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The probability distribution of a discrete random variable X is given below:

\begin{array}{|c|c|c|c|c|} \hline X & 2 & 3 & 4 & 5 \\ \hline P(X) & \frac{5}{k} & \frac{7}{k} & \frac{9}{k} & \frac{11}{k} \\ \hline \end{array}

The value of k is
A. 8

B. 16
C. 32
D. 48

Answers (1)

Given-

Probability distribution table

As we know \sum_{i=1}^{n} P_{i}=1
\\\Rightarrow \sum P_{i}=\left[\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}\right]=1$ \\$\Rightarrow\left[\frac{32}{\mathrm{k}}\right]=1$ \\$\mathrm{K}=32
Hence, the value of k is 32

Option C is correct.

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infoexpert22

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