Question about normal approximation Question: A class has $20$ students, each student tosses a fair coin $300$ times and records the number of heads. Approximately what is the probability that no one gets exactly $150$ heads? (Use normal Approximation)
Attempt: I let $X_i$ be number of heads in $300$ coin tosses for some student $i$ (for $i = 1,\ldots,20$), and $X_i \sim  \binom{300}{1/2}$.
I know that the formula for normal approximation for binomials is: $\Pr(\text{no one gets 150 heads}) = 1 - \Pr(X=150)$, where $\Pr(X = 150) = \Pr\left(\dfrac{149.5-\text{mean}}{\text{SD}} <X < \dfrac{150.5-\text{mean}}{\text{SD}}\right)$. But I am not sure how to get the mean and SD.
 A: You have a start, but you need to break it down a bit more coherently.
General Approach.  First, find the probability that a single student fails to record exactly $150$ heads.  As you say, this can be determined using the normal approximation by first determining the standard deviation $\sigma$ for the number of heads in $300$ flips.  You then consult a table (or a calculator/computer) to find the probability that a normally distributed quantity fails to fall within $(1/2)/\sigma$ standard deviations of the mean.  Call this probability $p$.
Second, you find the probability that every one of the $20$ students fails to record exactly $150$ heads.  Making the reasonable assumption that the students' records are independent, this can be obtained simply by multiplying the probabilities for each individual student, so the desired probability is just $p^{20}$.
You should obtain a number in the neighborhood of $0.4$; the exact answer, using the binomial theorem, is approximately $0.38969$, but that is not what you should be checking against.
A: For just one student you have
$$
\Pr(X=150)=\Pr\left(\dfrac{149.5-\text{mean}}{\text{SD}}\right. < \overbrace{\frac{X-\text{mean}}{\text{SD}}}^{\text{Note well.}} < \left.\dfrac{150.5-\text{mean}}{\text{SD}}\right). \tag 1
$$
You can't put just $X$ in the middle.
The mean and SD for $\mathrm{Binomial}\left(300,\frac 1 2\right)$ are $np = 300\cdot\frac 1 2 = 150$ and $\sqrt{npq} = \sqrt{300\cdot\frac1 2 \cdot\left(1 - \frac 1 2\right)}$.
In $(1)$, replace the thing in the middle, $(X-150)/{\text{SD}}$, with $Z$, the standard normal random variable.  Then a table or software will give you a probability.
However, that's for just one student. Say you get probability $r$ that that one student fails to get exactly $150$.  Then the probability that ALL of them fail to get exactly $150$ is $r^{20}$.
