JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 18)
If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to
582.5
507.5
586.5
509.5
Explanation
$$\overline x = 10 \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 50} $$
S.D. $$ = \sqrt {{{\sum\limits_{i = 1}^5 {x_i^2} } \over 5} - {{\left( {\overline x } \right)}^2}} = 8$$
$$ \Rightarrow \,\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} = 109$$
variance $$ = \,\,{{\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} + {{\left( { - 50} \right)}^2}} \over 6} - \left( {\sum\limits_{i = 1}^5 {{{{x_i} - 50} \over 6}} } \right)$$
$$ = \,\,507.5$$
S.D. $$ = \sqrt {{{\sum\limits_{i = 1}^5 {x_i^2} } \over 5} - {{\left( {\overline x } \right)}^2}} = 8$$
$$ \Rightarrow \,\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} = 109$$
variance $$ = \,\,{{\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} + {{\left( { - 50} \right)}^2}} \over 6} - \left( {\sum\limits_{i = 1}^5 {{{{x_i} - 50} \over 6}} } \right)$$
$$ = \,\,507.5$$
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