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 _________.

Question

If the mean and the variance of the data

are $\mu$ and 19 respectively, then the value of $\lambda + \mu$ is

Answer 1

Let $ X $ represent the count of defective items in the sample.

The mean, denoted by $ \mu $, is calculated as follows:

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

The sum of $ P(x_i^2) $ is:

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

The variance, denoted by $ \sigma^2 $, is determined by:

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

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

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

Finally, $ 96\sigma^2 $ is computed as:

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

\(x\)	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
\( P(x_i^2) \)	\( 0 \)	\( \frac{5}{12} \)	\( \frac{20}{12} \)	\( \frac{9}{12} \)

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 _________.