Question

An electronics firm claims that the proportion of defective units of a certain process is 5%. A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot. On a particular occasion, the bury found 5 items defective. a) What is the probability of this occurrence, given that the claim of 5% defective is correct

Answer 1

Answer:

0.0006 = 0.06% probability of this occurrence

Step-by-step explanation:

For each unit, there are only two possible outcomes. Either it is defective, or it is not. The probability of an unit being defective is independent of any other unit, which means that the binomial probability distribution is used 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.

An electronics firm claims that the proportion of defective units of a certain process is 5%.

This means that $p = 0.05$

A buyer has a standard procedure of inspecting 15 units selected randomly from a large lot.

This means that $n = 15$

a) What is the probability of this occurrence?

Probability of 5 defective items, whch is P(X = 5). So

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

$P(X = 5) = C_{15,5}.(0.05)^{5}.(0.95)^{10} = 0.0006$

0.0006 = 0.06% probability of this occurrence