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

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