Question

A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 20 randomly selected pens yields no more than two defective pens. find the probability that the shipment is accepted if 5% of the total shipment is defective?

Answer 1

Answer: p(def) = .05, p(good) = .95 , n = 20 Accept if x ≤ 2 P(x ≤ 2) = binomcdf(20, .05, 2) = .9245 0r 92.45%

Answer 2

Answer: Probability that the shipment is accepted is 92.45%.

Step-by-Step Explanation:

Since we have given that

Number of total pens = 20

Probability of defective pens = 5%

Probability of good pens = 95%

We need to find the probability that the shipment that yields no more than two defective pens.

So, we will use "Binomial distribution:

$P(x\leq 2)=P(x=0)+P(x=1)+P(x=2)\\\\P(x\leq 2)=0.95^{20}+^{20}C_1(0.5)(0.95)^{19}+^{20}C_2(0.05)^2(0.95)^{18}\\\\P(x\leq 2)=0.9245=92.45\%$

Hence, Probability that the shipment is accepted is 92.45%.