Question

22. Suppose that 10% of a certain type of com- ponent last more than 600 hours in opera- tion. For n = 200 components, let X de- note the number of those that last more than 600 hours. Approximate the probabil- ity P(15 < X < 25). A. 0.197. B. 0.238 C. 0.263 D. 0.737 E. 0.762

The correct answer is E, I need help on the process/reasoning to get to that answer, please.

Answer 1

mean = np = 20 * 0.10 = 20

std.dev = sqrt(npq)
= sqrt(20 * 0.10 * 0.90) = 4.2426

Here, μ = 20, σ = 4.2426, x1 = 15 and x2 = 25. We need to compute P(15<= X <= 25). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (15 - 20)/4.2426 = -1.18
z2 = (25 - 20)/4.2426 = 1.18

Therefore, we get
P(15 <= X <= 25) = P((25 - 20)/4.2426) <= z <= (25 - 20)/4.2426)
= P(-1.18 <= z <= 1.18) = P(z <= 1.18) - P(z <= -1.18)
= 0.881 - 0.119
= 0.762