SOLUTION: It is known that 10% of the items produced by a certain machine are defective. Find the probability that in a random sample of 900 items produced by this machine. A. At most 110 it

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Question 1009767: It is known that 10% of the items produced by a certain machine are defective. Find the probability that in a random sample of 900 items produced by this machine. A. At most 110 items will be defective. B. At least 82 items will be defective. C. Between 75 and 100 items inclusive will be defect.
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Question:
It is known that 10% of the items produced by a certain machine are defective. Find the probability that in a random sample of 900 items produced by this machine. A. At most 110 items will be defective. B. At least 82 items will be defective. C. Between 75 and 100 items inclusive will be defect.

Solution:
The problem is essentially that of the binomial distribution, namely:
Trials are Bernoulli and have constant probability throughout if we assume infinite population size.
The probability for x successes out of n trials and probability p is
P%28X=x%29=C%28n%2Cx%29%2Ap%5Ex%2A%281-p%29%5E%28n-x%29
The parameters are:
n = 900 = sample size
p = 0.1 = probability of defects

Part I: exact calculations using binomial distribution

A. At most 110 items will be defective.
P%28X%3C=110%29+=+sum%28P%28X=i%29%29+ for i=0 to 110
=0.98685

B. At least 82 items will be defective.
P%28X%3E=82%29+=1-sum%28P%28X=i%29%29+ for i=0 to 81
=1-0.17282
=0.82718

C. Between 75 to 100 items will be defective.
P%28100%3E=X%3E=75%29+=sum%28P%28X=i%29%29+ for i=75 to 100
=0.87737-0.039684
=0.83769

Part II: Using normal approximation with continuity correction.
mean, μ = np = 900*0.1 = 90
σ² = npq = 81
=>
σ = sqrt(81) = 9
and
Z(x) = (x-μ)/σ
= (x - 90)/9

A. At most 110 items will be defective.
Using continuity correction, x=110.5
P(x<=110.5)
=P(Z<=(110.5-90)/9)
=P(Z<=2.27778)
=0.98863

B. At least 82 items will be defective.
Using continuity correction,
P(x>=81.5)
=P(Z>=(81.5-90)/9)
=P(Z>=-0.94444)
=1-P(Z<=-0.94444)
=1-0.17247
=0.82753

C. Between 75 to 100 items will be defective.
Using continuity corrections,
P(74.5<=x<=100.5)
=P((74.5-90)/9 <= Z <= (100.5-74.5)/9)
=P(-1.722222<=Z<=1.1666667)
=P(Z<1.1666667)-P(Z<-1.72222)
=0.87833-0.042515
=0.83581
Note that values calculated using either methods are comparable.