The distribution of a random variable $X$ is given below [table] $\mathbf{X}$ | $- 2$ | $- 1$ | $0$ | $1$ | $2$ | $3$ ; $\mathbf{P(X)}$ | $\frac{1}{10}$ | k | $\frac{1}{5}$ | 2k | $\frac{3}{10}$ | k [/table] The value of $k$ is

Question

Dr. Abubacker · Accepted Answer

(a)
We know, total probability distribution is 1.
$	herefore\ \frac{1}{10} + k + \frac{1}{5} + 2k + \frac{3}{10} + k = 1$
$\Rightarrow \ \frac{6}{10} + 4k = 1$
$\Rightarrow \ k = \frac{1}{10}$

Q. The distribution of a random variable $X$ is given below [table] $\mathbf{X}$ | $- 2$ | $- 1$ | $0$ | $1$ | $2$ | $3$ ; $\mathbf{P(X)}$ | $\frac{1}{10}$ | k | $\frac{1}{5}$ | 2k | $\frac{3}{10}$ | k [/table] The value of $k$ is

Explanation

Reviewed by Dr. Abubacker

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