the definition of mathematical expectation using continous random variable.
Solution:
If X is a continuous random variable and fix) is the value of its probability density function at x, the expected value of X is
E(X) = \(\int_{ -∞ }^{∞}\) x f(x) dx …….(2)

Question 12.
In a business venture a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What is his expected, variance and standard deviation of profit?
Solution:
Let X be the random variable of getting profit in a business

X	2000	-1000
P(x = x)	0.4	0.6

E(x) = Σ_xxp_x(x)
= (0.4 × 2000) +[0.6 × (-1000)]
= 800 – 600
E(X) = 200
∴ Expected value of profit = ₹ 200
E(X²) = Σx² P_x(x)
= [(2000)² × 0.4] + [(-1000)² × 0.6]
= (4000000 × 0.4) + (1000000 × 0.6)
E(X²) = 2200000
Var(X) = E(X²) – [E(X)]²
= 22000000 – (200)²
= 2200000 – 40000
Var(X) = 21,60,000
Variance of his profit = ₹ 21,60,000
Standard deviation(S.D) = \(\sqrt { var (x)}\)
σ = \(\sqrt { 2160000}\)
σ = 1469.69
Standard deviation of his profit is ₹ 1,469.69

Question 13.
The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f.

Find the expected number of miles (in thousands) a tire would last until it reaches the critical tread wear point.
Solution:
We know that,

Question 14.
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.
Solution:
Let X denote the amount the person receives in a game
Then X takes values 4,-2 and
So P(X = 4) = P (of getting a head)
= \(\frac { 1 }{2}\)
P(X = – 2) = P (of getting a tail)
= \(\frac { 1 }{2}\)
Hence the Probability distribution is

X	4	-2
P(x = x)	1/2	1/2

E(x²) = 10
Var(x) = E(x²) – E(x²)
= 10 – (1)²
Var(x) = 9
∴ His expected gain = ₹ 1
His variance of gain = ₹ 9

Question 15.
Let X be a random variable and Y = 2X + 1. What is the variance of Y if variance of X is 5?
Solution:
Var(X) = 5
Var(Y) = var (2x + 1) {∴ v(ax + b) = a²v(x)}
= (2)² var(X)
= 4(5)
Var() = 20

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 1.
Construct cumulative distribution function for the given probability distribution.

Solution:
F(0) = P(x ≤ 0) = p(0) = 0.3
F(1) = P(x ≤ 1) = p(0) + p(1)
= 0.3 + 0.2 = 0.5
F(2) = P(x ≤ 2) = P(0) + P(1) + P(2)
= 0.3 + 0.2 + 0.4 = 0.9
F(3) = P(x ≤ 3) = P(0) + P(2) + P(3) + P(4)
= 0.3 + 0.2 + 0.4 + 0.1 = 1

Question 2.
Let X be a discrete random variable with the following p.m.f

Find and plot the c.d.f. of X.
Solution:
F(3) = P(x ≤ 3) = P(3) = 0.3
F(5) = P(x ≤ 5) = P(x = 3) + (x = 5)
= 0.3 + 0.2 = 0.5
F(8) = P(x ≤ 8) = P(3) + P(5) + P(8)
= 0.3 + 0.2 + 0.3
= 0.8
F(10) = P(x ≤ 10)
= P(3) + P(5) + P(8) + P(10)
= 0.3 + 0.2 + 0.3 + 0.2
= 1

Question 3.
The discrete random variable X has the following probability function.

where k is a constant. Show that k = \(\frac { 1 }{18}\)
Solution:
From the data
P(x = 2) = kx
= 2k
P(x = 4) = kx
= 4k
P(x = 6) = kx
= 6k
P(x = 8) = k(x – 2)
= k(8 – 2) = 6k
Since P(X = x) is a probability mass function
\(\sum_{x=2}^{8}\) P(X = x) = 1
\(\sum_{i=2}^{∞}\) P(xi) = 1
(i.e) P(x = 2) + P(x = 4) + P(x = 6) + P(x = 8) = 1
2k + 4k + 6k + 6k = 1
18k = 1
∴ k = \(\frac { 1 }{18}\)

Question 4.
The discrete random variable X has the probability function

Solution:
From the data
P(x = 1) = k
p(x = 2) = 2k
p(x = 3) = 3k
P(x = 4) = 4k
Since P(X = x) is a probability Mass function
\(\sum_{x=1}^{4}\) P(X = x) = 1
\(\sum_{i=1}^{∞}\) P(x_i) = 1
p(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = 1
P(x = 1) + P(x = = 2) + P(x = 3) + P(x = 4) = 1
k + 2k + 3k + 4k = 1
10k = 1
k = \(\frac { 1 }{10}\)
∴ k = 0.1

Question 5.
Two coins are tossed simultaneously. Getting a head is termed as success. Find the t probability distribution of the number of successes.
Solution:
Let X is the random variable which counts the Number of Heads when the coins are tossed the outcomes are stated below

Question 6.
The discrete random variable X has the probability function.

(i) Find k
(ii) Evaluate p( x < 6), p(x ≥ 6)and p(0 < x < 5) (iii) If P(X ≤ x) > 1, 2 then find the minimum value of x.
Solution:
(i) Since the condition of Probability Mass function
\(\sum_{i=2}^{∞}\) P(x_i) = 1
\(\sum_{i=0}^{7}\) P(x_i) = 1
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P P(x = 6) + p(x = 7) = 1
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
10k² + 9k – 1 = 0
10k² + 10k – k – 1 = 0
10k(k + 1) -1(k + 1) = 0
(k + 1)(10k – 1) = 0
k + 1 = 0; 10k – 1 = 0
k = -1 10k = 1
k = \(\frac { 1 }{10}\)
k = -1 is not possible

(ii) P(x < 6)
P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)
= 0 + k + 2k +2k + 3k + k²
= 8k + k²

P(x ≥ 6) = P(x = 6) + p(x = 7)
= 2k² + 7k² + k
= 9k² + k

P(0 < x < 5)
= P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)
= k + 2k + 2k + 3k
= 8k = 8(\(\frac { 1 }{10}\))
∴ P(0 < x < 5) = \(\frac { 8 }{10}\)

(iii) We want the minimum value of x for which P(X ≥ x) > \(\frac { 1 }{2}\)
Now P(X ≤ 0) = 0 < \(\frac { 1 }{2}\)
P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k = k
\(\frac { 1 }{10}\) < \(\frac { 1 }{2}\)
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k = 3k
\(\frac { 3 }{10}\) \(\frac { 1 }{2}\)
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)
= 0 + k + 2k + 2k = 5k
= \(\frac { 5 }{10}\) = \(\frac { 1 }{2}\)
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)
= 0 + k + 2k + 2k + 3k = 8k
= \(\frac { 8 }{10}\) > \(\frac { 1 }{2}\)
This Shows that the minimum value of X for which P(X ≤ x) > \(\frac { 1 }{2}\) is 4

Question 7.
The distribution of a continuous random variable X in range (-3, 3) is given by p.d.f.

Verify that the area under the curve is unity.
Solution:
Since(-3, 3) in the range of given p.d.f
Area under the curve A = \(\int_{-∞}^{∞}\) f(x)dx = \(\int_{-3}^{3}\) f(x)dx

Hence the Area under the curve is unity

Question 8.
A continuous random variable X has the following distribution function:

Find (i) k and (ii) the probability density function.
Solution:
(i) Here F(3) – F(1) = 1
K(3 – 1)⁴ – O = 1
K(2)⁴ = 1
K(16) = 1
k = \(\frac { 1 }{16}\)

(ii) The Probability density function
f(x) = \(\frac { d(F(x)) }{dx}\) = \(\left\{\begin{array}{l}
4 k(x-1)^{3}, 1<x \leq 3 \\
0 \text { else where }
\end{array}\right.\)

Question 9.
The length of time (in minutes) that a certain person speaks on the telephone is found to be random phenomenon, with a probability function specified by the probability density function f(x) as

(a) Find the value of A that makes f (x) a p.d.f.
(b) What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes.
Solution:
(a) Since f(x) is a probability density Function

b (i) more than 10 minutes

Question 10.
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function

(a) Is the distribution function continuous? If so, give its probability density function?
(b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes?
Solution:
(i) Yes, the distribution function is continuous on [0, 4]
The probability density function

(b) The probability that a person will have to wait
(i) more than 3 minutes

(iii) between 1 and 3 minutes

Question 11.
Define random variable.
Solution:
A random variable (r.v.) is a real valued function defined on a sample space S and taking values in (-∞, ∞) or whose possible values are numerical outcomes of a random experiment.

Question 12.
Explain What are the types of random variable.
Solution:
Random variables are classified into two types namely discrete and continuous random variables these are important for practical applications in the field of Mathematics and Statistics.

Question 13.
Define dicrete random Variable
Solution:
A variable which can assume finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.

Question 14.
What do you understand by continuous random variable?
Solution:
A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable.

Question 15.
Describe what is meant by a random variable
Solution:
If S is a sample space with a probability measure and x is a real valued function defined over the elements of S then x is called a random variable A random variable is also called a change variable.

Question 16.
Distinguish between discrete and continuous random variable
Solution:

Question 17.
Explain the distribution function of random variable
Solution:
The discrete cumulative distribution function or distribution function of a real valued discrete random variable X takes the countable number of points x₁, x₂, …. with corresponding probabilities p(x₁), p(x₂),… and then the cumulative distribution function is defined by
F_x(x) = P(X ≤ x), for all x ∈ R
i.e. F_x (x) = \(\sum_{x \leq x}\) p(x_i)

Question 18.
Explain the terms (i) probability Mass function (ii) probability density function and (iii) probability
Solution:
(i) If X is a discrete random variable with distinct values x₁, x₂, …. x_n, …, then the function, denoted by P_x(x) and defined by

This is defined to be the probability mass function or discrete probability function of X.

(ii) The probability that a random variable X takes a value in the interval [t₁, t₂] (open or closed) is given by the integral of a function called the probability density function f_x(x):
P(t₁ ≤ X ≤ t₂) = \(\int_{t_{1}}^{t_{2}}\) f_x(x)dx.

(iii)The probability distribution of a random variable X is defined only when we have the various values oft the various values of the random variable e.g. x₁, x₂ …… x_n togather with respective probabilities p₁, p₂, p₃ …… p₄ satisfying
\(\sum_{i=1}^{n}\) Pi = 1

Question 19.
What are the properties of (i) discrete random variable and (ii) Continuous random variable
Solution:
(i) The probability mass function p(x) must satisfy the following conditions
(i) p(x_i) ≥ 0 ∀ i,
(ii) \(\sum_{i=1}^{∞}\) p(x_i) = 1

(ii) The probability density functions/)(x) or simply ; by/(x) must satisfy the following conditions.
(i) f(x) ≥ 0 ∀ x and
(ii) \(\int_{-∞}^{∞}\) f(x) dx = 1

Question 20.
State the Properties of distribution function.
Solution:
The function F_x(x) or simply F(x) has the following properties.
(i) 0 ≤ F(x) ≤ 1, -∞ < x < ∞
(ii) F(-∞) = \(\lim _{x \rightarrow-\infty}\) F(x) = 0 and F(+∞) = \(\lim _{x \rightarrow ∞}\) F(x) = 1.
(iii) F(.) is a monotone, non-decreasing function; ; that is, F(a) < F(b) for a < b.
(iv)F(.) is continuous from the right; that is, \(\lim _{h \rightarrow 0}\) F(x + h) = F(x).
(v) F(x) = \(\frac { d }{dx}\) F(x) = f(x) ≥ 0;
(vi) F'(x) = \(\frac { d }{dx}\) F(x) = f(x) ⇒ dF(x) = f(x)dx
dF(x) is known as probability differential of X.
(vii) P(a ≤ x ≤ b) = \(\int_{a}^{b}\) f(x)dx = \(\int_{-∞}^{b}\) f(x)dx – \(\int_{-∞}^{a}\) f(x)dx
= P(X ≤ b) – P(X ≤ a)
= F(b) – F(a)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 11 Recent Developments in Physics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 11 Recent Developments in Physics

12th Physics Guide Recent Developments in Physics Text Book Back Questions and Answers

Part – I

TextBook Evaluation:

I. Multiple Choice Questions:

Question 1.
The particle size of ZnO material is 30 nm. Based on the dimension it is classified as
a) Bulk material
b) Nanomaterial
c) Soft material
d) Magnetic material
Answer:
b) Nanomaterial

Question 2.
Which one of the following is the natural nanomaterial.
a) Peacock feather
b) Peacock beak
c) Grain of sand
d) Skin of the Whale
Answer:
a) Peacock feather

Question 3.
The blue print for making ultra-durable synthetic material is mimicked from ________.
a) Lotus leaf
b) Morpho butterfly
c) Parrot fish
d) Peacock feather
Answer:
c) Parrot fish

Question 4.
The method of making nanomaterial by assembling the atoms is called ______.
a) Top down approach
b) Bottom up approach
c) Cross down approach
d) Diagonal approach
Answer:
b) Bottom up approach

Question 5.
“Sky wax” is an application of nano product in the field of ________.
a) Medicine
b) Textile
c) Sports
d) Automotive industry
Answer:
c) Sports

Question 6.
The materials used in Robotics are ________.
a) Aluminium and silver
b) Silver and gold
c) Copper and gold
d) Steel and aluminium
Answer:
d) Steel and aluminium

Question 7.
The alloys used for muscle wires in Robots are _______.
a) Shape memory alloys
b) Gold copper alloys
c) Gold silver alloys
d) Two-dimensional alloys
Answer:
a) Shape memory alloys

Question 8.
The technology used for stopping the brain from processing pain is ______.
a) Precision medicine
b) Wireless brain sensor
c) Virtual reality
d) Radiology
Answer:
c) Virtual reality

Question 9.
The particle which gives mass to protons and neutrons are _______.
a) Higgs particle
b) Einstein particle
c) Nanoparticle
d) Bulk particle
Answer:
a) Higgs particle

Question 10.
The gravitational waves were theoretically proposed by _______.
a) Conrad Rontgen
b) Marie Curie
c) Albert Einstein
d) Edward Purcell
Answer:
c) Albert Einstein

II. Short Answers Questions:

Question 1.
Distinguish between Nanoscience and Nanotechnology.
Answer:

Nanoscience	Nanotechnology
1. Nanoscience is the science of objects with a typical size of 1-100 nm. Nano means one-billionth of a metre that is 10-9m.	Nanotechnology is a technology involving design, production.
2. If matter is divided into such small objects the mechanical, electrical, optical magnetic and other properties change.	Characterization and application of nano structured materials.

Question 2.
What is the difference between Nano materials and Bulk materials?
Answer:

Nano Materials	Bulk Mateerials
1. Nano materials are particle that have their size in 1-100 nm range atleast in one dimension.	Bulk materials are particle that have their size above lOOnm in all dimensions.
2. We cannot see particles of nanomaterials from the naked eye.	We can see particle of most of the bulk materials from the naked eye.
3. The example of nanomaterials include nano zymes, titanium dioxide, nano particles, graphene, etc.	The example of bulk materials include plaster sand, gravel, cement, ore, slag, salts, etc.

Question 3.
Give any two examples for “Nano” in nature.
Answer:
1.  Single-strand DNA:
A single strand of DNA, the building block of all living things, is about three nanometers wide.

2. Morpho Butterfly:
The scales on the wings of a morpho butterfly contain nanostructures that change the way light waves interact with each other, giving the wings brilliant metallic blue and green hues. Mimic in laboratories – Manipulation of colours by adjusting the size of nanoparticles with which the materials are made.

Question 4.
Mention any two advantages and disadvantages of Robotics.
Answer:
Advantage:

1. The robots are much cheaper than humans.
2. Robots never get tired like humans.
3. Stronger and faster than humans.

Disadvantage:

1. Robots have no sense of emotions or conscience.
2. They lack empathy and hence create an emotionless workplace.
3. Unemployment problem will increase.

Question 5.
Why steel is preferred in making Robots?
Answer:
Steel is several times stronger. In any case, because of the inherent strength of metal, robot bodies are made using a sheet, bar, rod, channel, and other shapes.

Question 6.
What are black holes?
Answer:

1. Black holes are the end-stage of stars which are the highly dense massive objects. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun.
2. It has very strong gravitational force such that no particle or even light can escape from it.
3. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other stars. Every galaxy has a black hole at its center.

Question 7.
What are sub atomic particles?
Answer:

1. The three main subatomic particles that form an atom are protons, neutrons, and electrons.
2. Subatomic particles are particles that are smaller than the atom, proton and neutron are made up of quarks which is interact through gluons.
3. A subatomic particle having two types of particles, they are elementary particles and composite particles.

III. Long Answer Questions:

Question 1.
Discuss the applications of Nanomaterials in various fields.
Answer:

(i) Automotive industry:

• Lightweight construction
• Painting (fillers, base coat, clear coat)
• Catalysts
• Tires (fillers)
• Sensors
• Coatings for window screen and car bodies

(ii) Chemical industry:

• Fillers for paint systems
• Coating systems based on nanocomposites
• Impregnation of papers
• Switchable adhesives
• Magnetic fluids

(iii) Engineering

• Wear protection for tools and machines (anti blocking coatings, scratch resistant coatings on plastic parts, etc.)
• Lubricant – free bearings

(iv) Electronic industry

• Data memory
• Displays
• Laser diodes
• Glass fibres
• Optical switches
• Filters (IR-blocking)
• Conductive, antistatic coatings

(v) Construction:

• Construction materials
• Thermal insulation
• Flame retardants
• Surface – functionalised building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
• Facade coatings
• Groove mortar

(vi) Medicine:

• Drug delivery systems
• Contrast medium
• Prostheses and implants
• Agents in cancer therapy
• Active agents
• Medical rapid tests
• Antimicrobial agents and coatings

(vii) Textile / fabrics / non – wovens:

• Surface – processed textiles
• Smart clothes

(viii) Energy:

• Fuel cells
• Solar cells
• Batteries
• Capacitors

(ix) Cosmetics:

• Sun protection
• Lipsticks
• Skin creams
• Tooth paste

(x) Food and drinks:

• Package materials
• Additives
• Storage life sensors
• Clarification of fruit juices

(xi) Household:

• Ceramic coatings for irons
• Odors catalyst
• Cleaner for glass, ceramic, floor, windows

(xii) Sports / outdoor:

• Ski wax
• Antifogging of glasses/goggles
• Antifouling coatings for ships/boats
• Reinforced tennis rackets and balls.

Question 2.
What are the possible harmful effects of the usage of Nanoparticles? Why?
Answer:

1. The research on the harmful impact of the application of nanotechnology is also equally important and fast developing.
2. The major concern here is that the nanoparticles have the dimensions same as biological molecules such as protein.
3. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.
4. The interaction with living systems is also affected by the dimensions of the nanoparticles.
5. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules.
6. It is also possible for the inhaled nanoparticles to reach the blood to reach other sites such as the liver, heart, or blood cells.
7. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition, and surface characteristics.

Question 3.
Discuss the functions of key components in Robots?
Answer:

Power conversion unit:
Robots are powered by batteries, solar power, and hydraulics.

Actuators:

1. Converts energy into movement.
2. The majority of the actuators produce rotational (or) linear motion.

Most robots are composed of 3 main parts.

The Controller:

1. This is also known as the “brain” which is run by a computer program.
2. It gives commands for the moving parts to perform the job.

Most robots are composed of 3 main parts.

Mechanical parts:

1. Motors
2. Pistons
3. Grippers
4. Wheels and gears that make the robot move, grab, turn and lift.

Sensors:

1. To tell the robot about its surroundings.
2. It helps to determine the sizes and shapes of the objects around, the distance between the objects, and the directions as well.

Question 4.
Elaborate any two types of Robots with relevant examples.
Answer:
(i) Human-Robot: Certain robots are made to resemble humans in appearance and replicate human activities like walking, lifting, and sensing, etc.

1. Power conversion unit: Robots are powered by batteries, solar power, and hydraulics.
2. Actuators: Converts energy into movement. The majority of the actuators produce rotational or linear motion.
3. Electric motors: They are used to actuate the parts of the robots like wheels, arms, fingers,
   legs, sensors, camera, weapon systems etc. Different types of electric motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor, etc.
4. Pneumatic Air Muscles: They are devices that can contract and expand when air is pumped inside. It can replicate the function of a human muscle. The contract almost 40% when the air is sucked inside them.
5. Muscle wires: They are thin strands of wire made of shape memory alloys. They can contract by 5% when an electric current is passed through them.
6. Piezo Motors and Ultrasonic Motors: Basically, we use it for industrial robots.
7. Sensors: Generally used in task environments as it provides information of real-time knowledge.
8. Robot locomotion: Provides the types of movements to a robot.
   The different types are:
   • Legged
   • Wheeled
   • Combination of Legged and Wheeled Locomotion
   • Tracked slip/skid.

(ii) Industrial Robots:
Six main types of industrial robots:

1. Cartesian
2. SCARA (Selective Compliance Assembly Robot Arm)
3. Cylindrical
4. Delta
5.  Polar
6. Vertically articulated

Six-axis robots are ideal for:

1. Arc Welding
2. Spot Welding
3. Material Handling
4. Machine Tending
5. Other Applications

Question 5.
Comment on the recent advancement in medical diagnosis and therapy.
Answer:
The recent advancement in medical diagnosis and therapy:

1. Virtual reality
2. Precision medicine
3. Health wearables
4. Artificial organs
5.  3 – D printing
6. Wireless brain sensors
7. Robotic surgery
8. Smart inhalers

1. Virtual reality:
Medical virtual reality is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients. Virtual reality has enhanced surgeries by the use of 3D models by surgeons to plan operations. It helps in the treatment of Autism, Memory loss, and Mental illness.

2. Precision medicine:
Precision medicine is an emerging approach for disease treatment and prevention that takes into account individual variability in genes, environment, and lifestyle for each person. In this medical model, it is possible to customise healthcare, with medical decisions, treatments, practices, or products which are tailored to the individual patient.

3. Health wearables:
A health wearable is a device used for tracking a wearer’s vital signs or health and fitness-related data, location, etc. Medical wearables with artificial intelligence and big data provide an added value to healthcare with a focus on diagnosis, treatment, patient monitoring, and prevention.

Note Big Data: Extremely large data sets that may be analysed computationally to reveal patterns, trends, and associations, especially relating to human behavior and interactions.

4. Artificial organs:
An artificial organ is an engineered device or tissue that is implanted or integrated into a human. It is possible to interface it with living tissue or to replace a natural organ. It duplicates or augments a specific function or functions of human organs so that the patient may return to normal life as soon as possible.

5. 3D printing:
Advanced 3D printer systems and materials assist physicians in a range of operations in the medical field from audiology, dentistry, orthopedics, and other applications.

6. Wireless brain sensors:
Wireless brain sensors monitor intracranial pressure and temperature and then are
absorbed by the body. Hence there is no need for surgery to remove these devices.

7. Robotic surgery:
Robotic surgery is a type of surgical procedure that is done using robotic systems. Robotically-assisted surgery helps to overcome the limitations of pre – existing minimally invasive surgical procedures and to enhance the capabilities of surgeons performing open
surgery.

8. Smart inhalers:
Inhalers are the main treatment option for asthma. Smart inhalers are designed with health systems and patients in mind so that they can offer maximum benefit. Smart inhalers use Bluetooth technology to detect inhaler use, remind patients when to take their medication, and gather data to help guide care.

Part – II:

12th Physics Guide Recent Developments in Physics Additional Questions and Answers

I. Match the following:

Question 1.

I	II
1. Molecule	a. Located in the center of an atom
2. Nucleon	b. Electrons revolving around the atomic nucleus
3. Atom	c. Protons & Neutrons comprising the nucleus of an atom
4. Nucleus	d. Composed of two (or) more

Answer:

1. d
2. c
3. b
4. a

Question 2.

I	II
1. Geroge Devol	a. Audiology
2. Nano	b. No need surgery
3. Wireless brain sensors	c. One billionth of a meter (109 m)
4. 3D Printing	d. Unimate

Answer:

1. d
2. c
3. b
4. a

Question 3.

I	II
1. Higgs Particle	a. Slavic word
2. Albert Einstein	b. Mass of particle
3. Black Holes	c.Gravitational waves
4. Robot	d. End stage of stars

Answer:

1. b
2. c
3. d
4. a

Question 4.

I	II
1. Stephen Hawking	a. Rossum universal robots
2. Einstein’s theory	b. Twin mars rovers
3. Karl Capek	c. Field of black holes
4. Outer space	d. General relativity

Answer:

1. c
2. d
3. a
4. b

Question 5.

I	II
1. Inhalers	a. Aluminium and steel
2. 3D Printing	b. Mars pathfinder mission
3. Make robots	c. Dentistry
4. Outer Space	d. Asthma

Answer:

1. d
2. c
3. a
4. b

II. Fill in the blanks:

Question 1.
_____ existed in nature long before scientists began studying them in laboratories.
Answer:
Nanoscale structures

Question 2.
_____ and _____ is the interdisciplinary area covering its applications in various fields.
Answer:
Nanoscience, technology

Question 3.
Chinese scientists have created the world’s first autonomous DNA robots to combat _______.
Answer:
cancer tumours

Question 4.
________ is the fundamental entity of matter.
Answer:
Atom

Question 5.
There are two ways of preparing the nanomaterials _____ and _____ approaches.
Answer:
top-down, bottom-up

III. Choose the odd man out:

Question 1.
The robotic system mainly consists of ________.
a) sensors
b) power supplies
c) blood cells
d) control system
Answer:
c) blood cells

Question 2.
The key components of a robot are _______.
a)Power conversion unit
b) Actuators
c) Electric motors
d) Muscle wires
e) Delta
f) Sensors
Answer:
e) Delta

Question 3.
Six main types of industrial robots are _______.
a) Cartesian
b) SCARA
c) Cylindrical
d) Delta
e) Polar
f) Ultrasonic motors
Answer:
f) Ultrasonic motors

Question 4.
The recent advancement in medical technology includes ________.
a) Artificial organs
b) Precision medicine
c) Virtual reality
d) Pool cleaning
e) 3D printing etc
Answer:
d) Pool cleaning

Question 5.
Household robots are used as _______.
a) floor cleaners
b) gutter cleaners
c) Pool cleaning
d) Investigation of the rocks
e) Lawn mowing
Answer:
d) Investigation of the rocks

IV. Find the correct pair:

Question 1.
a) SCARA – Spot welding
b) George Devol and – First robot Joseph Engalberger company
c) Found rocks and soils – Pool cleaning
d) Size less than 100 nm – Bulk solid
Answer:
b) George Devol and Joseph Engalberger – First robot company

Question 2.
a) Top-down approach – Plasma etching
b) Chemical Industry – Glass Fibres
c) Medicine – Active agents
d) Human-Robot – Machine Tending
Answer:
c) Medicine – Active agents

V. Find the incorrect pair:

Question 1.
a) 3D Printing – Audiology
b) Karel Capek – Rossum universal robots
c) Inhalers – Asthma
d) Higgs particle – End stage of stars
Answer:
d) Higgs particle – End stage of stars

Question 2.
a) Higgs particles – God particles
b) Stephen Hawking – Field of Black holes
c) Molecule – Fundamental entity of matter
d) George Devol – Unimate
Answer:
c) Molecule – Fundamental entity of matter

VI. Choose the incorrect statement:

Question 1.
Statement 1: Nanotechnology is a technology involving the design, production, characterization, and applications of nanostructured materials.
Statement 2: Nanoparticles of a few micrometers size may reach well inside biomolecules, which is not possible for larger nanoparticles.
a) statement 1
b) statement 2
c) statement 1 and 2
d) None of these
Answer:
b) Statement 2:
Correct Sentence:
Nanoparticles of a few nanometer-size may reach well inside biomolecules, which is not possible for larger nanoparticles.

Question 2.
a) Statement 1: Five major fields of robotics are human-robot interface, mobility, manipulation, programming, and sensors.
b) Statement 2: Aluminum is a softer metal and is, therefore, easier to work with.
c) Statement 3: Industrial robots are used for exploring stars, planets, etc.
d) Statement 4: The robotic system mainly consists of sensors, power supplies, control systems, manipulators and
necessary software.
Answer:
c) Statement 3
Correct Sentence:
Industrial robots are used for welding, cutting, robotic water jet cutting, lifting, etc.

Question 3.
a) Statement 1: Accelerated mass emits gravitational force which is very week.
b) Statement 2: Black holes are the strongest source of gravitational waves.
c) Statement 3: Cosmology is the branch that involves the origin and evolution of the universe.
Answer:
a) Statement 1:
Correct Sentence:
Accelerated mass emits gravitational wave which is very week.

VII. Choose the correct statement:

Question 1.
a) Statement 1: Human is a mechanical device.
b) Statement 2: Nanoparticles can also cross cell membranes.
c) Statement 3: Top up and Bottom down are the two ways to preparing the nanomaterials.
d) Statement 4: It is possible to deliver a drug directly to a specific cell in the body by designing the surface of bulk particles.
Answer:
b) Statement 2: Nanoparticles can also cross cell membranes.

Question 2.
a) Statement 1: Manipulation of colours is found in laboratories by Morpho butterfly in nature.
b) Statement 2: Similar nanostructures are made in a lab to glow in different colors from peacock feathers in nature.
c) Statement 3: Water repellent nano paints are made from a lotus leaf surface idea.
d) Statement 4: All statements are correct.
Answer:
d) Statement 4: All statements are correct

Question 3.
Nanomaterial-based products in different fields.
a) Statement 1: Lightweight construction in medicine.
b) Statement 2: Displays nanomaterial application in the engineering field.
c) Statement 3: Antimicrobial agents and coating in the medical field.
d) Statement 4: None of these
Answer:
c) Statement 3: Antimicrobial agents and coating in the medical field.

VIII. Assertion and Reason:

a) Assertion is correct, the reason is correct; the reason is a correct explanation for the assertion.
b) Assertion is correct, the reason is correct; the reason is not a correct explanation for the assertion.
c) Assertion is correct, reason is incorrect.
d) Assertion is incorrect, reason is correct.

Question 1.
Assertion (A):
Nanoparticles can also cross cell membranes.
Reason (R):
It is also possible for the inhaled nanoparticles to reach the blood, other sites as the liver, heart.
Answer:
a) Assertion is correct, the reason is correct; the reason is a correct explanation for the assertion.

Question 2.
Assertion (A):
A drug delivery system is a medical application of a nano-based product.
Reason (R):
The mechanical parts of Robotics are motors, pistons, grippers, wheels, and gears.
Answer:
b) Assertion is correct, the reason is correct; the reason is not a correct explanation for the assertion.

Question 3.
Assertion (A):
Human robots replicate human activities like walking, lifting, and sensing, etc.
Reason (R):
Electric motors are not used in robots.
Answer:
c) Assertion is correct, reason is incorrect.

Question 4.
Assertion (A):
Face recognition is a natural intelligence of robots.
Reason (R):
Robot can translate words from one language to another.
Answer:
d) Assertion is incorrect, reason is correct.

IX. Choose the correct answer:

Question 1.
An automatic apparatus or device that performs functions ordinarily ascribed to human or operate with what appears to be almost human intelligence is called ……………..
(a) Robot
(b) Human
(c) Animals
(d) Reptiles.
Answer:
(a) Robot.

Question 2.
If the particle of a solid is of size less than 100 nm, it is said to be a .
a) nano solid
b) bulk solid
c) nano & bulk solids
d) None of these
Answer:
a) nano solid

Question 3.
The basic components of the robot are ……………..
(a) mechanical linkage
(b) sensors and controllers
(c) user interface and power conversion unit
(d) All the above.
Answer:
(d) All the above.

Question 4.
Nano means ________.
a) 10^-33 m
b) 10^-6 m
c) 10^-9 m
d) 10^-12 m
Answer:
c) 10^-9 m

Question 5.
_______ and _______ are the two important phenomena that govern nano properties.
a) Quantum confinement effects
b) Surface effects
c) Quantum confinement (or) Surface light
d) Quantum confinement effect and Surface confinement effects
Answer:
d) Quantum confinement effect and Surface confinement effects

Question 6.
Which of the following atoms do not move from each other ……………..
(a) Shape memory alloys
(b) Nanomaterials
(c) Dielectrics
(d) Static materials.
Answer:
(b) Nanomaterials

Question 7.
A health wearable is a device used for ________.
a) health and fitness-related data, location, etc.
b) to replace a natural organ
c) (a) and (b)
d) (a) or (b)
Answer:
a) health and fitness-related data, location, etc.

Question 8.
Atom is made up of ________.
a) electrons
b) protons
c) neutrons
d) above all
Answer:
d) above all

Question 9.
For nanometers whose diameters less than …………….are used for welding purposes.
(a) 10 nm
(b) 20 nm
(c) 30 nm
(d) 40 nm.
Answer:
(a) 10 nm

Question 10.
The Strongest source of gravitational waves is _______.
a) black holes
b) accelerated mass
c) sun
d) stars
Answer:
a) black holes

Question 11.
This is the black hole at the centre of the milky way galaxy.
a) Sagittarius A*
b) Sagittarius B*
c) Sagittarius C*
d) Sagittarius D*
Answer:
a) Sagittarius A*

Question 12.
Who worked in the field of black holes.
a) Marie curie
b) Stephen Hawking
c) Conrad Rontgen
d) Edward Purcell
Answer:
b) Stephen Hawking

Question 13.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956 ……………..
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell.
Answer:
(c) Engelberger

Question 14.
Slavic word robota means ________.
a) labour
b) work
c) labour (or) work
d) None of these
Answer:
c) labour (or) work

Question 15.
These robots are used for vacuum cleaners, floor cleaners, pool cleaning, etc.,
a) Household robots
b) Industrial robots
c) Space robots
d) None of these
Answer:
a) Household robots

Question 16.
Similar nanostructures are made in a lab to glow in different colors.
a) Morpho butterfly
b) Peacock feathers
c) DNA
d) Parrotfish
Answer:
b) Peacock feathers

Question 17.
Manipulation of colours by adjusting the size of nanoparticles from this idea is _______.
a) Morpho butterfly
b) Peacock feathers
c) Parrotfish
d) Lotus leaf surface
Answer:
a) Morpho butterfly

Question 18.
The phenomenon of artificial radioactivity was invented by ……………..
(a) Joliot and Irene curie
(b) Felix Bloch and Edward Purcell
(c ) Connick and Hounsfield
(d) Wilhelm Conrad – Rontgen.
Answer:
(a) Joliot and Irene curie

Question 19.
Bottom up approach example is _______.
a) ball milling
b) sol-gel
c) lithography
d) plasma etching
Answer:
d) plasma etching

Question 20.
Coatings for wind-screen and car bodies
a) Chemical industry
b) Electronic industry
c) Automotive industry
d) Medicine
Answer:
c) Automotive industry

Question 21.
This is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients.
a) Virtual reality
b) Precision medicine
c) Health wearables
d) Artificial organs
Answer:
a) Virtual reality

Question 22.
This is an emerging approach for disease treatment.
a) Artificial organs
b) Health wearables
c) Precision medicine
d) Virtual reality
Answer:
c) Precision medicine

Question 23.
This is possible to interface it with living tissue (or) to replace a natural organ.
a) Precision medicine
b) Health wearables
c) Virtual reality
d) Artificial organs
Answer:
d) Artificial organs

Question 24.
This device used for tracking a wearer’s vital signs.
a) 3D printing
b) Health wearables
c) Artificial organs
d) Robotic surgery
Answer:
b) Health wearables

Question 25.
This monitor intracranial pressure and temperature and then are absorbed by the body.
a) 3D printing
b) Health wearables
c) wireless brain sensors
d) Robotic surgery
Answer:
c) wireless brain sensors

Question 26.
Composed of two or more atoms.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
a) Molecule

Question 27.
Electrons revolving around the atomic nucleus.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
b) Atom

Question 28.
Composed of protons and neutrons located in the center of an atom.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
c) Nucleus

Question 29.
Protons and neutrons comprising the nucleus of an atom.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
d) Nucleon

Question 30.
Smart inhaler uses this technology to detect inhaler use.
a) system technology
b) Bluetooth technology
c) both (a) and (b)
d) None of these
Answer:
b) Bluetooth technology

X. Two mark questions:

Question 1.
What is physics?
Answer:
Physics is the basic building block for Science, Engineering, Technology, and Medicine.

Question 2.
What is Robotics?
Answer:
Robotics is an integrated study of mechanical engineering, electronic engineering, computer engineering, and science.

Question 3.
What are Robots?
Answer:
The robot is a mechanical device designed with electronic circuitry and programmed to perform a specific task.

Question 4.
What is meant by ‘Robot’? Write its uses?
Answer:
Robot is a mechanical device designed with electronic circuitry and programmed to perform a specific task. These automated machines are highly significant in this robotic era where they can take up the role of humans in certain dangerous environments that are hazardous to people like defusing bombs, finding survivors in unstable ruins, and exploring mines and shipwrecks.

Question 5.
Name main types of industrial robots.
Answer:
Six main types of industrial robots are,

1. Cartesian
2. SCARA
3. Cylindrical
4. Delta
5. Polar
6. Vertically articulated

Question 6.
Name the axis robots.
Answer:
Six-axis robots are ideal for

1. Arc Welding
2. Spot Welding
3. Material Handling
4. Machine Tending
5. Other applications

Question 7.
What is the aim of artificial intelligence?
Answer:
The aim of artificial intelligence is to bring in human-like behavior in robots.

Question 8.
Define cosmology?
Answer:
Cosmology is the branch that involves the origin and evolution of the universe. It deals with the formation of stars, galaxy, etc.

Question 9.
What are the uses of outer space robots?
Answer:
In outer space, robots are used for exploring stars, planets, etc., investigation of the mineralogy of the rocks and soils on Mars, analysis of elements found in rocks and soils.

Question 10.
Name some outer space robots.
Answer:

1. Mars Rovers of NASA
2. Twin Mars Rovers
3. Mars Pathfinder Mission

Question 11.
What are the uses of household robots?
Answer:
Household robots are used as,

1. Vacuum cleaners
2. Floor cleaners
3. Gutter cleaners
4. Lawn mowing
5. Pool cleaning
6. To open and close doors

Question 12.
What are the developments in Nano-robots?
Answer:
Nano-robots are being developed to be in the bloodstream to perform small surgical procedures, to fight against bacteria, repairing individual cells in the body.

Question 13.
Define Particle physics.
Answer:
Particle physics deals with fundamental particles of nature. Protons and neutrons are made of quarks.

Question 14.
What is cosmology?
Answer:
Cosmology is the branch that involves the origin and evolution of the universe.

Question 15.
What is physics?
Answer:

1. Physics is the basic building block for Science, Engineering, Technology and Medicine.
2. Nanoscience is the science of objects with typical sizes of 1-100 nm.

Question 16.
What is Nano?
Answer:
Nano means one-billionth of a meter that is 10^-9 m.

Question 17.
What is nano solid?
Answer:
If the particle of a solid is of size less than lOOnm, it is said to be a nano solid.

Question 18.
What is bulk solid?
Answer:
When the particle size exceeds 100 nm, it forms a bulk solid.

Question 19.
Name the two ways of preparing the nanomaterials.
Answer:

• Top-down approaches
• Bottom-up approaches

Question 20.
What are the major fields of robotics?
Answer:
Five major fields of robotics are

1. Human-robot interface
2. Mobility
3. Manipulation
4. Programming
5. Sensors

Question 21.
Is the accelerated mass emits gravitational waves?
Answer:
Yes, the accelerated mass emits gravitational waves which are very weak.

Question 22.
What is the important phenomena of nano properties?
Answer:
Quantum confinement effects and surface effects are the two important phenomena that govern nano properties.

Question 23.
Is the nano form of the material the same as its bulk counterpart?
Answer:
No, The nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Question 24.
Can nanoparticles get absorbed?
Answer:
Yes, Nanoparticles can easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

Question 25.
What is the function of inhaled nanoparticles?
Answer:
The inhaled nanoparticles reach the blood and that may also reach other sites such as the liver, heart, and blood cells.

Question 26.
Can nanoparticles cross the cell membrane?
Answer:
Yes, nanoparticles can also cross cell membranes.

Question 27.
Is the larger nanoparticles can reach inside the biomolecules?
Answer:

1. The absorbing nature depends on the surface of the nanoparticle.
2. Nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles.

Question 28.
What is Unimate?
Answer:
In 1954, George Devol invented the first digitally operated programmable robot called “Unimate”.

Question 29.
Write a short note on Human-Robot.
Answer:
Certain robots are made to resemble humans in appearance and replicate human activities like walking, lifting and sensing, etc.

XI. Three mark questions:

Question 1.
List out the nanomaterial-based products in the automotive industry.
Answer:

1. Lightweight construction
2. Painting (fillers, base coat, clear coat)
3. Catalysts
4. Tires (fillers)
5. Sensors
6. Coatings for wind-screen and car bodies

Question 2.
Give five examples of nanomaterial-based products in the chemical industry.
Answer:

1. Fillers for paint systems
2. Coating systems based on nanocomposites
3. Impregnation of papers
4. Switchable adhesives
5. Magnetic fluids

Question 3.
Write the application of nanomaterial-based products in construction.
Answer:

1. Construction materials
2. Thermal insulation
3. Flame retardants
4. Surface-functionalized building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
5. Facade coatings
6. Groove mortar

Question 4.
What are the applications of nanomaterial-based products in the following areas?
i) Textile
ii) Energy
iii) cosmetics
Answer:

Textile/fabrics/ non-wovens	Energy	Cosmetics
1. Surface Processed textiles 2. Smart clothes	1. Fuel cells 2. Solar cells 3. Batteries 4. Capacitors	1. Sun protection 2. Lipsticks 3. Skin creams 4. toothpaste

Question 5.
List the nanomaterial-based products in Engineering.
Answer:

1. Wear protection for tools and machines (anti-blocking coatings, scratch-resistant coatings on plastic parts, etc.)
2. Lubricant-free bearings.

Question 6.
Write about the top-down approach of nanomaterials synthesis.
Answer:
Top-down approach:

• Nanomaterials are synthesised by breaking down bulk solids into nano sizes.
• Ex: Ball milling, sol-gel, lithography.

Question 7.
What is bottom-up approach?
Answer:

1. Nanomaterials are synthesised by assembling the atoms/molecules together.
2. Selectively atoms are added to create structures.
3. Example: Plasma etching and chemical vapour deposition.

Question 8.
Short note on nano in laboratories.
Answer:

1. The nanostructures made in the laboratory mimic some of nature’s amazing nanostructures.
2. As the nanostructures are so small, specialized methods are needed to manufacture objects in this size range.
3. There are two ways of preparing the nanomaterials, top-down and bottom-up approaches.

Question 9.
Mention the uses of wireless brain sensors.
Answer:

1. Wireless brain sensors monitor intracranial pressure and temperature and then are absorbed by the body.
2. Hence there is no need for surgery to remove these devices.

Question 10.
What is robotic surgery?
Answer:

1. Robotic surgery is a type of surgical procedure that is done using robotic systems.
2. Robotically assisted surgery helps to overcome the limitations of pre-existing minimally-invasive surgical procedures and to enhance the capabilities of surgeons performing open surgery.

Question 11.
What do you know about smart inhalers?
Answer:

1. Inhalers are the main treatment option for asthma.
2. Smart inhalers are designed with health systems and patients in mind so that they can offer maximum benefit.
3. Smart inhalers use BlueTooth technology to detect inhaler use, remind patients when to take their medication, and gather data to help guide care.

Question 12.
What is the role of physics in medical diagnosis?
Answer:

1. Medical science very much revolves around physics principles.
2. Medical instrumentation has widened the life span due to the technology integrated diagnosis and treatment of most of the diseases.
3. This modernisation in all fields is possible due to the efficient application of fundamental physics.

Question 13.
What is gravitational waves? How are they produced?
Answer:

1. Gravitational waves are the disturbances in the curvature of space-time and it travels with the speed of light.
2. Any accelerated charge emits an electromagnetic wave. Any accelerated mass emits gravitational waves but these waves are very weak even for masses like earth. The strongest source of gravitational waves is black holes.
3. The recent discoveries of gravitational waves are emitted by two black holes when they merge into a single black hole.

Question 14.
Can we completely replace humans with robots? Give any three reasons.
Answer:
No, we cannot completely replace humans with robots. Following are the reasons:

1. Robots have no sense of emotions or conscience.
2. Robots cannot handle unexpected situations.
3. Humans cannot be replaced by robots in decision-making.

Question 15.
What is a quark?
Answer:

1. A quark is a type of elementary particle and a fundamental constituent of matter.
2. Quarks combine to form composite particles called hadrons, the most stable of which are protons and neutrons, the components of atomic nuclei.

Question 16.
How nano are found in mimic in laboratories from i) Morpho butterfly ii) Peacock feathers.
Answer:

1. Morpho butterfly: Manipulation of colours by adjusting the size of nanoparticles with which the materials are made.
2. Peacock feathers: Similar nanostructures are made in a lab to glow in different colors.

Question 17.
From parrotfish, how nano are found in laboratories?
Answer:
The natural structure provides a blueprint for creating ultra-durable synthetic materials that could be useful for mechanical components in electronics and in other devices that undergo repetitive movement, abrasion, and contact stress.

Question 18.
How the lotus leaf surface (Nano in nature) found in laboratories?
Answer:

1. Water repellant nano paints are made.
2. Coating with such nano paints gives durability, protection against stains and dirt also enhances fuel efficiency when coated on ships.

Question 19.
What do you know about animate?
Answer:

1. In 1954, George Devol invented the first digitally operated programmable robot called Unimate.
2. George Devol and Joseph Engelberger, the father of the modern robotics industry formed the world’s first robot company in 1956.
3. In 1961, Unimate was operated in a General Motors automobile factory for moving car parts around in New Jersey.

XII. Five mark questions:

Question 1.
Write the advantage of Robotics.
Answer:

1. Robots are much cheaper than humans.
2. Robots never get tired like humans. It can work 24 x 7. Hence absenteeism in the workplace can be reduced.
3. Robots are more precise and error-free in performing the task.
4. Stronger and faster than humans.
5. Robots can work in extreme environmental conditions: extreme hot (or) cold, space (or) underwater.
6. In dangerous situations like bomb detection and bomb deactivation.
7. In warfare, robots can save human lives.
8. Robots are significantly used in handling materials in chemical industries especially in nuclear plants which can lead to health hazards in humans.

Question 2.
List out the disadvantage of Robotics.
Answer:

1. Robots have no sense of emotions or conscience.
2. They lack empathy and hence create an emotionless workplace.
3. If ultimately robots would do all the work, and the humans will just sit and monitor them, health hazards will increase rapidly.
4. The unemployment problem will increase.
5. Robots can perform defined tasks and can’t handle unexpected situations
6. The robots are well programmed to do a job and if a small thing goes wrong it ends up in a big loss to the company.
7. If a robot malfunctions, it takes time to identify the problem, rectify it, and even program if necessary. This process requires significant time.
8. Humans can’t be replaced by robots in decision-making.
9. Till the robot reaches the level of human intelligence, the humans in the workplace will exit.

Question 3.
Give an example of “Nano” in nature.
Answer:

1. A single strand of DNA the building block of all living things is about three nanometers wide.
2. The scales on the wings of a morpho butterfly contain nanostructures that change the way light waves interact with each other, giving the wings brilliant metallic blue and green hues.
3. Peacock feathers get their iridescent coloration from light interacting with 2-dimensional photonic crystal structures just tens of nanometers thick.
4. Parrotfish crunches up coral all day. The source of the parrot fish’s powerful bite is the interwoven fiber nanostructure.
5. Crystals of a mineral called fluorapatite are woven together in a chain mail-like arrangement. This structure gives parrotfish teeth incredible durability.
6. Lotus leaf surface scanning electron micrograph (SEM) showing the nanostructures on the surface of a leaf from a lotus plant. This is the reason for the self-cleaning process in lotus leaf.

Question 4.
List out the applications of Nanotechnology in various fields.
Answer:

1. Optical engineering and communication
2. Electronics
3. Metallurgy and materials
4. Defense and security
5. Energy storage
6. Biomedical and drug delivery
7. Agriculture and food
8. Cosmetics and paints
9. Biotechnology
10. Textile

Question 5.
Explain about Nanorobots.
Answer:

1. The size of the nanorobots is reduced to a microscopic level to perform a task in very small spaces.
2. However, it is only in the developmental stage.
3. The future prospects of it are much expected in the medical field:
   • Nano-robots in the bloodstream to perform small surgical procedures.
   • To fight against bacteria
   • Repairing individual cells in the body.
4. It can travel into the body and once the job is performed it can find its way out.
5. Chinese scientists have created the world’s first autonomous DNA robots to combat cancer tumours.

Question 6.
List out the recent advancement in medical technology.
Answer:

1. Virtual reality
2. Precision medicine
3. Health wearables
4. Artificial organs
5. 3D printing
6. Wireless brain sensors
7. Robotic surgery
8. Smart inhalers

Question 7.
Discuss the applications of nanomaterial-based products in
i) Electronic Industry
ii) Medicine.
Answer:

Electronic industry	Medicine
1. Data memory 2. Displays 3. Laser diodes 4. Glass fibers 5. Optical switches 6. Filters 7. Conductive, antistatic coatings	1. Drug delivery system 2. Active agents 3. Contrast medium 4. Medical rapid tests 5. Prostheses and implants 6. Antimicrobial agents and coatings 7. Agents in cancer therapy

Question 8.
Write about artificial intelligence briefly.
Answer:
The aim of artificial intelligence is to bring in human-like behaviour in robots. It works on

1. Face recognition
2. Providing a response to player’s actions in computer games
3. Taking decisions based on previous actions
4. To regulate the traffic by analyzing the density of traffic on roads.
5. Translate words from one language to another

Question 9.
What are the uses of industrial robots?
Answer:
Industrial robots are used for

1. Welding
2. Cutting
3. Lifting
4. Packing
5. Transport
6. Sorting
7. Bending
8. Assembling
9. Manufacturing
10. Weaponry
11. Industrial goods
12. Laboratory research
13. Mass production of consumer
14. Robotic water jet cutting
15. Robotic laser cutting
16. Handling hazardous materials like nuclear waste.