JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 15)
A student scores the following marks in five tests
:
45, 54, 41, 57, 43.
His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is
45, 54, 41, 57, 43.
His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is
$$100 \over {\sqrt 3}$$
$$10 \over {\sqrt 3}$$
$$10 \over3$$
$$100 \over3$$
Explanation
Let the score in the sixth test = x
Given, Mean ($$\overline x $$) = 48
$$ \Rightarrow $$ $${{45 + 54 + 41 + 57 + 43 + x} \over 6}$$ = 48
$$ \Rightarrow $$ x = 48
Standard deviation (SD)
= $$\sqrt {{{\sum\limits_{i = 1}^N {{{\left( {{x_i} - \overline x } \right)}^2}} } \over N}} $$
= $$\sqrt {{\matrix{ {\left( {45 - 48} \right)^2} + {\left( {54 - 48} \right)^2} \hfill \cr + {\left( {41 - 48} \right)^2} + {\left( {57 - 48} \right)^2} \hfill \cr + {\left( {43 - 48} \right)^2} + {\left( {48 - 48} \right)^2} \hfill \cr} \over 6}} $$
= $$\sqrt {{{9 + 36 + 49 + 81 + 25} \over 6}} $$
= $$\sqrt {{{200} \over 6}} $$
= $$\sqrt {{{100} \over 3}} $$
= $${{10} \over {\sqrt 3 }}$$
Given, Mean ($$\overline x $$) = 48
$$ \Rightarrow $$ $${{45 + 54 + 41 + 57 + 43 + x} \over 6}$$ = 48
$$ \Rightarrow $$ x = 48
Standard deviation (SD)
= $$\sqrt {{{\sum\limits_{i = 1}^N {{{\left( {{x_i} - \overline x } \right)}^2}} } \over N}} $$
= $$\sqrt {{\matrix{ {\left( {45 - 48} \right)^2} + {\left( {54 - 48} \right)^2} \hfill \cr + {\left( {41 - 48} \right)^2} + {\left( {57 - 48} \right)^2} \hfill \cr + {\left( {43 - 48} \right)^2} + {\left( {48 - 48} \right)^2} \hfill \cr} \over 6}} $$
= $$\sqrt {{{9 + 36 + 49 + 81 + 25} \over 6}} $$
= $$\sqrt {{{200} \over 6}} $$
= $$\sqrt {{{100} \over 3}} $$
= $${{10} \over {\sqrt 3 }}$$
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