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Question

(A) The random variable $X$ follows a hypergeometric distribution with parameters $N=12$ (total items),$K=3$ (defective items),and $n=5$ (sample size)

Vedclass · Accepted Answer

$71$

Vedclass · Answer

(A) The random variable $X$ follows a hypergeometric distribution with parameters $N=12$ (total items),$K=3$ (defective items),and $n=5$ (sample size).The probability mass function is given by $P(X=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$.The variance of a hypergeometric distribution is given by $\sigma^2 = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1}$.Substituting the values: $\sigma^2 = 5 \cdot \frac{3}{12} \cdot \frac{12-3}{12} \cdot \frac{12-5}{12-1}$.$\sigma^2 = 5 \cdot \frac{1}{4} \cdot \frac{9}{12} \cdot \frac{7}{11} = 5 \cdot \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{7}{11} = \frac{105}{176}$.Given $\sigma^2 = \frac{m}{n} = \frac{105}{176}$,where $\operatorname{gcd}(105, 176) = 1$.Thus,$m = 105$ and $n = 176$.The value of $n-m = 176 - 105 = 71$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

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