Statistics problem...?

Question

Suppose that at a soda bottling plant it is known that 20% of the bottles filled require some “rework” because of improper labeling. If the quality control manager takes a random sample of 25 filled bottles off the mass production process: Estimate the probability that at least 3 defective labels appear in the sample?

The proportion of defective labels in the sample can be approximated by normal distribution.

Please help me. I have no idea what to do on this problem...

Answer 1

You are working with a binomial probability distribution with probability p = 0.20
http://en.wikipedia.org/wiki/Binomial_distribution

Your problem is actually on the borderline for this, but it explicitly allows you to use the Gaussian (aka Normal) distribution approximation (see the wiki page) if you wish.

In the context of a Gaussian distribution, all that counts is the mean and the standard deviation.

http://en.wikipedia.org/wiki/Normal_distribution

Suppose your mean is M and your standard deviation is S, then erf(n/sqrt(2)) gives the fraction of the population within n standard deviations of the mean.

Thus, erf(1, sqrt(2) = 0.68 so 68% of the population falls between M-S and M+S. Similarly, 95% falls between M-2S and M+2S.

The rest of the population is evenly distributed below and above the central portion.

So in your case, the boundary line is at 3 defective labels. The mean will be 20% of 25 or 5, so 3 is below the mean.

First determine how many standard deviations below the mean 3 is. Call it x.

Then determine what percentage has between 3 and 7 defective labels. This will be erf(x/sqrt(2))

The remaining will be split evenly between those with less than 3 and those greater than 7. So divide this remainder and compute the desired result.