Thread: Sampling of the proportions - Question

According to a Gallup Poll 60%, of adults say they experience heartburn. Assume this proportion is true for the population of all adults. If a random sample of 300 adults is selected from this population, then determine the probability that:
a) Less than 172 adults within the sample experience heatburn

this is what I believe is:

mean of the Sampling distrib of the proportion = .60
standard deviation of the sampling distr of the proportion = .0283

But I am not sure what to do after this. The answer in back of the book is .1727. I can not get that.

Any help would greatly be appreciated.

Thanks

Originally Posted by onemin

According to a Gallup Poll 60%, of adults say they experience heartburn. Assume this proportion is true for the population of all adults. If a random sample of 300 adults is selected from this population, then determine the probability that:
a) Less than 172 adults within the sample experience heatburn

this is what I believe is:

mean of the Sampling distrib of the proportion = .60
standard deviation of the sampling distr of the proportion = .0283

But I am not sure what to do after this. The answer in back of the book is .1727. I can not get that.

Any help would greatly be appreciated.

Thanks

X ~ Binomial(n = 300, p = 0.6).

Calculate Pr(X < 172).

Normal approximation to Binomial distribution could be used.

The book's answer is wrong.