Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Miscellaneous Problems

Question 1.
Using three yearly moving averages, Determine the trend values from the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 1
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 2

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 2.
From the following data, calculate the trend values using fourly moving averages.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 3
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 4

Question 3.
Fit a straight line trend by the method of least squares to the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 5
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 6
Therefore, the required equation of the straight line trend is given by
y = a + bx
y = 55.9875 + 0.830 x
⇒ y = 55.9875 + 0.83 (\(\frac { x-1983.5 }{0.5}\))
The trend values can be obtained by
When x = 1980
y = 55.9875 + 0.83 (\(\frac { 1980-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-7)
= 55.9875 – 5.81
= 50.1775
When x = 1981
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-5)
= 55.9875 – 4.15
= 51.8375
When x = 1982
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-3)
= 55.9875 – 2.49
= 53.4975
When x = 1983
y = 55.9875 + 0.83 (\(\frac { 1983-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-1)
= 55.9875 – 0.83
= 55.1575
When x = 1984
y = 55.9875 + 0.83 (\(\frac { 1984-1983.5 }{0.5}\))
55.9875 + 0.83 (1)
= 56.8175
when x = 1985
y = 55.9875 + 0.83 (\(\frac { 1985-1983.5 }{0.5}\))
= 55.9875 + 0.83 (3)
= 55.9875 + 2.49
= 58.4775
when x = 1986
y = 55.9875 + 0.83 (\(\frac { 1986-1983.5 }{0.5}\))
= 55.9875 + 0.83 (5)
= 55.9875 + 4.15
= 60.1375
when x = 1987
y = 55.9875 + 0.83 (\(\frac { 1987-1983.5 }{0.5}\))
= 55.9875 + 0.83 (7)
= 55.9875 + 5.81
= 61.7975

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 4.
Fit a straight line trend by the method of least squares to the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 7
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 8
Lasperyre’s price Index number
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 9
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 10
Hence Fisher’s Ideal Index satisfies Time reversal test

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 5.
Using the following data, construct Fisher’s Ideal Index Number and Show that it satisfies Factor Reversal Test and Time Reversal Test?
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 11
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 12
Factor reversal test
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 13
Hence Fisher’s Ideal Index satisfies Factor reversal test.

Question 6.
Compute the consumer price index for 2015 on the basis of 2014 from the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 14
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 15

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 7.
An Enquiry was made into the budgets of the middle class families in a city gave the following information.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 16
What changes in the cost of living have taken place in the middle class families of a city?
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 17
Conclusion:
The cost of living has increased up to 26.10% in 2011 as compared to 2010.

Question 8.
From the following data, calculate the control limits for the mean and range chart.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 18
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 19
UCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 51 + 0.577(6.5)
= 51 + 3.7505
= 54.7505
= 54.75
CL = \(\bar { \bar x}\) = 51
UCL = \(\bar { \bar x}\) – A2\(\bar { R}\)
= 51 – 0.577(6.5)
= 51 – 3.7505
= 47.2495
= 47.25
The control limits for Range chart is
UCL = D4\(\bar { R}\)
= 2.114(6.5)
= 13.741
CL = \(\bar { R}\) = 6.5
LCL = D3\(\bar { R}\) = 0(6.5) = 0

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 9.
The following data gives the average life(in hours) and range of 12 samples of 5 lamps each. The data are
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 20
Construct control charts for mean and range Comment on the control limits.
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 21
UCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.6675
= 1614.1675
= 1614.17
CL = \(\bar { \bar x}\) = 1367.5
LCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 1367.5 – 0.577(427.5)
= 1367.5 – 246.6675
= 1120.8325
= 1120.83
The control limits for Range chart is
UCL = D4\(\bar { R}\)
= 2.115(427.5)
= 904.1625
= 904.16
CL = \(\bar { R}\) = 427.5
LCL = D3\(\bar { R}\)
= 0(427.5)
= 0

Question 10.
The following are the sample means and I ranges for 10 samples, each of size 5. Calculate ; the control limits for the mean chart and range chart and state whether the process is in control or not.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 22
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 23
UCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 4.982 + 0.577(0.36)
= 4.982 + 0.20772
= 5.18972
= 5.19
CL = \(\bar { \bar x}\) = 4.982
LCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 4.982 – 0.577(0.36)
= 4.982 – 0.20772
= 4.77428
= 4.774
The control limits for range chart is
UCL = D2\(\bar { R}\) = 2.115(3.6)
= 7.614
CL = \(\bar { R}\) = 3.6
LCL = D3\(\bar { R}\)
= 0(0.36) = 0

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems