A random variable $X$ has the following probability distribution [table] $\mathbf{X}$ | $1$ | $2$ | $3$ | $4$ ; $\mathbf{P(X)}$ | $k$ | $2k$ | $3k$ | $4k$ [/table] Then, the mean of $X$ is

Question

Dr. Abubacker · Accepted Answer

(a)
We know, sum of probability distribution is 1
$	herefore\ k + 2k + 3k + 4k = 1\  \Rightarrow \ k = \frac{1}{10}$
Now, mean
$\overline{X} = k 	imes 1 + 2k 	imes 2 + 3k 	imes 3 + 4k 	imes 4$
$= k + 4k + 9k + 16k = 30k$
$\Rightarrow \ \overline{X} = 30 	imes \frac{1}{10} = 3$

Q. A random variable $X$ has the following probability distribution [table] $\mathbf{X}$ | $1$ | $2$ | $3$ | $4$ ; $\mathbf{P(X)}$ | $k$ | $2k$ | $3k$ | $4k$ [/table] Then, the mean of $X$ is

Explanation

Reviewed by Dr. Abubacker

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