Question

A random variable $X$ has the probability distribution given below. Find its variance.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$k$	$2k$	$3k$	$2k$	$k$

• A. $\frac{16}{3}$
• B. $\frac{4}{3}$
• C. $\frac{5}{3}$
• D. $\frac{10}{3}$

Answer 1

The correct option is B

$\frac{4}{3}$

Explanation for the correct option:

Step 1. Find the variance of the given data:

As we know, $\sum _{x=1}^{5}P\left(X\right)=1$

$\Rightarrow$$k+2k+3k+2k+k=1$

$\Rightarrow$ $9k=1$

$\Rightarrow$ $k=\frac{1}{9}$

$\because$Variance, ${\sigma }^2=\sum {x_i}^{2}P\left(x_i\right)–(x_{i}P{\left(x_i\right))}^2$

$=1\left(k\right)+2^2\left(2k\right)+3^2\left(3k\right)+4^2\left(2k\right)+5^2\left(k\right)-{\left[1\right(k)+2(2k)+3(3k)+4(2k)+5(k\left)\right]}^2$

$=93k–{\left(27k\right)}^2$

Step 2. Substitute $k=\frac{1}{9}$ in above equation:

$=93\times \frac{1}{9}–{\left(27\times \frac{1}{9}\right)}^2$

$=\frac{31}{3}-9$

$=\frac{31-27}{3}$

$=\frac{4}{3}$

$\therefore$variance $=\frac{\mathbf{4}}{\mathbf{3}}$

Hence, Option ‘B’ is Correct.