JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 2)
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $$\sqrt {13.44} $$, then the standard deviation of the second sample is :
8
6
4
5
Explanation
n1 = 100
m = 250
$$\overline X $$1 = 15
$$\overline X $$ = 15.6
V1(x) = 9
Var(x) = 13.44
$${\sigma ^2} = {{{n_1}\sigma _1^2 + {n_2}\sigma _2^2} \over {{n_1} + {n_2}}} + {{{n_1}{n_2}} \over {{{({n_1} + {n_2})}^2}}}{({\overline x _1} - {\overline x _2})^2}$$
n2 = 150, $${\overline x _2}$$ = 16, V2(x) = $$\sigma$$2
$$13.44 = {{100 \times 9 + 150 \times \sigma _2^2} \over {250}} + {{100 \times 150} \over {{{(250)}^2}}} \times 1$$
$$\Rightarrow$$ $${\sigma _2}^2$$ = 16 $$\Rightarrow$$ $$\sigma$$2 = 4
m = 250
$$\overline X $$1 = 15
$$\overline X $$ = 15.6
V1(x) = 9
Var(x) = 13.44
$${\sigma ^2} = {{{n_1}\sigma _1^2 + {n_2}\sigma _2^2} \over {{n_1} + {n_2}}} + {{{n_1}{n_2}} \over {{{({n_1} + {n_2})}^2}}}{({\overline x _1} - {\overline x _2})^2}$$
n2 = 150, $${\overline x _2}$$ = 16, V2(x) = $$\sigma$$2
$$13.44 = {{100 \times 9 + 150 \times \sigma _2^2} \over {250}} + {{100 \times 150} \over {{{(250)}^2}}} \times 1$$
$$\Rightarrow$$ $${\sigma _2}^2$$ = 16 $$\Rightarrow$$ $$\sigma$$2 = 4
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