Question

The diameter of a pipe is normally distributed, with a mean of 0.4 inch and a variance of 0.0009. What is the probability that the diameter of a randomly selected pipe will exceed 0.418 inch

Answer 1

Answer:

P(X > 0.418) = 0.27425

Step-by-step explanation:

We are given;

Population mean; μ = 0.4 inch

Variance; Var = 0.0009

Sample mean; x¯ = 0.418

First of all let's get the standard deviation which is;

σ = √Var

σ = √0.0009

σ = 0.03

Now, z - score formula is;

z = (x¯ - μ)/σ

z = (0.418 - 0.4)/0.03

z = 0.6

We want to find the probability that the diameter of a randomly selected pipe will exceed 0.418 inch. Thus;

P(X > 0.418) = 1 - P(X ≤ 0.418)

From z-distribution table attached, P(X ≤ 0.418) = P(Z) = 0.72575

Thus;

P(X > 0.418) = 1 - 0.72575

P(X > 0.418) = 0.27425