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  • Let X be a discrete random variable whose probability distribution is defined as follows: P(X=x) = k(x+1) where k is a constant. Calculate (i) the value of k (ii) E (X) (iii) Standard deviation of X.

Let X be a discrete random variable whose probability distribution is defined as follows:

P(X=x)=\left\{\begin{array}{ll} k(x+1) \text { for } x=1,2,3,4 \\ 2 k x & \text { for } x=5,6,7 \\ 0 & \text { otherwise } \end{array}\right.

where k is a constant. Calculate
(i) the value of k (ii) E (X) (iii) Standard deviation of X.

Answers (1)

Given-

P(X=x)=\left\{\begin{array}{ll} k(x+1) \text { for } x=1,2,3,4 \\ 2 k x & \text { for } x=5,6,7 \\ 0 & \text { otherwise } \end{array}\right.

Therefore, we get the probability distribution of X as

 \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text { otherwise } \\ \hline \mathrm{P}(\mathrm{X}) & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} & 0 \\ \hline \end{array}

(i) the value of k
As we know, Sum of the probabilities  =1 
i.e. \sum_{i=1}^{n} p_{i}=1$
\\\therefore 2 k+3 k+4 k+5 k+10 k+12 k+14 k=1$ \\$\Rightarrow 50 k=1$ \\$\Rightarrow k=\frac{1}{50}$ \\$\Rightarrow \mathrm{k}=0.02$

To find: E(X)

The probability distribution of X is:

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} \text { or } \mathrm{x}_{i} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text { Otherwise } \\ \hline \mathrm{P}(\mathrm{X}) \text { or } \mathrm{p}_{\mathrm{i}} & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} & 0 \\ \hline \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} & 2 \mathrm{k} & 6 \mathrm{k} & 12 \mathrm{k} & 20 \mathrm{k} & 50 \mathrm{k} & 72 \mathrm{k} & 98 \mathrm{k} & 0 \\ \hline \end{array}

\begin{aligned} &\text { Therefore, } \mu=\mathrm{E}(\mathrm{X})\\ &\because E(X)=\sum_{i=1}^{n} x_{i} p_{i}\\ &\therefore E(X)=2 k+6 k+12 k+20 k+50 k+72 k+98 k+0\\ &=260 \mathrm{k}\\ &=260 \times \frac{1}{50}\left[\because k=\frac{1}{50}\right]\\ &=\frac{26}{5}\\ &=5.2 \ldots(i) \end{aligned}

(iii) To find: Standard deviation of X

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} \text { or } \mathrm{x}_{\mathrm{i}} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text { Otherwise } \\ \hline \mathrm{P}(\mathrm{X}) \text { or } \mathrm{pi} & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} & 0 \\ \hline \mathrm{XP}(\mathrm{X}) & 2 \mathrm{k} & 6 \mathrm{k} & 12 \mathrm{k} & 20 \mathrm{k} & 50 \mathrm{k} & 72 \mathrm{k} & 98 \mathrm{k} & 0 \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 2 \mathrm{k} & 12 \mathrm{k} & 36 \mathrm{k} & 80 \mathrm{k} & 250 \mathrm{k} & 432 \mathrm{k} & 686 \mathrm{k} & 0 \\ \hline \end{array}

As we know,
\\\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}$ \\$=\Sigma X^{2} P(X)-[\Sigma\{X P(X)]]^{2}$ \\$=[2 k+12 k+36 k+80 k+250 k+432 k+686 k+0]-[5.2]^{2}=1498 k-27.04$ \\$=\left[1498 \times \frac{1}{50}\right]-27.04$ \\$=29.96-27.04$ \\$=2.92$
As we know,
standard deviation of \mathrm{X}=\sqrt{\operatorname{Var}(\mathrm{X})}=\sqrt{2.92}=1.7088$ $\cong 1.7$

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