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Published on: 05/06/2021
QB365 provides detailed and simple solution for every Book back Questions in class 12 Business Maths Subject. It will helps to get more idea about question pattern in every book back questions with solution.
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Questions + Answers key
Take MCQ Business Maths and Statistics Test

1.
The following data gives the readings for 8 samples of size 6 each in the production of a certain product. Find the control limits using mean chart.
| Sample | 1 | 2 | 3 | 4 | 5 | 6 |
| Mean | 300 | 342 | 351 | 319 | 326 | 333 |
| Range | 25 | 37 | 20 | 28 | 30 | 22 |
Given for n = 6, A2 = 0.483,
2.
The following table shows the number of salesmen working for a certain concern:
| Year | 1992 | 1993 | 1994 | 1995 | 1996 |
| No. of salesmen | 46 | 48 | 42 | 56 | 52 |
Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997.
3.
Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2002, 2003 and 2004.
| 2002 | 15 | 18 | 17 | 19 | 16 | 20 | 21 | 18 | 17 | 15 | 14 | 18 |
| 2003 | 20 | 18 | 16 | 13 | 12 | 15 | 22 | 16 | 18 | 20 | 17 | 15 |
| 2004 | 18 | 25 | 21 | 11 | 14 | 16 | 19 | 20 | 17 | 16 | 18 | 20 |
4.
Determine the equation of a straight line which best fits the following data
| Year | 2000 | 2001 | 2002 | 2003 | 2004 |
| Sales(Rs.000) | 35 | 36 | 79 | 80 | 40 |
Compute the trend values for all years from 2000 to 2004
5.
Compute the average seasonal movement for the following series
| Year | Quarterly Production | |||
| I | II | III | IV | |
| 2002 | 3.5 | 3.8 | 3.7 | 3.5 |
| 2003 | 3.6 | 4.2 | 3.4 | 4.1 |
| 2004 | 3.4 | 3.9 | 3.7 | 4.2 |
| 2005 | 4.2 | 4.5 | 3.8 | 4.4 |
| 2006 | 3.9 | 4.4 | 4.2 | 4.6 |
1.
| Sample | 1 | 2 | 3 | 4 | 5 | 6 | Total |
| Mean | 300 | 342 | 351 | 319 | 326 | 333 | 1971 |
| Range | 25 | 37 | 20 | 28 | 30 | 22 | 162 |
\(\overset { = }{ X } =\frac { \sum { \overset { - }{ X } } }{ number\ of\ samples } =\frac { 1971 }{ 6 } =328.5\quad \quad \overset { - }{ R } =\frac { \sum { R } }{ n } =\frac { 162 }{ 6 } =27\)
The control limits for \(\overset {-}{X}\) chart is
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\overset { - }{ R } =328.5+0.483(27)=341.54\)
\(CL=\overset { = }{ X } =328.5\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\overset { - }{ R } =328.5+0.483(27)=315.45\)
2.
| Year (X) | No. of Salesmen Y | X = x -1994 | X2 | XY |
| 1992 | 46 | -2 | 4 | -92 |
| 1993 | 48 | -1 | 1 | -48 |
| 1994 | 42 | 0 | 0 | 0 |
| 1995 | 56 | 1 | 1 | 56 |
| 1996 | 52 | 2 | 4 | 104 |
| 244 | 0 | 10 | 20 |
Smce \(\sum\)X = 0, a =\(\frac {\sum Y}{n}\) = \(\frac {244}{5}\) = 48.8
b = \(\frac {\sum XY}{\sum X^2}\) = \(\frac {20}{10}\) = 2
∴ The required equation of the straight line trend is given by
Y = a + bX \(\Rightarrow \) Y = 48.8 + 2X
\(\Rightarrow \) Y = 48.8 +2 (X - 1994) .... (1)
∴ Number of salesmen in 1997 is put X = 1997 in (1)
∴ Y = 48.8 + 2 (1997 - 1994)
= 48.8 + 2 (3)
= 48.8 + 6 = 54.8
∴ Number of salesmen in 1997 is 54.8
3.
| Year | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| 2002 | 15 | 18 | 17 | 19 | 16 | 20 | 21 | 18 | 17 | 15 | 14 | 18 |
| 2003 | 20 | 18 | 16 | 13 | 12 | 15 | 22 | 16 | 18 | 20 | 17 | 15 |
| 2004 | 18 | 25 | 21 | 11 | 14 | 16 | 19 | 20 | 17 | 16 | 18 | 20 |
| Monthly Total | 53 | 61 | 54 | 43 | 42 | 51 | 62 | 54 | 52 | 51 | 49 | 53 |
| Monthly Average | 17.6 | 20.3 | 18 | 14.3 | 14 | 17 | 20.6 | 18 | 17.3 | 17 | 16.3 | 17.6 |
Grand Average = \(= \frac {17.6+ 20.3+ 18+ 14.3 + 14+ 17 + 20.6+ 18+ 17.3 + 17 + 16.3+ 17.6}{12}\) = \(\frac {208}{12}\) = 17.33
S.I for Jan = \(\frac {17.6}{17.33}\) × 100 = 101.56
S.I for Feb = \(\frac {20.3}{17.33}\) × 100 = 117.14
S.I for Mar = \(\frac {18}{17.33}\) × 100 = 103.87
S.I for April = \(\frac {14.3}{17.33}\) × 100 = 82.51
S.I for May = \(\frac {14}{17.33}\) × 100 = 80.78
S.I for June = \(\frac {17}{17.33}\) × 100 = 98.10
S.I for July = \(\frac {20}{17.33}\) × 100 = 118.87
S.I for Aug = \(\frac {20.6}{17.33}\) × 100 = 103.87
S.I for Sep = \(\frac {17.3}{17.33}\) × 100 = 99.83
S.I for Oct = \(\frac {17}{17.33}\) × 100 = 98.10
S.I for Nov = \(\frac {16.3}{17.33}\) × 100 = 94.06
S.I for Dec = \(\frac {17.6}{17.33}\) × 100 = 101.56
4.
| Year | Sales | X=x- 2002 | X2 | XY |
| 2000 | 35 | -2 | 4 | -70 |
| 2001 | 36 | -1 | 1 | -36 |
| 2002 | 79 | 0 | 0 | 0 |
| 2003 | 80 | 1 | 1 | 80 |
| 2004 | 40 | 2 | 4 | 80 |
| 270 | 0 | 10 | 54 |
Since \(\sum X = 0, a = \frac{\sum Y}{n}\) = \(\frac {270}{5}\) = 54
b = \(\frac {\sum XY}{\sum X^2} = \frac {54}{10}\) = 5.4
∴ The required equation of the straight line trend is given by Y = a +bX
\(\Rightarrow \) Y = 54 + 5.4 (x - 2002)
The trend values can be obtained as follows:
When X = 2000, Yt = 54 + 5.4 (2000 - 2002)
= 54 + 5.4 (-2) = 43.2
When X = 2001, Yt = 54 + 5.4 (2001 - 2002)
= 54 + 5.4(-1) = 48.6
When X = 2002, Yt = 54 + 5.4 (2002 - 2002)
= 54
When X = 2003, Yt = 54 + 5.4 (2003 - 2002)
= 54 + 5.4 = 59.4
When X = 2004, Yt = 54 + 5.4 (2004 - 2002)
= 54 + 5.4 (2) = 64.8
5.
| Year | Quarterly production | |||
| I | II | III | IV | |
| 2002 | 3.5 | 3.8 | 3.7 | 3.5 |
| 2003 | 3.6 | 4.2 | 3.4 | 4.1 |
| 2004 | 3.4 | 3.9 | 3.7 | 4.2 |
| 2005 | 4.2 | 4.5 | 3.8 | 4.4 |
| 2006 | 3.9 | 4.4 | 18.8 | 20.8 |
| Quarterly Total | 18.6 | 20.8 | 18.8 | 20.8 |
| Average | 3.72 | 4.16 | 3.76 | 20.8 |
Grand average = (3.72 + 4.16 + 3.76 + 4.16)/4 = 3.95
Seasonal Index (S.I) for I quarter = (Average of I quarter)/(Grand Average) \(\times\) 100
S.I. for I quarter = 3.72/3.95 \(\times\) 100 = 94.1772
S.I. for II quarter = 4.16/3.95 \(\times\) 100 = 105.3165
S.I. for III quarter = 3.76/3.95 \(\times\) 100 = 95.1899
S.I. for IV quarter = 4.16/3.95 \(\times\) 100 = 105.3165
Thus we obtain the average seasonal movement.
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