From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at r…

Question

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $$\sigma^2$$

Accepted Answer

$x$
 0
 1
 2
 3

$P(x)$
 $\frac{{}^{7}{C}_{5}}{{}^{10}{C}_{5}}=\frac{1}{12}$
 $\frac{{C}_{4}\cdot {}^{3}{C}_{1}}{{}^{10}{C}_{5}}=\frac{5}{12}$
 $\frac{{}^{7}{C}_{3}\cdot {}^{3}{C}_{2}}{{}^{10}{C}_{5}}=\frac{5}{12}$
 $\frac{{}^{7}{C}_{2}\cdot {}^{3}{C}_{3}}{{}^{10}{C}_{5}}=\frac{1}{12}$

$xP(x)$
 0
 $\frac{5}{12}$
 $\frac{10}{12}$
 $\frac{3}{12}$

$\begin{gathered} \mu =\sum xP(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \ {\sigma }^{2}=\sum (x-\mu )P(x)=\sum {(x-\frac{3}{2})}^{2}P(x) \ =\frac{9}{4}	imes \frac{1}{12}+\frac{1}{4}	imes \frac{5}{12}+\frac{1}{4}	imes \frac{5}{12}+\frac{9}{4}	imes \frac{1}{12}=\frac{7}{12} \end{gathered}$

$\Rightarrow {\sigma }^{2}\cdot 96=8	imes 7=56$

Solutions · 1