Binomial stats problem help

Question

A poll reported that 63% of adults were satisfied with the job the major airlines were doing. Suppose 15 adults are selected at random and the number who are satisfied is recorded.

-Would it be unusual to find more than 12 who are satisfied with the job the major airlines were doing?

The result is unusual, because P(xgreater than>12)equals=.0449 under the assumption that the proportion of adults that are satisfied with the airlines is 63%. Thus, in 100 random samples of size 15, this result is expected in about 4.4990 of the random samples.

I know how to find P(x>12) which equal .0449. I do not really understand how you figure out that the result is expected in about 4.4990 of the random samples. Can you please explain how to find that answer.

Answer 1

Hello Ashley, 

 

As you calculated, the likelihood of a 15-person sample to have 12 or more people satisfied, is about 4.49%. You could take many different 15-person samples and see how many people said they were satisfied. It would be unlikely (only 4.49%) that any one 15-person sample would have 12 or more people satisfied. If you were to take 100, 15-person samples, you would find that somewhere between 4 and 5 (4.49) of them would have 12 or more people satisfied. It's a little confusing because it's hard to imagine .49 of a group of people. Another way of thinking of it is that if you took 100 samples, about 4 or 5 of them would have 12 or more people reporting they are satisfied.