This is a binomial probability question with n=100, p=.3 and q=.7.
The probability you seek is the sum of 100Ck * (.3)k * (.7)100-k for k=0 to 20.
You can also use the normal approximation to the binomial with np = 30 and npq =21.
Natali D.
asked 03/02/19A telephone survey is conducted to determine the average number of pets in the typical American family. Past experience has shown that 30% of those telephoned will refuse to respond to the survey. If 100 randomly selected families are called then what is the probability that no more than 20 of these families will refuse to respond to the survey. Give your answer to four decimal points
This is a binomial probability question with n=100, p=.3 and q=.7.
The probability you seek is the sum of 100Ck * (.3)k * (.7)100-k for k=0 to 20.
You can also use the normal approximation to the binomial with np = 30 and npq =21.
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