X | 0 | 1 | 2 |
---|---|---|---|
P(X) | $\frac{188}{221}$ | $\frac{32}{221}$ | $\frac{1}{221}$ |
अब माध्य X = E(X) = $\sum \limits_{i=1}^{n} x_{i} p\left(x_{i}\right)$
= 0$ \times \frac{188}{221}$$+1 \times \frac{32}{221}$$+2 \times \frac{1}{221}$=$\frac{34}{221}$
साथ ही $\mathrm{E}\left(\mathrm{X}^{2}\right)$ = $\sum \limits_{i=1}^{n} x_{i}^{2} p\left(x_{i}\right)$$=0^{2} \times \frac{188}{221}+1^{2}$$ \times \frac{32}{221}+2^{2} $$\times \frac{1}{221}=\frac{36}{221}$
अब Var(X) = E$\left(X^{2}\right)-[E(X)]^{2}$
= $\frac{36}{221}-\left(\frac{34}{221}\right)^{2}$$=\frac{6800}{(221)^{2}}$
इसलिए
$\sigma_{x}$= $\sqrt{\operatorname{Var}(\mathrm{X})}$$=\frac{\sqrt{6800}}{(221)}$ = 0.37 (लगभग)
बॉक्स | सगंमरमर की टुकड़ियों का रंग | ||
लाल | सफेद | काला | |
$A$ | $1$ | $6$ | $3$ |
$B$ | $6$ | $2$ | $2$ |
$C$ | $8$ | $1$ | $1$ |
$D$ | $0$ | $6$ | $4$ |