Question

A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 16 randomly selected pens yields no more than two defective pens. (a) Find the probability that this shipment is accepted if 5% of the total shipment is defective. (Use 3 decimal places.)
(b) Find the probability that this shipment is not accepted if 20% of the total shipment is defective. (Use 3 decimal places.)

Answer 1

Answer:

(a) The probability that the shipment is accepted if 5% of the total shipment is defective is 0.957.

(b) The probability that the shipment is not accepted if 20% of the total shipment is defective is 0.648.

Step-by-step explanation:

Let X = Number of defective pens.

The random variable X follows a binomial distribution with parameters n (sample size) = 16 and P (Success) = p.

The probability function of a Binomial distribution is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x}$ ; x = 0, 1, 2, 3...

It is provided that the shipment is accepted if there are no more than 2 defective pens in the lot.

(a)

The probability of defective pens in the shipment is, p = 0.05.

The probability that this shipment is accepted is:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              $={16\choose 0}(0.05)^{0}(1-0.05)^{16-0}+{16\choose 1}(0.05)^{1}(1-0.05)^{16-1}+\\{16\choose 2}(0.05)^{2}(1-0.05)^{16-2}\\=0.4401+0.3706+0.1463\\=0.957$

Thus, the probability that the shipment is accepted if 5% of the total shipment is defective is 0.957.

(b)

The probability of defective pens in the shipment is, p = 0.20.

The probability that this shipment is not accepted is:

P (X > 2) = 1 - P (X ≤ 2)

              = 1 - [P (X = 0) + P (X = 1) + P (X = 2)]

              $=1-[{16\choose 0}(0.20)^{0}(1-0.20)^{16-0}+{16\choose 1}(0.20)^{1}(1-0.20)^{16-1}+\\{16\choose 2}(0.20)^{2}(1-0.20)^{16-2}]\\=0.0281+0.1126+0.2111\\=1-0.3518\\=0.6482\approx0.648$

Thus, the probability that the shipment is not accepted if 20% of the total shipment is defective is 0.648.