3 Marks Question

Question 13 Marks

Find the mean variance and standard deviation for the following data:
2, 4, 5, 6, 8, 17.

Answer

x	d = (x - Mean)	d²
2	-5	²⁵
4	-3	⁹
5	-2	4
6	-1	1
8	1	1
17	10	100
42		140

$\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}=\frac{1}{6}[42]=7$

$\text{var}(\text{x})=\frac{1}{\text{n}}\Big\{\sum(\text{x}_\text{i}-\overline{\text{x}})^2\Big\}=\frac{1}{6}\big\{140\big\}=23.33$

$\text{S.D}(\text{x})=\sqrt{\text{var}(\text{x})}=\sqrt{23.33}=4.8$

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Question 23 Marks

Find the mean variance and standard deviation for the following data:
15, 22, 27, 11, 9, 21, 14, 9.

Answer

x_i	d_i=x_i	d_i²
15	0	0
22	7	49
27	12	144
11	-4	16
9	-6	36
21	6	36
14	-1	1
9	-6	36
	Total = 8	Total = 318

$\text{Mean}=15+\frac{8}{8}=16$

$\text{Var}=\frac{318}{8}-1=38.75$

$\text{SD}=\sqrt{38.75}=6.22$

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Question 33 Marks

Two plants A and B of a factory show following results about the number of workers and the wages paid to them:

	Plant A	Plant B
No. of workers	5000	6000
Average monthly wages	₹ 2500	₹ 2500
Variance of distribution of wages	81	100

In which plant A or B is there greater variability in individual wages?

Answer

Variance of the distribution of wages in plant $\text{A}\big(\sigma^2\big)=81$
Standard deviation of the distribution of wages in plant $\text{A}(\sigma)=9$
Variance of the distribution of wages in plant $\text{B}\big(\sigma^2\big)=100$
Standard deviation of the distribution of wages in plant $\text{B}(\sigma)=10$
Average monthly wages in both the plants are Rs 2500.
Thus, the plant with greater value of SD will have more variability in salary.
Plant B has more variability in individual wages than plant A.

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Question 43 Marks

The means and standard deviations of heights ans weights of 50 students of a class are as follows:

	Weights	Heights
Mean	63.2kg	63.2 inch
Standard deviation	5.6kg	11.5 inch

Which shows more variability, heights or weights?

Answer

Coeffient of variations (CV) in weights $=\frac{\text{SD}}{\text{Mean}}\times100$
$=\frac{5.6}{63.2}\times100$
$=8.86$
Coefficient of variations (CV) in heights $=\frac{11.5}{63.2}\times100$
$=18.19$
CV in heights is greater than CV in weights.
Thus, heights will show more variability than weights.

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Question 53 Marks

Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

Answer

Coeffient of variations $=\frac{\sigma}{\overline{\text{x}}}\times100$
So, we have:
$60\%=\frac{21}{\overline{\text{x}}}\times100\Rightarrow\overline{\text{x}}=\frac{21}{0.60}\times100=35$
$70\%=\frac{16}{\overline{\text{x}}}\times100\Rightarrow\overline{\text{x}}=\frac{16}{0.70}\times100=22.85$

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