Calculate the probability of acceptance of a lot containing 3% defective by a sampling plan with acceptance?

Question

number 2 and sample size 80.

Answer 1

Your question is incomplete, but what I think you are saying is that if we sample 80 parts and find 2 or more defective parts we will not accept the shipment.

Let X be the number of defective parts.  X has the binomial distribution with n = 80 trials and success probability p = 0.03



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 80 , p = 0.03 )

the mean of the binomial distribution is n * p = 2.4

the variance of the binomial distribution is n * p * (1 - p) = 2.328

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.525778

The first few elements of the Probability Mass Function, PMF,

f(X) = P(X = x) is:

P( X =  0 ) =  0.08744576

P( X =  1 ) =  0.2163606

P( X =  2 ) =  0.2643169

P( X =  3 ) =  0.2125435

we know that P( X ≥ 2) = 1 - P(X < 2)

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

= 1 - (0.08744576 + 0.2163606)

= 0.6961936

P(X ≥ 2) = 0.6961936 this is the probability we reject the lot.

the probability we accept is:

P(X < 2)

= 0.08744576 + 0.2163606

= 0.3038064