JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 15)
The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :
$${{92} \over 5}$$
$${{134} \over 5}$$
$${{536} \over {25}}$$
$${{112} \over 5}$$
Explanation
Let 8, 16, x1, x2, x3, x4, x5 be the observations.
Now, $${{{x_1} + {x_2} + .... + {x_5} + 14} \over 7} = 8$$
$$ \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 42} $$ .... (1)
Also, $${{x_1^2 + x_2^2 + ...x_5^2 + {8^2} + {6^2}} \over 7} - 64 = 16$$
$$ \Rightarrow \sum\limits_{i = 1}^5 {x_i^2 = 560 - 100 = 460} $$ ..... (2)
So variance of x1, x2, ......., x5
$$ = {{460} \over 5} - {\left( {{{42} \over 5}} \right)^2} = {{2300 - 1764} \over {25}} = {{536} \over {25}}$$
Now, $${{{x_1} + {x_2} + .... + {x_5} + 14} \over 7} = 8$$
$$ \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 42} $$ .... (1)
Also, $${{x_1^2 + x_2^2 + ...x_5^2 + {8^2} + {6^2}} \over 7} - 64 = 16$$
$$ \Rightarrow \sum\limits_{i = 1}^5 {x_i^2 = 560 - 100 = 460} $$ ..... (2)
So variance of x1, x2, ......., x5
$$ = {{460} \over 5} - {\left( {{{42} \over 5}} \right)^2} = {{2300 - 1764} \over {25}} = {{536} \over {25}}$$
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