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  • A discrete random variable X has the probability distribution given as below: X 0.5 P(x) K (i) Find the value of k (ii) Determine the mean of the distribution.

A discrete random variable X has the probability distribution given as below:
\begin{array}{|l|l|l|l|l|} \hline \mathrm{x} & 0.5 & 1 & 1.5 & 2 \\ \hline \mathrm{P}(\mathrm{x}) & \mathrm{k} & \mathrm{k}^{2} & 2 \mathrm{k}^{2} & \mathrm{k} \\ \hline \end{array}
(i) Find the value of k
(ii) Determine the mean of the distribution.

Answers (1)

Solution

Given-

\begin{array}{|l|l|l|l|l|} \hline \mathrm{x} & 0.5 & 1 & 1.5 & 2 \\ \hline \mathrm{P}(\mathrm{x}) & \mathrm{k} & \mathrm{k}^{2} & 2 \mathrm{k}^{2} & \mathrm{k} \\ \hline \end{array}

\begin{aligned} &\text { (i) As we know that } \sum_{i=1}^{n} P_{i}=1, \text { where } p i \geq 0\\ &\Rightarrow \mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}+\mathrm{P}_{4}=1\\ &\Rightarrow k+k^{2}+2 k^{2}+k=1\\ &\Rightarrow 3 k^{2}+2 k-1=0\\ &\Rightarrow 3 k^{2}+3 k-k-1=0 \end{aligned}

\\\Rightarrow 3 \mathrm{k}(\mathrm{k}+1)-1(\mathrm{k}+1)=0 \\\Rightarrow \mathrm{k}=\frac{1}{3} and \mathrm{k}=-1

Since, k \geq 0, we take k=\frac{1}{3}

(ii) Mean of distribution

 \\(\mu)=E(X)=\sum_{i=1}^{n} x_{i} P_{i}=1 \\=0.5(\mathrm{k})+1\left(\mathrm{k}^{2}\right)+1.5\left(2 \mathrm{k}^{2}\right)+2(\mathrm{k}) \\=4 k^{2}+2.5 k \quad\left[\because k=\frac{1}{3}\right] \\=4 \times \frac{1}{9}+2.5 \times \frac{1}{3} \\=\frac{4+7.5}{9}=\frac{23}{18}

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