JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 16)
If the data x1, x2,......., x10 is such that the mean of first four of these is 11, the mean of the remaining six is
16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :
$$\sqrt 2 $$
2
2$$\sqrt 2 $$
4
Explanation
$${\sigma ^2} = {{\sum {x_i^2} } \over {10}} - {\left( {{{\sum {{x_i}} } \over {10}}} \right)^2} \to (i)$$
Now x1 + x2 + x3 + x4 = 44 & x5 + x6 + ......... + x10 = 96
Hence $${\sigma ^2}$$ = $${{2000} \over {10}} - {\left( {{{140} \over {10}}} \right)^2}$$ = 200 - 196 = 4
Hence $$\sigma $$ = 2
Now x1 + x2 + x3 + x4 = 44 & x5 + x6 + ......... + x10 = 96
Hence $${\sigma ^2}$$ = $${{2000} \over {10}} - {\left( {{{140} \over {10}}} \right)^2}$$ = 200 - 196 = 4
Hence $$\sigma $$ = 2
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