If the mean of the following probability distribution of a radam variable $\mathrm{X}$ :
| $\mathrm{X}$ | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| $\mathrm{P(X)}$ | $a$ | $2a$ | $a+b$ | $2b$ | $3b$ |
is $\frac{46}{9}$, then the variance of the distribution is
Solution
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<th class="tg-baqh">0</th>
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<p>$$\begin{aligned}
& \text { Mean }=\sum x_i P\left(x_i\right) \\
& \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\
& \frac{46}{9}=8 a+40 b \\
& \frac{23}{9}=4 a+20 b \\
& 36 a+180 b=23 \quad \text{.... (1)}
\end{aligned}$$</p>
<p>Sum of probability is 1</p>
<p>$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$</p>
<p>$$\begin{aligned}
& \text { Solving (1) and (2) } \\
& a=\frac{1}{12}, b=\frac{1}{9} \\
& \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\
& =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\
& =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\
& =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\
& =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\
& =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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