JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 9)
If the mean of the data : 7, 8, 9, 7, 8, 7, $$\lambda $$, 8 is 8, then the variance of this data is :
$${7 \over 8}$$
1
$${9 \over 8}$$
2
Explanation
$$\overline x $$ = $${{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8} \over 8}$$ = 8
$$ \Rightarrow $$$$\,\,\,$$ $${{54 + \lambda } \over 8}$$ = 8 $$ \Rightarrow $$ $$\lambda $$ = 10
Now variance = $$\sigma $$2
= $${{{{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {9 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {10 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2}} \over 8}$$
$$ \Rightarrow $$ $$\sigma $$2 = $${{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0} \over 8}$$ = $${8 \over 8}$$ = 1
Hence, the variance is 1.
$$ \Rightarrow $$$$\,\,\,$$ $${{54 + \lambda } \over 8}$$ = 8 $$ \Rightarrow $$ $$\lambda $$ = 10
Now variance = $$\sigma $$2
= $${{{{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {9 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {10 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2}} \over 8}$$
$$ \Rightarrow $$ $$\sigma $$2 = $${{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0} \over 8}$$ = $${8 \over 8}$$ = 1
Hence, the variance is 1.
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