Let X X X denote the number of defective items in the sample.
The mean μ \mu μ is given by:
μ=∑P(xi)⋅xi=1812=32 \mu = \sum P(x_i) \cdot x_i = \frac{18}{12} = \frac{3}{2} μ=∑P(xi)⋅xi=1218=23
Calculate ∑P(xi2) \sum P(x_i^2) ∑P(xi2):
∑P(xi2)=3412 \sum P(x_i^2) = \frac{34}{12} ∑P(xi2)=1234
The variance σ2 \sigma^2 σ2 is given by:
σ2=∑P(xi2)−μ2 \sigma^2 = \sum P(x_i^2) - \mu^2 σ2=∑P(xi2)−μ2
σ2=3412−(32)2 \sigma^2 = \frac{34}{12} - \left( \frac{3}{2} \right)^2 σ2=1234−(23)2
σ2=3412−94=3412−2712=712 \sigma^2 = \frac{34}{12} - \frac{9}{4} = \frac{34}{12} - \frac{27}{12} = \frac{7}{12} σ2=1234−49=1234−1227=127
Now, calculate 96σ2 96\sigma^2 96σ2:
96σ2=96×712=56 96\sigma^2 = 96 \times \frac{7}{12} = 56 96σ2=96×127=56
If dydx+2ysec2x=2sec2x+3tanx⋅sec2x \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x dxdy+2ysec2x=2sec2x+3tanx⋅sec2x and
and f(0)=54 f(0) = \frac{5}{4} f(0)=45, then the value of 12(y(π4)−1e2) 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) 12(y(4π)−e21) equals to: