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A random variable $X$ has the probability distribution given below.
Its variance is
Options:
Its variance is
Solution:
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Verified Answer
The correct answer is:
$\frac{4}{3}$
Given distribution is
$\begin{aligned} & \therefore \text { Variance }=\Sigma x_i^2 p-\left(\Sigma x_i p\right)^2 \\ & =(1 k+8 k+27 k+32 k+25 k) \\ & -(k+4 k+9 k+8 k+5 k)^2 \\ & =(93 k)-(27 k)^2=\left(93 \times \frac{1}{9}\right)-\left(27 \times \frac{1}{9}\right)^2 \\ & \left(\because \Sigma p=1, \therefore k=\frac{1}{9}\right) \\ & =\frac{93}{9}-9=\frac{93-81}{9}=\frac{12}{9}=\frac{4}{3} \\ & \end{aligned}$
$\begin{aligned} & \therefore \text { Variance }=\Sigma x_i^2 p-\left(\Sigma x_i p\right)^2 \\ & =(1 k+8 k+27 k+32 k+25 k) \\ & -(k+4 k+9 k+8 k+5 k)^2 \\ & =(93 k)-(27 k)^2=\left(93 \times \frac{1}{9}\right)-\left(27 \times \frac{1}{9}\right)^2 \\ & \left(\because \Sigma p=1, \therefore k=\frac{1}{9}\right) \\ & =\frac{93}{9}-9=\frac{93-81}{9}=\frac{12}{9}=\frac{4}{3} \\ & \end{aligned}$
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