Question

Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items, and if there are 0 or 1 de- fective cases in the group of 6, the process is said to be in control. If the number of defects is more than 1, the process is out of control. Suppose that the true proportion of defective items is 0.15. What is the probability that there will be 0 or 1 defects in a sam- ple of 6 if the true proportion of defects is 0.15

Answer 1

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that $p = 0.15$

Sample of 6:

This means that $n = 6$

What is the probability that there will be 0 or 1 defects in a sample of 6?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771$

$P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764$

77.64% probability that there will be 0 or 1 defects in a sample of 6.