JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 15)
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :
16
22
20
18
Explanation
Average height of 5 students,
$$\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150$$
$$ \Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750$$
We know,
Variance $$\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$$
given that,
$${{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18$$
$$ \Rightarrow \,\,\,\sum {x_i^2} = 112590$$
Height of new student, x6 $$=$$ 156 cm
New average height $$\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151$$
New variance $$ = {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}$$
$$ = {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}$$
$$ = 22821 - 22801$$
$$ = 20$$
$$\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150$$
$$ \Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750$$
We know,
Variance $$\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$$
given that,
$${{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18$$
$$ \Rightarrow \,\,\,\sum {x_i^2} = 112590$$
Height of new student, x6 $$=$$ 156 cm
New average height $$\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151$$
New variance $$ = {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}$$
$$ = {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}$$
$$ = 22821 - 22801$$
$$ = 20$$
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