Question

Suppose that 80% of widgets produced by a factory that generates thousands of widgets every year are of acceptable quality.

(a) If a quality control inspector selects 3 widgets independently and at random, what is the probability that at least one will not be of acceptable quality?

(b) Suppose that a quality control inspector selects a random sample of 60 widgets to determine the proportion that have an acceptable level of quality. If Pˆ is the proportion of acceptable widgets he will find in his sample, what is the sampling distribution for Pˆ.

(c) What is the probability that more than 47 of the selected 60 widgets are acceptable?

(d) If the inspector finds that exactly 47 of the widgets selected are acceptable and calculates a 95% confidence interval for the true proportion of widgets from the factory that are acceptable, what would the corresponding margin of error be?

Answer 1

a) X= number of widgets of non acceptable quality out of the 3 selected.

So, X~Binomial(3,0.2) distribution

Pdf of X: P(X=x) = 3Cx*(0.2)^x*(0.8)^3-x

We want, P(X>=1)= 1- P(X=0) = 1- 3C0*0.8^3= 0.488

b) From the Central limit theorem,

We know, ~N(p, √p(1-p)/n) distribution

Now, p=0.8, n=60

So, ~N(0.8, 0.0516) distribution

c) 47/60= 0.7833

So, we want, P(>0.7833)

Using standard normal approximation,

P(z>0.7833-0.8/0.0516) = P(z>-0.323) = P(z<0.323) = 0.37335 (from the standard normal distribution tables)

d) Margin of Error= z0.025*√(1-)/n

Here, = 47/60= 0.7833, n=60, z0.025= 1.96

So required margin of error= 0.0732