Question

A random variable $X$ has the probability distribution given below. Its variance is

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 $\frac{4}{3}$

We know that

$\sum P\left(x\right)=1$

$k+2k+3k+2k+k=1$

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

Now, var ${\sigma }^2=\sum x{i}^{2}P\left(xi\right)-(\sum xiP(xi){)}^2$

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

$=93k-(27k{)}^2$

$=93\left(\frac{1}{9}\right)-\left[27\right(\frac{1}{9}){]}^2$

$=\frac{31}{3}-3^2$

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

$=\frac{4}{3}$