Question

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

X:	0.5	1	1.5	2
P(X):	k	k2	2k2	k

(i) Find the value of k.

(ii) Determine the mean of the distribution. [NCERT EXEMPLAR]

Answer 1

The probability distribution of X is given as:

X:	0.5	1	1.5	2
P(X):	k	k2	2k2	k

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{As},\sum p_i=1 \\ \Rightarrow k+k^2+2k^2+k=1 \\ \Rightarrow 3k^2+2k-1=0 \\ \Rightarrow 3k^2+3k-k-1=0 \\ \Rightarrow 3k\left(k+1\right)-1\left(k+1\right)=0 \\ \Rightarrow \left(3k-1\right)\left(k+1\right)=0 \\ \Rightarrow k=\frac{1}{3}\mathrm{or}k=-1 \\ \mathrm{but}k\mathrm{cannot}\mathrm{be}\mathrm{negative} \\ \therefore k=\frac{1}{3} \\ \left(\mathrm{ii}\right)\mathrm{Mean}=\sum p_{\mathit{i}}{x}_{\mathit{i}}=0.5\times k+1\times k^2+1.5\times 2k^2+2\times k \\ =0.5\times \frac{1}{3}+1\times {\left(\frac{1}{3}\right)}^2+1.5\times 2{\left(\frac{1}{3}\right)}^2+2\times \frac{1}{3} \\ =\frac{0.5}{3}+\frac{1}{9}+\frac{3}{9}+\frac{2}{3} \\ =\frac{1.5+1+3+6}{9} \\ =\frac{11.5}{9} \\ =\frac{115}{90} \\ =\frac{23}{18} \end{gathered}$$