Question

A random variable has the following probability distribution:

$X$	1	2	3	4
$p\left( X \right)$	$k$	$2k$	$2k$	$4k$

Then the mean of X is
A. 3
B. 1
C. 4
D. 2

Answer 1

Hint: We use the known probability distribution theorems like ${{\mu }_{x}}=\sum{X\times p\left( X \right)}$ and $\sum{p\left( X \right)}=1$. We take the values and simplify to find the value of $k$. We put the value in the equation of ${{\mu }_{x}}=\sum{X\times p\left( X \right)}$ to find the mean.

Complete step by step answer:
For the given probability distribution of the random variable

$X$	1	2	3	4
$p\left( X \right)$	$k$	$2k$	$2k$	$4k$

In the given table we denote the given distribution as the expectations of the variables. The theorem varies for discrete variables.
We use some theorem where if ${{\mu }_{x}}$ is the mean then ${{\mu }_{x}}=\sum{X\times p\left( X \right)}$ and $\sum{p\left( X \right)}=1$.
Using the second theorem we get $k+2k+2k+4k=1$.
On simplification this gives $9k=1\Rightarrow k=\dfrac{1}{9}$.
Now we try to find the mean value which gives
${{\mu }_{x}}=\sum{X\times p\left( X \right)}=1\times k+2\times 2k+3\times 2k+4\times 4k=27k$.
We put the value of $k$ in the expression of ${{\mu }_{x}}=27k$ to get ${{\mu }_{x}}=\dfrac{27}{9}=3$
Therefore, the mean value is 3.

So, the correct answer is “Option A”.

Note: Usually the average is very different from the mean or average with probability distribution. The respective ratio gives the probability instead of values. Distribution represents the results from a simple experiment where there is “success” or “failure.”