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અહી $\mathrm{X}$ એ વિતરણનું યાર્દચ્છિક ચલ છે.
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
જો મધ્યક $X$ એ $2.3$ અને $X$ નું વિચરણ $\sigma^{2}$ હોય તો $100 \sigma^{2}$ ની કિમંત મેળવો.
$781$
$100$
$529$
$1310$
Solution
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
$\bar{X}=2.3$
$-a+6 b=\frac{9}{10} \ldots (1)$
$\sum P_{i}=\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$
$a+b=\frac{4}{15} \ldots (2)$
From equation $(1)$ and $(2)$
$a=\frac{1}{10}, b=\frac{1}{6}$
$\sigma^{2}=\Sigma p_{i} x_{i}^{2}-(\bar{X})^{2}$
$\frac{1}{5}(4)+a(1)+\frac{1}{3}(9)+\frac{1}{5}(16)+b(36)-(2.3)^{2}$
$=\frac{4}{5}+a+3+\frac{16}{5}+36 b-(2.3)^{2}$
$=4+a+3+36 b-(2.3)^{2}$
$=7+a+36 b-(2.3)^{2}$
$=7+\frac{1}{10}+6-(2.3)^{2}$
$=13+\frac{1}{10}-\left(\frac{23}{10}\right)^{2}$
$=\frac{131}{10}-\left(\frac{23}{10}\right)^{2}$
$=\frac{1310-(23)^{2}}{100}$
$=\frac{1310-529}{100}$
$=\frac{781}{100}$
$100 \sigma^{2}=781$
