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13.Statistics
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If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is
A
$400$
B
$380$
C
$525$
D
$480$
(JEE MAIN-2019)
Solution
Mean $\left( \mu \right) = \frac{{\sum {{x_i}} }}{{50}} = 16$
Standard deviation $\left( \sigma \right) = \sqrt {\frac{{\sum {x_i^2} }}{{50}} – {{\left( \mu \right)}^2}} = 16$
$ \Rightarrow \left( {256} \right) \times 2 = \frac{{\sum {x_i^2} }}{{50}}$
$\Rightarrow$ New mean
$ = \frac{{\sum {{{\left( {{x_i} – 4} \right)}^2}} }}{{50}} = \frac{{\sum {x_i^2 + 16 \times 50 – 8\sum {{x_i}} } }}{{50}}$
$ = \left( {256} \right) \times 2 + 16 – 8 \times 16 = 400$
Standard 11
Mathematics
Similar Questions
From the data given below state which group is more variable, $A$ or $B$ ?
Marks | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
Group $A$ | $9$ | $17$ | $32$ | $33$ | $40$ | $10$ | $9$ |
Group $B$ | $10$ | $20$ | $30$ | $25$ | $43$ | $15$ | $7$ |
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