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13.Statistics
hard
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
Find the mean, variance and standard deviation using short-cut method
| Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
| No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
hard
