# question 8 and 9 for midterm study guide

%(1508+1541+232)/3%

for the mean and the standard deviation is just

%sqrt((1508-1093.66)^2+(1541-1093.66)^2+(232-1093.66)^2)/(n-1)?%

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%(1508+1541+232)/3%

for the mean and the standard deviation is just

%sqrt((1508-1093.66)^2+(1541-1093.66)^2+(232-1093.66)^2)/(n-1)?%

## Comments

First off, I edited your post to include mathematical typesetting - which is basically just a matter of including percentage symbols around your formulae. One thing that's nice about that is that it forces you to use correct parentheses. For example, you had typed

If you wrap that exactly in percentage symbols, you get

Which typesets like so: %1508+1541+232/3%

I suspect, though, that you want all three data points over 3. To get that you should type

Having said all that, you're treating this like numerical data when it's actually categorical data. I would treat it like a binomial problem. That is, you pick 50 students and you know that the probability that any one of them is

notfrom WNC is$$\frac{1541+232}{3281} \approx 0.540384.$$

That gives you a single random variable that could call $X$. I think that the direct answer to your question is that the mean of $X$ is $0.540384$. You find the variance using the formula $p(1-p)$.

If we now let

$$S = X_1 + X_2 + \cdots + X_{50}$$

denote the sum of 50 independent copies of $X$, then the question is "What is the probability that $S$ is greater than 28?" Or $P(S>28) = ?$

You can answer that question with a normal distribution whose mean and variance are the same as the mean and variance of $S$. You find that mean and variance by multiplying the corresponding terms for $X$ by 50.