Probability question : Normal distribution

Question

An auto parts company, produces cylinder liners for engines of 1.2 inches in average diameter with a standard deviation of 0.1 inches. Every piece has a diameter less than an inch or more than 1.4 inches, it will be rejected. Taking into account that the random variable "diameter of the jacket" is normally distributed, it is desired that you determine:

a) The probability that a shirt taken at random will be rejected.

b) If 60 shirts are produced in one day, what is the probability of finding more than 4 defects?

I was able to resolve the point a), which gave me a result of 0,0455. But I honestly don't know how to resolve the point b). I am asking the question because I want to learn from the answer, and not for the sake of having the answer. Thank you.

Answer 1

Let $p$ denote the probability from a). The probability of no flaws is $(1-p)^{60}$, probability of exactly one flaw is $60p(1-p)^{59}$ etc. You can also use CLT to get an approximate result.

Answer 2

Whether or not each shirt meets the criterion of acceptability is the outcome of a Bernoulli trial with a probability of success of $\ p = 1-0,0455\ $ and a probability of failure of $\ 1-p=0,0455\ $. The number, $\ F\ $, of failures occurring in a sequence of $\ n\ $ Bernoulli trials, with probability of failure $\ 1-p\ $, follows a Binomial distribution: $\ \mathrm{Pr}\left(F=k\right)={n\choose k}\left(1-p\right)^kp^{n-k}\ $. In your case, $\ n=60\ $, and you want the probability of $\ 5\ $ or more defects: \begin{eqnarray} \mathrm{Pr}\left(F\ge 5\right)&=& 1-\mathrm{Pr}\left(F\le 4\right)\\ &=&1-\sum_\limits{k=0}^4{60\choose k}0,0455^k\,0,9545^{60-k}\ . \end{eqnarray} You can get a reasonable estimate of the result by using the Poisson approximation. Because the skewness of the distribution (namely $\ 0,56\ $) is so large, the normal approximation is rather less accurate. The value of $\ \mathrm{Pr}\left(F\le 4\right)\ $ obtained by evaluating the binomial sum to $5$ decimal places is $\ 0,86294\ $, the Poisson approximation gives $\ 0,85796\ $, and the normal approximation gives $\ 0,89208\ $.