SOLUTION: The probability of a defective pen in a large office building is 0.08. If a random sample of 30 telephones is selected, find the following probabilities, assuming this experiment i

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Question 414068: The probability of a defective pen in a large office building is 0.08. If a random sample of 30 telephones is selected, find the following probabilities, assuming this experiment is a binomial experiment.
A. P[exactly 3 defective pens]
B. P[at most 3 defective pens]
C. P[no defective pens]
D. Determine mean and variance
For P[exactly 3] I came up with 0.219 and for P[at most 3] I came up with 0.784. Are these correct? Need help figuring out C and D. Thanks!
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Note: The probability of x successes in n trials is: 

P = nCx* p%5Ex%2Aq%5E%28n-x%29 where p and q are the probabilities of success and failure respectively. 

In this case p= .08 and q = .92

nCx = n%21%2F%28x%21%28n-x%29%21%29

A. P[exactly 3 defective pens] = .2188    |Good work

B. P[at most 3 defective pens] = .7842    |Good work

C. P[no defective pens] = (.92)^30 = .082

D. Determine mean and variance 

    mean = np = 30*.08 = 2.4

 variance = npq = 30*.08*.92 = 2.208