Question

If a random variable $X$ has the probability distribution given by $P(X=0)=3C^3,P(X=2)=5C-10C^2$ and $P(X=4)=4C-1$, then the variance of that distribution is

• A. $\frac{68}{9}$
• B. $\frac{22}{9}$
• C. $\frac{612}{81}$
• D. $\frac{128}{81}$

Answer 1

Answer: (D)

Given,
$P(X=0)=3C^3$
$P(X=2)=5C-10C^2$
and $P(X=4)=4C-1$
We know that,
$\sum P(X)=1$
$\Rightarrow 3C^3+\left(5C-10C^2\right)+(4C-1)=1$
$\Rightarrow 3C^3-10C^2+9C-2=0$
$\Rightarrow (C-1)\left(3C^2-7C+2\right)=0$
$\Rightarrow (C-1)(3C-1)(C-2)=0$
$\Rightarrow C=1,\frac{1}{3},2$
$\therefore C=\frac{1}{3}$
Now,

Hence, variance $=\Sigma X_P^2-{\left(\Sigma X_P\right)}^2$
$=\left(0^2\times \frac{1}{9}+4\times \frac{5}{9}+16\times \frac{1}{3}\right)-{\left(\frac{10}{9}+\frac{4}{3}\right)}^2$
$=\left(\frac{20}{9}+\frac{16}{3}\right)-{\left(\frac{66}{27}\right)}^2$
$=\frac{60+144}{27}-\frac{484}{81}$
$=\frac{204}{27}-\frac{484}{81}$
$=\frac{612-484}{81}=\frac{128}{81}$