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 \( 96\sigma^2 \):
\[ 96\sigma^2 = 96 \times \frac{7}{12} = 56 \]
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: