Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1

Question 1.
What is population?
Solution:
The group of individuals considered under study is called as population. The word population here refers not only to people but to all items that have been chosen for the study.

Question 2.
What is sample?
Solution:
A selection of a group of observation/individuals/population in such a way that is represents the population is called as sample.

Question 3.
What is statistic?
Solution:
Statistic: Any statistical measure computed from sample is known as statistic.

Question 4.
Define parameter.
Solution:
Parameter: The statistical constants of the population like mean (µ),variance (σ²) are referred as population parameters.

Question 5.
What is sampling distribution of a statistic?
Solution:
Sampling distribution of a statistic is the frequency distribution which is formed with various of a statistic computed from different samples of the same size drawn from the same population.

Question 6.
What is standard error?
Solution:
The standard deviation of the sampling distribution of a statistic is known as its Standard Error abbreviated as S.E.

Question 7.
Explain in detail about simple random sampling with a suitable example.
Solution:
In this technique the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. Simple random sampling may be done, with or without replacement of a samples selected. In a simple random sampling with replacement there is a possibility of selecting the same sample any number of times. So, simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units ,the probability of drawing any unit at the first draw is \(\frac { 1 }{N}\), the probability of drawing any unit in the second draw from among the available (N – 1) units is \(\frac { 1 }{(N-1)}\), and so on.

For example, if we want to select 10 students, out of 100 students, then we must write the names/roll number of all the 100 students on slips of the same size and mix them, then we make a blindfold selection of 10 students. This method is called unrestricted random sampling, because units are selected from the population without any restriction. This method is mostly used in lottery draws. If the population or universe is infinite, this method is inapplicable.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
In stratified random sampling, first divide the population into sub-populations, which are called strata. Then,the samples are selected from each of the strata through random techniques. The collection of all the samples from all strata gives the stratified random samples.

When the population is heterogeneous or different segments or groups with respect to the variable or characteristic under study, then stratified Random sampling methos is studied. First, the population is divided into homogeneous number of sun-groups of strata before the sample is drawn. A sample from each stratum at random. Following steps are involved for selecting random sample in a stratified random sampling method.

(a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.

(b) After the population is stratified,a sample of a specified size is drawn at random from each stratum using Lottery Method or table of random number method.

Question 9.
Explain in detail about systematic random sampling with example.
Solution:
In a systematic sampling, randomly select the first sample from the first k units. Then every k th members, starting with the first selected sample, is included in the sample.

Systematic sampling is a commonly used technique,if the complete and up-to-date list of the sampling units is available. We can arrange the items in numerical, alphabetical, selecting the first at random, the rest being automatically selected according to some pre-determined pattern. A systematic is formed by selecting every item from the population, where k refers to the sample interval. The sampling interval can be determined by divided the size of the population by the size of the sample to be chosen.

That is K = \(\frac { N }{n}\), where k is an integer.
k = sampling interval, size of the population, sample size
Procedure for selection of samples by systematic sampling method

(i) If we want to select a sample of 10 students from a class of 100 students,the sampling interval is Calculated as k = \(\frac { N }{n}\) = \(\frac { 100 }{10}\) = 10
Thus sampling interval = 10 denotes that for every 10 samples one sample

(ii) The first sample is selected from the first 10(sampling interval) samples through selection procedures.

(iii) If the selected first random sample is 5, then the rest of the samples are automatically selected by incriminating the value of the sampling interval 9k = 10. i.e, 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.

Question 10.
Explain in detail about sampling error.
Solution:
Sampling Error:
Error, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors, sampling errors are inherent in the method of sampling. They may arise accidentally without any bias or prejudice. Sampling Errors arise primarily due to the following reasons:
(a) Faulty selection of the sample instead of correct sample by defective sampling technique.
(b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
(c) In area surveys,while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as faulty demarcation of sampling units.

Question 11.
Explain in detail about non-sampling error.
Solution:
Non-sampling Errors:
The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments (tape, scale) are called Non¬sampling errors. It may arise in the following ways:
(a) Due to neglience and carelessness of the part of either investigator or respondents
(b) Due to lack of trained and qualified investigators.
(c) Due to framing of a wrong questionnaire.
(d) Due to apply wrong statistical measure.
(e) Due to incomplete investigation and sample survey.

Question 12.
State any two merits of simple random sampling.
Solution:
Merits:
1. Personal bias is completely eliminated.
2. This method is economical as it saves time, money and labour.
3. The method requires minimum knowledge about the population in advance.

Question 13.
State any three merits of stratified random sampling.
Solution:
Merits:
(a) A random stratified sample is superior to a sample random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
(b) A stratified random sample can be kept small in size without losing its accuracy
(c) it is easy to administer, if the population under study is sub divided
(d) It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had the geographical areas.

Question 14.
State any two demerits of systematic random sampling.
Solution:
Demerits:
1. Systematic samples are not random samples.
2. If N is not multiple of n-then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

Question 15.
State any two merits for systematic random sampling.
Solution:
Merits:
1. This is simple and convenient method.
2. This method distributes the sample more evenly over the entire listed population.
3. The time and work is reduced much.

Question 16.
Using the following Tippet’s random number table

Draw a sample of 10 three digit numbers which are even numbers.
Solution:
There are many ways to select 10 random samples from the given Tippets random number number table since the population size is three digit numbers, Here the door numbers must be even (ie) the unit digit must be even. Here we consider column wise selection of random numbers starting from first column.

So the first sample is 416 and other 9 samples are 056, 664, 952, 748, 524, 914, 154, 340 and 140.
Tippets random number Table

Question 17.
A wholesaler in apples claims that only 4% of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning of good apples.
Solution:
sample size = 600; Number of success = 600 – 36
= 564
sample proportion p = \(\frac { 564 }{600}\) = 0.94
600
population proportion (p) = probability of getting good apple
= 96%
= \(\frac { 96 }{100}\) {∵ 4% of the apples 100 are defective}
P = 0.96
Q = 1 – p = 1 – 0.96
Q = 0.04
The S.E for a sample proporation is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.96)(0.04) }{600}}\)
\(\sqrt{\frac { 0.0384 }{600}}\) = \(\sqrt{0.000064}\)
∴ S.E = 0.008
Hence the standard error foe sample proportion is S.E = 0.008

Question 18.
A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Tamil Nadu State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. Calculate standard error of mean.
Solution:
Given n = 1000; \(\bar{x}\) = 119 lbs (pounds)
s = 30 lbs is known in this problem.
since σ is unknown, so we consider \(\bar{σ}\) = s and µ = 120 lbs
S.E = \(\frac { \bar{σ} }{√n}\) = \(\frac { s }{√n}\) = \(\frac { 30 }{\sqrt{1000}}\)
= \(\frac { 30 }{31.623}\) = 0.9487
Therefore the standard error for the average weight of large group of students of 120 lbs is 0.9487

Question 19.
A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. Calculate the suitable standard error that this sample is taken from a population with standard deviation 3?
Solution:
Sample size n = 60
Sample S.D S = 2.5
population S.D a = 3
The standard error for sample S.D is given by
\(\sqrt{\frac { σ^2 }{2n}}\) = \(\sqrt{\frac { (3)^2 }{2(60)}}\) = \(\frac { 3 }{\sqrt{120}}\)
= \(\frac { 3 }{10.954}\) = 0.27387
= 0.2739
Thus standard error for sample S.D = 0.2739

Question 20.
In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village?
Solution:
sample size n = 400
case (i):
sample proporation of vegetarian p = \(\frac { 3 }{10.954}\) = \(\frac { 230 }{400}\)
p = 0.575
q = 1 – p
= 1 – 0.575
q = 0.425
Sample error S.E= \(\sqrt{\frac { pq }{n}}\)
= \(\sqrt{\frac { 0.575×0.425 }{400}}\) = \(\sqrt{\frac { 0.223125 }{400}}\)
\(\sqrt{0.0005578125}\)
S.E = 0.2361

Case(ii):
sample size n = 400
since both vegetarian and non- vegetarian foods are equally popular in that village
sample proparation of vegetarian p = \(\frac { 1 }{2}\) = 0.5
q = 1 -p ⇒ q = 1 – 0.5
q = 0.5

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)

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Samacheer Kalvi 12th Business Maths Book Solutions Answers Guide

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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

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• 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

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• 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

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• 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

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• 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

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• 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

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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

(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

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:

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

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

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)