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12th Standard English Medium Business Maths Subject Applied Statistics Creative 3 Mark Questions with Solution Part - I

Published on: 23/06/2021

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Questions:

Generic 3

5 x 3 = 15

Compute Fisher's price index number for the following data.

Commodity	Base Year		Current Year
Commodity	Price	Quantity	Price	Quantity
A	10	12	12	15
B	7	15	5	20
C	5	24	9	20
D	16	5	14	5

Calculate the seasonal indices for the following data by the method of simple average.

Year	Quarters
Year	I	II	III	IV
1994	78	66	84	80
1995	76	74	82	78
1996	72	68	80	70
1997	74	70	84	74
1998	76	74	86	82

Find a trend line to the following data by the method of sami-averages.

Years	1980	1981	1982	1983	1984	1985	1986
Sales	102	105	114	110	108	116	112

Fit a trend line to the following data by graphic method.

Year	1978	1979	1980	1981	1982	1983	1984	1985	1986
Production of steel	20	22	24	21	23	25	23	26	25

Fit a straight line trend for the following data using the method of least squares.

x	0	1	2	3	4
y	1	1	3	4	6

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Answers Key:

Generic 3

5 x 3 = 15

Commodity	Base Year		Current Year
Commodity	p₀	q₀	p₁	q₁
A	10	12	12	15
B	7	15	5	20
C	5	24	9	20
D	16	5	14	5

p₁q₀	p₀q₀	p₁q₁	p₀q₁
144	120	180	150
75	105	100	140
216	120	180	100
70	80	70	80
505	425	530	470

Fisher's ideal index = \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}} \times 100\)
= \(\sqrt{\frac{505}{425} \times \frac {430}{470}} \times 100\)
\(P^{F}_{01}\) = 115.75

Year	Quarters
Year	I	II	III	IV
1994	78	66	84	80
1995	76	74	82	78
1996	72	68	80	70
1997	74	70	84	74
1998	76	74	86	82
Total	376	352	416	384
Average	75.2	70.4	83.2	76.8

Grand average = \(\frac {75.2+70.4+83.2+76.8}{4}\)
= \(\frac {305.6}{4}\) = 76.4
Seasonal index S.I = \(\frac {Quarterly average}{Grand average} \times 100\)
Hence, S.I for I quarter = \(\frac {75.2}{76.4} \times 100\) = 98.4
S.I for II quarter =  \(\frac {70.4}{76.4} \times 100\) = 92.14
S.I for III quarter =  \(\frac {83.2}{76.4} \times 100\) = 108.9
S.I for IV quarter =  \(\frac {76.8}{76.4} \times 100\) = 100.5

No of years = 7 (odd), by omitting the middle year we have

x	y	x²	xy
0	1	0	0
1	1	1	1
2	3	4	6
3	4	9	12
4	6	16	24
10	15	30	43

Let y= ax + b be the best line of fit.
The normal equations are
a\(\sum\)x + nb = \(\sum\)y
a\(\sum\)x² + b\(\sum\)x = \(\sum\)xy
10a+ 5b = 15
⇒ 2a+b = 3 ...(1)
30a + 10b = 43
⇒ 3a+b = 4.3 ....(2)
(1)-(2) ⇒ - a = -1.3
⇒ a = 1.3
Substituting a = 1.3 in (1) we get,
2(1.3) + b = 3
⇒ 2.6 + b = 3
⇒ b = 3 - 2.6 = 0.4
∴ The line of best fit isy = 1.3x + 0.4