Question

A manufacturer of metal pistons finds that on the average, 15% of his pistons are rejected because they are either oversized or undersized. Nine pistons were randomely selected. Please show work.

1. What is the mean number of pistons that will be rejected?

2. What is the probablilty that exactly two pistons will be rejected? Round to the nearest thousandth.

3. What is the probablilty that less than three pistons will be rejected?

Answer 1

This is a direct application of binomial distribution,

let X be a number of pistons that will be rejected.

Here, X ~ Binomial (n = 9, p = 0.15)

probability mass function of X is,

P(X = x) = ⁿC_x p^x (1-p)^n-x

1) Mean of X is,

E(X) = np = 9 * 0.15 = 1.35

Mean = 1.35

2)

P(X = 2)

= ⁹C₂ * (0.15)² * (1-0.15)^9-2

= 36 * (0.15)² * (0.85)⁷

= 0.260

P(X = 2) = 0.260

3)

P(X < 3)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0.232 + 0.368 + 0.260

= 0.860

P(X < 3) = 0.860

Note: Using Excel command we find P(X <3)

P(X < 3) = P(X <=2) = BINOMDIST(2,9,0.15,1) = 0.859

P(X < 3) = 0.859