# How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$?

I'm having trouble solving this problem:

From past experience, a professor knows that the test score of a student taking her final examination is a random variable with mean $75$.
How many students would have to take the examination to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? Use the central limit theorem.

The professor knows that the variance of a student's test score is $25$.

I'm not entirely sure on how to solve this problem.

Right now this is what I have:

We know: $\mu = 75$ and $\sigma^2 = 25$.

This is what I set up (by defn C.L.T): $\mathbb{P}(\frac{X_1+\cdots+X_n - n75}{5\sqrt{n}}\le\frac{.9-n75}{5\sqrt{n}}) = 1-\mathbb{P}(\frac{X_1+\cdots+X_n - n75}{5\sqrt{n}}>\frac{.9-n75}{5\sqrt{n}})$

I'm not sure how to solve for $n$. Thanks.

• You're missing some information -- you need to have the variance of the final exam of the students. Apr 28, 2016 at 15:32
• @Batman Just updated Apr 28, 2016 at 15:34
• You want to write the probability that $P(|S_n/n - 75| \le 5) \ge 0.9$ using the information from the CLT; looking cursorily at what you wrote, I think there might be an incorrect substitution. Apr 28, 2016 at 15:41
• This exercise is terrible. To the attention of non probabilist readers, let us mention that it is the logical equivalent of the following: Consider some real valued sequence $(x_n)$ such that $nx_n\to1$, find some $n$ such that $x_k\leqslant\frac1{10}$ for every $k\geqslant n$. Scary, no?
– Did
Apr 28, 2016 at 15:47
• @Did How would the exercise need to be changed? May 29, 2016 at 23:45

If the students have mean $\mu$ and variance $\sigma^2$, then the distribution for the average of $n$ students is approximately normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$.

In this case, you want to know the probability that the sample mean $\bar{X}$ is within $\pm r$ of the true mean $\mu$. So, $$P(\mu-r \leq \bar{X} \leq \mu-r) = P(-r \leq \bar{X} -\mu \leq r) = P(\frac{-r}{ \sqrt{\frac{\sigma^2}{n}}} \leq \frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}} \leq \frac{r}{ \sqrt{\frac{\sigma^2}{n}}} ) \approx P(\frac{-r \sqrt{n}}{\sigma} \leq N(0,1) \leq \frac{r \sqrt{n}}{\sigma})$$ since $\frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$ is approximated to be standard normal.

By symmetry of the normal distribution, $$P(\frac{-r \sqrt{n}}{\sigma} \leq N(0,1) \leq \frac{r \sqrt{n}}{\sigma}) = 1- 2 Q(\frac{r \sqrt{n}}{\sigma}) = 1-2 \Phi(\frac{-r \sqrt{n}}{\sigma})$$

where $Q$ is the complementary standard normal CDF and $\Phi$ is the standard normal CDF.

In this case, you want $0.9=1-2 \Phi(\frac{-r \sqrt{n}}{\sigma})$. You can solve this for $\Phi(\frac{-r \sqrt{n}}{\sigma})=0.05$, look up the value $z$ such that $\Phi(z)=0.05$ and then solve $z=\frac{-r \sqrt{n}}{\sigma}$ for $n$.

Since you have to have an integer number of students, round the resulting $n$ up (which will give you a higher probability of being within the interval than $0.9$ if it is not an integer).

• "since $\frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$ is standard normal" IS IT? In fact, nothing in the hypotheses implies that.
– Did
Apr 28, 2016 at 15:44
• @Did I agree with you that it isn't. Statisticians always assume everything is approximately normal, even when they shouldn't -- my guess would be that the problem is assuming the same fallacy. Apr 28, 2016 at 15:48
• "since $\frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$ is approximated to be standard normal" Approximated how? This is the whole point, which makes the question a terrible exercise.
– Did
Apr 28, 2016 at 15:48
• Its probably a high school stat problem, not some precise problem for a mathematical statistics class. Apr 28, 2016 at 15:49
• Let me follow you and let us assume this is not a maths question. Then, since it falsely masquerades as one, our urgent task as mathematicians should be to explain why this is not mathematics, instead of supporting the belief that it is. No?
– Did
Apr 28, 2016 at 15:51