Probability of proportion Assume that the proportion of students at forks high school who regularly workout is .25. Suppose Emmett interviews 100 students to discuss their habits. What is the probability that the proportion of the students interviewed who workout is between .2 and .3
 A: The number of students at Forks High School has not been specified. Perhaps it is a small school, and Emmett interviewed everyone. Then the required probability is $1$.
Or perhaps the school is huge, and $100$ is a tiny proportion of the students. Or perhaps, whatever the size of the school, Emmett chose the people he talked to with replacement. In that case, the number of students in the sample who exercise has binomial distribution, and the probability that the number is between $20$ and $30$ (inclusive) is 
$$\sum_{k=20}^{30} \binom{100}{k}(0.25)^k(0.75)^{100-k}.$$
Nowadays, there are are calculators that will evaluate the above sum exactly. There are also many computer programs that will do it, including the free to use Wolfram Alpha.
Chances are fairly good, however, that you are expected to use the normal approximation. Let $Y$ be the sample proportion of students who exercise. Then $Y$ has approximately normal distribution, mean $0.25$, and standard deviation 
$\frac{\sqrt{(0.25)(0.75)}}{\sqrt{100}}$. Call this number $s$. Then
$$\Pr(0.2\le Y\le 0.3)\approx \Pr\left(\frac{0.2-0.25}{s}\le Z\le \frac{0.3-0.25}{s}\right),$$
where $Z$ is standard normal.  This can be calculated using software or tables. 
Remark: As a further complication, you may be expected to apply the continuity correction to the normal approximation of the binomial. If that term has not been used in your course, don't worry about it. 
A: For the probability of the proportion of students that workout to be $0.2$ we need 20 students who work out and 80 who don't. Probability that a random student works out is $0.25$. Thus the probability that 20 of the 100 work out is $$\binom{100}{20} \cdot (0.25)^{20} \cdot (0.75)^{80}$$ Now you need to do the same for 21 students, 22 students and so on till 30 students. Then you add up the probabilities for your answer so $$\sum_{k=20}^{30} \binom{100}{k} (0.25)^{k} (0.75)^{100-k}\approx 0.797$$
I guess there's a more elegant way to do this problem. If I find one, I'll let you know. 
