Question

A discrete random variable X has the probability distribution as given below

$\begin{array}{ccccc}X& 0.5& 1& 1.5& 2\\ P\left(X\right)& k& k^2& 2k^2& k\end{array}$

(i) Find the value of k.

(ii) Determine the mean of the distribution.

Answer 1

We have,

$\begin{array}{ccccc}X& 0.5& 1& 1.5& 2\\ P\left(X\right)& k& k^2& 2k^2& k\end{array}$

We know that, ${\sum }_{i=1}^n{P}_i=1$, wehere $P_i\ge 0$

$\Rightarrow P_1+P_2+P_3+P_4=1\Rightarrow k+k^2+2k^2+k=1\Rightarrow 3k^2+2k-1=0\Rightarrow 3k^2+3k-k-1=0\Rightarrow 3k(k+1)-1(k+1)=0\Rightarrow (3k-1)(k+1)=0\Rightarrow k=\frac{1}{3}\Rightarrow k=-1$
Since, k is $\ge 0\Rightarrow k=\frac{1}{3}$

Mean of the distribution $\left(\mu \right)=E\left(X\right)={\sum }_{i=1_i}^n{x}_i{P}_i$

$0.5\left(k\right)+1\left(k^2\right)+1.5\left(2k^2\right)+2\left(k\right)=4k^2+2.5k$
$=4.\frac{1}{9}+\mathrm{2.5.}\frac{1}{3}$ $[\because k=\frac{1}{3}]$
$=\frac{4+7.5}{9}=\frac{23}{18}$