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

Correct Answer: 56

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 \( 96\sigma^2 \):

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

Answer 2

Step 1: Identify the distribution
We are drawing a sample of 5 items from a lot of 10, which contains 3 defective and 7 non-defective items. The random variable \( X \) = number of defective items in the sample follows a hypergeometric distribution.

Step 2: Recall the parameters of the hypergeometric distribution
For a hypergeometric distribution: \[ E(X) = n \frac{K}{N}, \quad Var(X) = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \frac{N - n}{N - 1}, \] where \[ N = 10, \quad K = 3, \quad n = 5. \]

Step 3: Substitute the values
\[ \sigma^2 = Var(X) = 5 \times \frac{3}{10} \times \left(1 - \frac{3}{10}\right) \times \frac{10 - 5}{10 - 1}. \] Simplify step by step: \[ = 5 \times \frac{3}{10} \times \frac{7}{10} \times \frac{5}{9} = 5 \times \frac{105}{900} = \frac{525}{900} = \frac{7}{12}. \]

Step 4: Compute \( 96\sigma^2 \)
\[ 96\sigma^2 = 96 \times \frac{7}{12} = 8 \times 7 = 56. \]

Final answer

56