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

Answer 1

Let $X$ denote the number of defective items in the sample.

$x$	0	1	2	3
$P(x)$	$\frac{7}{15}$	$\frac{5}{12}$	$\frac{5}{12}$	$\frac{1}{12}$
$x_i^2$	0	1	4	9
$P(x_i^2)$	$0$	$\frac{5}{12}$	$\frac{20}{12}$	$\frac{9}{12}$

The mean $\mu$ is given by:

$\mu = \sum P(x_i) \cdot x_i = \frac{18}{12} = \frac{3}{2}$

Calculate $\sum P(x_i^2)$:

$\sum P(x_i^2) = \frac{34}{12}$

The variance $\sigma^2$ is given by:

$\sigma^2 = \sum P(x_i^2) - \mu^2$

$\sigma^2 = \frac{34}{12} - \left( \frac{3}{2} \right)^2$

$\sigma^2 = \frac{34}{12} - \frac{9}{4} = \frac{34}{12} - \frac{27}{12} = \frac{7}{12}$

Now, calculate $96\sigma^2$:

$96\sigma^2 = 96 \times \frac{7}{12} = 56$