If X is a random-variable with probability distribution as given below: X = xi : 0 1 2 3 P (X = xi) : k 3 k 3 k k The value of k and its variance are

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KollegeApply · Accepted Answer

$\frac{1}{8}$, $\frac{3}{4}$ We know that the sum of probabilities in a probability distribution is always 1. $\therefore$ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1 \Rightarrow k + 3k + 3k + k = 1 \Rightarrow 8k = 1 \Rightarrow k = $\frac{1}{8}$ Now, xi pi pixi pixi2 0 k = $\frac{1}{8}$ 0 0 1 3k =$\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ 2 3k = $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$ 3 k = $\frac{1}{8}$ $\frac{3}{8}$ $\frac{9}{8}$ \sum\nolimits_{}^{}pixi = $\frac{12}{8}$ =$\frac{3}{2}$ \sum\nolimits_{}^{} pixi2 = $\frac{24}{8}$ \text{ Mean } = \sum p_i x_i = $\frac{3}{2}$ \text{ Variance } = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean } \right)^2 = $\frac{24}{8}$ - \left( $\frac{3}{2}$ \right)^2 = $\frac{24}{8}$ - $\frac{9}{4}$ = 24 - $\frac{18}{8}$ = $\frac{6}{8}$ = $\frac{3}{4}$

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