A sampling plan states that if 20 incoming bolts are checked and 2 or less defective bolts are discovered, the lot will be rejected. If an incoming lot is 10 percent defective, what is the probability of rejecting the lot?

Answer: the probability of rejecting the lot is 0.68

Step-by-step explanation:

Assuming a binomial probability distribution, the formula for binomial distribution is expressed as

P (x = r) = nCr * q^ (n - r) * p^r

Where

n = number of samples

p = probability that an event will happen.

q = probability that an event will not happen.

From the information given,

n = 20

If an incoming lot is 10 percent defective, it means that

p = 10/100 = 0.1

q = 1 - p = 1 - 0.1 = 0.9

If 2 or less defective bolts are discovered, the lot will be rejected. The probability of rejecting the lot would be

P (x lesser than or equal to 2)

P (x lesser than or equal to 2) =

P (x = 0) + P (x = 1) + P (x = 2)

P (x = 0) = 20C0 * 0.9^ (20 - 0) * 0.1^0 = 0.12

(x = 1) = 20C1 * 0.9^ (20 - 1) * 0.1^1 = 0.12 = 0.27

(x = 2) = 20C2 * 0.9^ (20 - 2) * 0.1^2 = 0.12 = 0.29

P (x lesser than or equal to 2) = 0.12 + 0.27 + 0.29 = 0.68