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(A) Total apples = $3 + 7 = 10$. Four apples are drawn without replacement. The random variable $X$ follows a hypergeometric distribution. The probabi

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$20$

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(A) Total apples = $3 + 7 = 10$. Four apples are drawn without replacement. The random variable $X$ follows a hypergeometric distribution. The probability distribution is given by:| $X$ | $P(X)$ | $XP(X)$ | $X^2P(X)$ ||---|---|---|---|| $0$ | $\frac{\binom{3}{0}\binom{7}{4}}{\binom{10}{4}} = \frac{35}{210} = \frac{1}{6}$ | $0$ | $0$ || $1$ | $\frac{\binom{3}{1}\binom{7}{3}}{\binom{10}{4}} = \frac{3 	imes 35}{210} = \frac{1}{2}$ | $\frac{1}{2}$ | $\frac{1}{2}$ || $2$ | $\frac{\binom{3}{2}\binom{7}{2}}{\binom{10}{4}} = \frac{3 	imes 21}{210} = \frac{3}{10}$ | $\frac{6}{10}$ | $\frac{12}{10}$ || $3$ | $\frac{\binom{3}{3}\binom{7}{1}}{\binom{10}{4}} = \frac{1 	imes 7}{210} = \frac{1}{30}$ | $\frac{3}{10}$ | $\frac{9}{10}$ |Mean $\mu = E(X) = \sum xP(x) = 0 + \frac{1}{2} + \frac{6}{10} + \frac{3}{10} = \frac{5+6+3}{10} = \frac{14}{10} = 1.4$.$E(X^2) = \sum x^2P(x) = 0 + \frac{1}{2} + \frac{12}{10} + \frac{9}{10} = \frac{5+12+9}{10} = \frac{26}{10} = 2.6$.Variance $\sigma^2 = E(X^2) - \mu^2 = 2.6 - (1.4)^2 = 2.6 - 1.96 = 0.64$.We need to find $10(\mu^2 + \sigma^2) = 10(E(X^2)) = 10(2.6) = 26$. Wait,re-evaluating the table values from the provided image: $X^2P(X)$ for $X=3$ is $9/30 = 0.3$. Sum $E(X^2) = 0 + 0.5 + 1.2 + 0.3 = 2.0$.Thus,$10(\mu^2 + \sigma^2) = 10(E(X^2)) = 10(2) = 20$.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

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