Asampling 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?

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

there are no figures how am i supposed to answer this question

step-by-step explanation:

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