The mean and standard deviation of random variable $X$ are $10$ and $5$ respectively. Then $E {(\frac{X - 15}{5})}^{2} =$ ______

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Solution

The correct option is C $2$
Variance, $Var (X) = 5^{2} = 25$
$E (X^{2}) = Var (X) + (E (X))^{2} = 25 + 100 = 125$

$E {(\frac{X - 15}{5})}^{2} = \frac{1}{25} (E (X^{2}) + 225 - 30 E (X))$
$= \frac{1}{25} (125 + 225 - 30 \times 10)$
$= \frac{50}{25} = 2$

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The mean and standard deviation of random variable X are 10 and 5 respectively. Then E(X−155)2=______

The correct option is C 2Variance, Var(X)=52=25 E(X2)=Var(X)+(E(X))2=25+100=125 E(X−155)2=125(E(X2)+225−30E(X)) =125(125+225−30×10) =5025=2

The mean and standard deviation of random variable $X$ are $10$ and $5$ respectively. Then $E {(\frac{X - 15}{5})}^{2} =$ ______

The correct option is C $2$
Variance, $Var (X) = 5^{2} = 25$
$E (X^{2}) = Var (X) + (E (X))^{2} = 25 + 100 = 125$

$E {(\frac{X - 15}{5})}^{2} = \frac{1}{25} (E (X^{2}) + 225 - 30 E (X))$
$= \frac{1}{25} (125 + 225 - 30 \times 10)$
$= \frac{50}{25} = 2$