SOLUTION: I need help solving this problem using the normal approximation of the binomial probability distribution. It is known that one out of 3 people entering a large department st

Question 182666: I need help solving this problem using the normal approximation of the binomial probability distribution.

It is known that one out of 3 people entering a large department store will make at least one purchase. If a random sample of 81 people is selected, what is the approximate probability that thirty or more of them will make at least one purchase? What is the probability that at most 40 of them will make at least one purchase?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
It is known that one out of 3 people entering a large department store will make at least one purchase.
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p = !/3 is therefore a population proportion
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If a random sample of 81 people is selected, what is the approximate probability that thirty or more of them will make at least one purchase?
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The distribution is binomial with p = 1/3, np = 81*(1/3) = 27, and sqrt(npq)
= sqrt(27(2/3)) = sqrt(18) = 3sqrt(2)
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Using the Normal Approximation:
Find the z-value of 30:
z(30) = (30-27)/[18/sqrt(81)] = 1.5
Then P(x > 30) = P(z > 1.5) = 0.067 or 6.7%
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What is the probability that at most 40 of them will make at least one purchase?
z(40) = (40-27)/[18/sqrt(81)] = 6.5
P(x <= 40) = 1 - P(x > 40) = 1-P(z > 6.5) = 1 - 0.00000000004036 is approx "1".
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Cheers,
Stan H.
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