According to an industry report 26 percent of all households

According to an industry report, 26 percent of all households have at least one cell phone. Further, of those that do have a cell phone, the mean monthly bill is $55.90 with a standard deviation equal to $9.60. Recently, a random sample of 400 households was selected. Of these households, 88 indicated that they had cell phones. The mean bill for these 88 households was $57.00. What is the probability of getting 88 or fewer households with cell phones if the numbers provided by the industry report are correct?

Solution

Solution :

If N=400 and p=.26
Pr[N<=88]=.0369
I used a powerful statistical tool that calculates cumulative Binomial Probabilities
You can use Excel or some other online Binomial calculator
if you wanted to use a Normal Approximation
mean=400(.26)=104
Var=104(1-.26)=76.96
Pr[Z<88.5-104/sqrt(76.96)]

=Pr[Z<-1.76685]=.0386, not a bad approximation considering the exact probability
is .0369. In fact, to two decimal places, the Normal Approximation and The exact
Binomial probability both yield .04.