Mathematical Statistics How do I find the answers to this question?
State Tech’s basketball team, the Fighting Loga-
rithms, have a 70% foul-shooting percentage.
(a) Write a formula for the exact probability that out of
their next one hundred free throws, they will make
between seventy-ﬁve and eighty, inclusive.
(b) Approximate the probability asked for in part (a).
 A: Let $p$ be the probability of sinking a foul shot. We will assume that $p=0.7$, and that success/failure on the next foul shot is independent of previous history, like tossing a coin or die. Probably neither assumption is correct.
But under the assumptions, the probability of $k$ "successes" in $n$ trials is equal to $\dbinom{n}{k}p^k(1-p)^{n-k}$.  For the number of successes has binomial distribution.
In our particular situation, we want the probability of $75$ or $76$ or $\dots$ or $80$ successes in $100$ trials. This is
$$\binom{100}{75}(0.7)^{75}(0.3)^{25}+\binom{100}{76}(0.7)^{76}(0.3)^{24}+\cdots +\binom{100}{80}(0.7)^{80}(0.3)^{20}.$$
This is somewhat unpleasant to calculate by hand. But nowadays there are plenty of pieces of software that compute this virtually instantly. However, particularly in the old days, it was important to have a reasonably quick way to approximate the difficult-to-compute exact answer. 
For an approximation, we use the approximation by the normal.  The mean number of successes in $100$ trials is "$np$", which is $70$. The variance in the number of successes is "$np(1-p)$", which is $21$. So the standard deviation is $\sqrt{21}$.
For the normal approximation, we probably should use the continuity correction, whiich I am not sure you have been taught. So we want to compute $\Pr(Y\le 80.5) -\Pr(Y\le 74.5)$, where $Y$ is normal with mean $70$ and standard deviation $\sqrt{21}$. I assume you know how to calculate these two probabilities. If you do not, please leave a message and I will complete the calculation. Take advantage of the symmetry! 
If you have not been taught the continuity correction, you are probably expected to compute $\Pr(Y\le 80)-\Pr(Y\le 74)$. 
A: This is a binomial distribution.  The probability of getting $75$ successes comes from the number of ways to pick $75$ of the $100$ throws to make, times $0.7^{75}).3^{25}$ for the chance of making 75 and missing 25.  Now add over  the range 75 to 80.
A: For a) are you familiar with the binomial distribution?
$$p=0.70, \  n=100$$
$$\sum_{k=75}^{80} \binom{n}{k} \cdot (\frac{7}{10})^{k}(1-\frac{7}{10})^{n-k} \approx 0.1542$$
For b) you could use the normal approximation of a binomial distribution
with $k$ as the number of trials, $x$ the number of trials as approximation and Z for the standard normal distribution
$$\mathbb{P}(75 \le k \le 80) \approx \mathbb{P}(75 \le x \le 80)=\mathbb{P}(\frac{X_{\text{min}}-0.5-np}{\sqrt{np(1-p)}} \le Z \le \frac{X_{\text{max}}+0.5-np}{\sqrt{np(1-p)}} )$$
So $$\mathbb{P}(\frac{74.5-70}{\sqrt{21}} \le Z \le \frac{80.5-70}{\sqrt{21}})=\mathbb{P}(\frac{4.5}{\sqrt{21}} \le Z \le \frac{10.5}{\sqrt{21}})$$
Now you can look this up in the table of Z distributions.
And your probability appears to be approximately $0.4890-0.3365=0.1525$
