Question

Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X denote the number of parts in the sample of 20 that require rework. A process problem is suspected if X exceeds its mean by more than 3 standard deviations.
a. If the percentage of parts that require rework remains at 1%, what is the probability that X exceeds its mean by more than 3 standard deviations?
b. If the rework percentage increases to 4%, what is the probability that X exceeds 1?
c. If the rework percentage increases to 4%, what is the probability that X exceeds 1 in at least one of the next five hours of samples?

Answer 1

Answer:

a) P(X>np+3$\sqrt{np(1-p)}$=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= $\sqrt{0.198}$ ≈ 0.445

a) P(X>np+3$\sqrt{np(1-p)}$=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

$\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}$

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - $\frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}$-$\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}$

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - $\frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}$

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=$\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}$=0.349

P(Y>1)=1-0.349=0.651