"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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$x$	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	$\frac{5}{12}$	$\frac{10}{12}$	$\frac{3}{12}$

$$\begin{aligned} & \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\ & \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\ & =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12} \end{aligned}$$

$\Rightarrow \sigma^2 \cdot 96=8 \times 7=56$

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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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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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