Question

A manufacturer produces a component for use in the automotive industry. It is known that 1% of the items produced are defective. Suppose a random sample of 20 items is examined. (a) Find the probability that (i) no defectives are found in the sample, (ii) one or fewer defectives are found in the sample. (b) Find: (i) the mean (expected) number of defectives in the sample, (ii) the variance and standard deviation.

Answer 1

Let X be a random variable which denotes the number of defective items in the sample of 20 items.

X follows binomial distribution with

n = 20, p = 0.01. Thus, q = 0.99

(a)

(i) P(no defectives are found in the sample) = P(X = 0)

= $0.99^{20}$ = 0.8179

(ii) P(one or fewer defectives are found in the sample)

= P(X <= 1) = P(X=0) + P(X=1)

= 0.8179 + $\binom{20}{1}*0.01*0.99^{19}$

= 0.98314

(b)

(i) Mean number of defectives in the sample = np = 20*0.01 = 0.2

(ii) Variance = npq = 20*0.01*0.99 = 0.198

Standard deviation = $\sqrt{0.198}$ = 0.445