2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experience shows that the probability of failure was 5%. The production line manufactures 1000 units daily. (a) What is the expected number of nondefective units? (b) what is the COV of the number of non-defective units? (c) What is the probability of having more than 980 non-defective units (Formulate pmf)?

(a)

(b)

(c)

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

Where the standard deviation is given by

So the coefficient of variance is

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

The z-score corresponding to 4.28 is 0.99999

So it means that it is very unlikely that there will be more than 980 non-defective units.