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Published on: 13/05/2022
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Take MCQ Business Maths and Statistics Test

1.
The following figures relates to the profits of a commercial concern for 8 years
| Year | 1986 | 1987 | 1988 | 1989 | 1990 | 1991 | 1992 | 1993 |
| Profit (Rs.) | 15,420 | 15,470 | 15,520 | 21,020 | 26,500 | 31,950 | 35,600 | 34,900 |
Find the trend of profits by the method of three yearly moving averages.
2.
Explain the method of fitting a straight line.
3.
Calculate four-yearly moving averages of number of students studying in a higher secondary school in a particular city from the following data.
| Year | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 |
| Sales | 124 | 120 | 135 | 140 | 145 | 158 | 162 | 170 | 175 |
4.
Fit a trend line by the method of semi-averages for the given data.
| Year | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 |
| Production | 105 | 115 | 120 | 100 | 110 | 125 | 135 |
5.
You are given below the values of sample mean ( \(\bar{X}\) ) and the range ( R ) for ten samples of size 5 each. Draw mean chart and comment on the state of control of the process.
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| \(\overset{-}{X}\) | 43 | 49 | 37 | 44 | 45 | 37 | 51 | 46 | 43 | 47 |
| R | 5 | 6 | 5 | 7 | 7 | 4 | 8 | 6 | 4 | 6 |
Given the following control chart constraint for : n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115
1.
| Year X | Profit (Rs) Y | 3-yearly moving total | 3-yearly moving average |
| 1986 | 15420} | - | - |
| 1987 | {15470} | 46410 | 15470 (∴ 46410 / 3) |
| 1988 | {{11520} | 52010 | 17336.666 |
| 1989 | {{21020 | 63040 | 21013.333 |
| 1990 | {{26500 | 79470 | 26490 |
| 1991 | 31950 | 94050 | 31350 |
| 1992 | 35600 | 102450 | 34150 |
| 1993 | 34900 | - | - |
2.
The line of least fit is a line from which the sum of the deviations of various points is zero. This is the best method for obtaining the trend values.
(i) The straight line trend is represented by
Y = a + b X ...(1)
where Y is the actual value and X is time
(ii) The constants 'a' and 'b' are estimated by solving the following two normal Equations
\(\sum\) Y = n a + b \(\sum\) X
\(\sum\) XY = a\(\sum\) X + b \(\sum\) X2 where n is the number of years given in the data.
(iii) By substituting the values of 'a' and 'b' in the trend equation (1), we get the line of best fit.
3.
Computation of four- yearly moving averages.
| Year | Sales | 4-yearly centered moving total | 4-yearly moving Average | 4-yearly centered moving Average |
| 2001 | 124 | --- | -- | -- |
| 2002 | 120 | -- | -- | -- |
| 519 | 129.75 | |||
| 2003 | 135 | -- | 139.37 | |
| 540 | 135 | |||
| 2004 | 140 | -- | 139.75 | |
| 578 | 144.50 | |||
| 2005 | 145 | -- | 147.87 | |
| 605 | 151.25 | |||
| 2006 | 158 | -- | 162.50 | |
| 635 | 158.75 | |||
| 2007 | 162 | -- | 162.50 | |
| 665 | 166.25 | |||
| 2008 | 170 | -- | -- | - |
| 2009 | 175 | -- | -- | - |
4.
Since the number of years is odd(seven), we will leave the middle year’s production value and obtain the averages of first three years and last three years.

| Year | Production | Average |
| 2000 | 105 | \(\frac{105+115+120}{3}=113.33\) |
| 2001 | 115 | |
| 2002 | 120 | |
| 2003 | 100(left out) | |
| 2004 | 110 | \(\frac{110+125+135}{3}=123.33\) |
| 2005 | 125 | |
| 2006 | 135 |
5.

\(\overset { = }{ X } =\frac { \sum { \bar { X } } }{ 10 } =\frac { 442 }{ 10 } =44.2\)
\(\bar { R } =\frac { \sum { R } }{ n } =\frac { 58 }{ 10 } =5.8\)
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\bar { R } \)
= 44.2 + 0.483(5.8) = 47.00
\(CL=\overset { = }{ X } =44.2\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\bar { R } =44.2-0.483(5.8)=41.39\)
The above diagram shows all the three control lines with the data points plotted, since four points falls out of the control limits, we can say that the process is out of control.
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