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(A) The random variable $X$ follows a hypergeometric distribution where $N=10$,$K=3$,and $n=5$.The probability mass function is given by $P(X=x) = \fr

Vedclass · Accepted Answer

$56$

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(A) The random variable $X$ follows a hypergeometric distribution where $N=10$,$K=3$,and $n=5$.The probability mass function is given by $P(X=x) = \frac{\binom{K}{x} \binom{N-K}{n-x}}{\binom{N}{n}}$.Calculating probabilities for $x \in \{0, 1, 2, 3\}$:$P(X=0) = \frac{\binom{3}{0} \binom{7}{5}}{\binom{10}{5}} = \frac{1 	imes 21}{252} = \frac{21}{252} = \frac{1}{12}$$P(X=1) = \frac{\binom{3}{1} \binom{7}{4}}{\binom{10}{5}} = \frac{3 	imes 35}{252} = \frac{105}{252} = \frac{5}{12}$$P(X=2) = \frac{\binom{3}{2} \binom{7}{3}}{\binom{10}{5}} = \frac{3 	imes 35}{252} = \frac{105}{252} = \frac{5}{12}$$P(X=3) = \frac{\binom{3}{3} \binom{7}{2}}{\binom{10}{5}} = \frac{1 	imes 21}{252} = \frac{21}{252} = \frac{1}{12}$Mean $\mu = E[X] = \sum x P(x) = 0(\frac{1}{12}) + 1(\frac{5}{12}) + 2(\frac{5}{12}) + 3(\frac{1}{12}) = \frac{5+10+3}{12} = \frac{18}{12} = \frac{3}{2}$.$E[X^2] = \sum x^2 P(x) = 0^2(\frac{1}{12}) + 1^2(\frac{5}{12}) + 2^2(\frac{5}{12}) + 3^2(\frac{1}{12}) = \frac{0+5+20+9}{12} = \frac{34}{12} = \frac{17}{6}$.Variance $\sigma^2 = E[X^2] - (E[X])^2 = \frac{17}{6} - (\frac{3}{2})^2 = \frac{17}{6} - \frac{9}{4} = \frac{34-27}{12} = \frac{7}{12}$.Thus,$96 \sigma^2 = 96 	imes \frac{7}{12} = 8 	imes 7 = 56$.

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 $\sigma^2$,then $96 \sigma^2$ is equal to....................

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