Here's the question: According to the CDC, the proportion of American women, aged 18-24, that binge drink is $p=0.24$. Your college surveys a SRS of 200 female students and finds that 56 binge drink.
If the proportion of women on your campus who binge drink is the same as the national proportion, what is the probability that the number of binge drinkers in a SRS of $200$ students is as equal or larger than the result of the college's sample? (Round to 3 decimal places)
I got $.918$ as my answer but it's wrong.
How do I get the correct answer? I'm using the =binom.idst() function in Excel to solve the problem.
In excel BINOMDIST$(k;200;0.24;FALSE)$ tells the probability that in $200$ experiment $k$ are successful given that the probability of a success is $0.24$.
I've got:
$$\sum_{k=56}^{200} \text{ BINOMDIST}(k;200;0.24;FALSE)=0.10843$$
for the probability that in $200$ cases $56$ or more are binge drinkers.
Also, BINOMDIST$(k;200;0.24;TRUE)=P(\text{number of successes }\le k).$
So,
$$P(\text{number of successes }\ge 56)=1-\text{BINOMDIST}(55;200;0.24;TRUE)=0.10843.$$