JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 5)
The mean and variance of 20 observations are
found to be 10 and 4, respectively. On
rechecking, it was found that an observation 9
was incorrect and the correct observation was
11. Then the correct variance is
3.98
3.99
4.01
4.02
Explanation
Let 20 observation be x1
, x2
,....., x20
Mean =
$$ \Rightarrow $$ x1 + x2 +, .....+ x20 = 200
Variance = $${{\sum\limits_{i = 1}^{i = n} {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$$
$$ \Rightarrow $$ 4 = $${{x_1^2 + x_2^2 + ... + x_{20}^2} \over {20}}$$ - 102
$$ \Rightarrow $$ $${x_1^2 + x_2^2 + ... + x_{20}^2}$$ = 2080
Also x1 + x2 +, .....+ x20 - 9 + 11 = 202
new variance will be
= $${{x_1^2 + x_2^2 + ... + x_{20}^2 - 81 + 121} \over {20}}$$ - $${\left( {{{202} \over {20}}} \right)^2}$$
= 3.99
Mean =
x1
+ x2
+, .....+ x20
20
= 10
$$ \Rightarrow $$ x1 + x2 +, .....+ x20 = 200
Variance = $${{\sum\limits_{i = 1}^{i = n} {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$$
$$ \Rightarrow $$ 4 = $${{x_1^2 + x_2^2 + ... + x_{20}^2} \over {20}}$$ - 102
$$ \Rightarrow $$ $${x_1^2 + x_2^2 + ... + x_{20}^2}$$ = 2080
Also x1 + x2 +, .....+ x20 - 9 + 11 = 202
new variance will be
= $${{x_1^2 + x_2^2 + ... + x_{20}^2 - 81 + 121} \over {20}}$$ - $${\left( {{{202} \over {20}}} \right)^2}$$
= 3.99
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