Let the mean and the standard deviation of the probability distribution be and , respectively. If then is equal to

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Solution

Q.
Let the mean and the standard deviation of the probability distribution

X $α$ 1 0 $- 3$

P(X) $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$

be $μ$ and $σ$ , respectively. If $σ - μ = 2,$ then $σ + μ$ is equal to _____________. [2024]

Ans.
(5)

We have, $\sum P (X) = 1$

$\Rightarrow \frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1 \Rightarrow \frac{8 + 4 + 6}{24} + K = 1$

$\Rightarrow K = 1 - \frac{18}{24} = \frac{6}{24} = \frac{1}{4}$

Now, $μ = \frac{α}{3} + \frac{1}{4} - \frac{3}{4} = \frac{α}{3} - \frac{1}{2}$

and $σ = \sqrt{(\frac{α^{2}}{3} + \frac{1}{4} + \frac{9}{4}) - {(\frac{α}{3} - \frac{1}{2})}^{2}}$ $= \sqrt{(\frac{2 α^{2}}{9} + \frac{α}{3} + \frac{9}{4})}$

Given, $σ - μ = 2$

$\Rightarrow σ = μ + 2$

$\Rightarrow σ^{2} = {(μ + 2)}^{2}$ $\Rightarrow \frac{2 α^{2}}{9} + \frac{α}{3} + \frac{9}{4} = \frac{α^{2}}{9} + \frac{9}{4} + α$

$\Rightarrow \frac{α^{2}}{9} - \frac{2 α}{3} = 0$

$\Rightarrow α = 0, 6$ $[α \neq 0$ $∵$ $X = 0$ is already given $]$

$\Rightarrow α = 6 \Rightarrow μ = \frac{3}{2}$ and $σ = \frac{7}{2}$

$So μ + σ = 5$

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