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13.Statistics
hard
यदि $50$ प्रेक्षणों $x _{1}, x _{2} \ldots, x _{50}$ का माध्य तथा मानक विचलन दोनों $16$ है, तो $\left(x_{1}-4\right)^{2},\left(x_{2}-4\right)^{2}, \ldots \cdots$ $\left( x _{50}-4\right)^{2}$ का माध्य है
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
