Question

Consider a process where 15% of the parts produced have a defect.
If we have a sample of 250 parts, we want to find the probability that there are between
30 and 45 defective parts in this sample.
i) Calculate the probability that there are exactly 30 defective parts in the sample.
ii) Write out (but don't calculate) the expression for finding the probability that between
30 and 45 parts are defective. Hint: use the binomial distribution.
iii) use the normal distribution to estimate the answer for part ii), stating the assumptions
you use for the mean and variance

Answer 1

This is a binomial distribution question with
n = 250
p = 0.15
q = 1 - p = 0.85
where

ii) P(30< X <45) = P(X = 31) + P(X = 32) + P(X = 33) + P(X = 34) + P(X = 35) + P(X = 36) + P(X = 37) + P(X = 38) + P(X = 39) + P(X = 40) + P(X = 41) + P(X = 42) + P(X = 43) + P(X = 44)

iii) This binomial distribution can be approximated as Normal distribution since
np > 5 and nq > 5
Since we know that

P(30.0 < x < 45.0)=?

This implies that
P(30.0 < x < 45.0) = P(-1.3284 < z < 1.3284) = 0.816
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