A professor knows that the test score of a student taking her exam is a random variable with mean $75$ and variance $25$. Find the $n$

Question

A professor knows that the test score of a student taking her exam is a random variable with mean $75$ and variance $25$. Find the $n$, the number of students to ensure that with probability $0.9375$ the class average would be within $5$ to $75$.

My Working:

As far as I have understood the problem required the use of Chebyshev's inequality, and I am sure of that. But I am unable to apply it properly. Can anyone guide and help. I will really appreciate it.

Answer 1

Let $\bar{X}$ denote the class average, i.e. $$\bar{X} = \frac{1}{n}\sum_{i=1}^{n}X_i,$$ where $X_i$ denotes the s.v. describing the score of student $i$. Assume that the $X_i$ are i.i.d:s with $\langle X_i \rangle = 75$ and $\mathrm{Var(X_i)}=25$. We then have $\mu = \langle \bar{X} \rangle = 75$ and $\sigma^2=\mathrm{Var(\bar{X})} = 25/n$, i.e. $\sigma = 5/\sqrt{n}$.

I understand the question as follows. Find $n$ such that the inequality $$ \mathrm{Pr}(|\bar{X}-75|\leq 5) \geq 0.9375$$ holds. Taking complements, this inequality is equivalent to $$1) \ \mathrm{Pr}(|\bar{X}-\mu|\geq r\sigma) \leq 1-0.9375 = 0.0625,$$ where $r= \sqrt{n}$. By Chebyshev's inequality we have $$2) \ \mathrm{Pr}(|\bar{X}-\mu|\geq r\sigma) \leq 1/r^2.$$ Thus, the inequality 1) is guaranteed by 2) if $1/n=1/r^2 \leq 0.0625$, or if $n \geq 16$.

Answer 2

For this problem you also have to use Fisher's Lemma, since the random variable we are interested in is $\overline{X}$, the sample mean for a sample of $n$ students. If we assume that the score of every student is a random variable with mean $\mu=75$ and variance $\sigma^2=25$, and that a sample consists of $n$ independent identically distributed grades (corresponding to $n$ different students), then Fisher's lemma guarantees that $\overline{X}$ has expectation (mean) $\mu=75$ and variance $\sigma^2/n = 25/n$

Chebyshev's inequality states that, for a random variable $X$ with expectation $\mu$ and variance $\sigma^2$, and for any $\epsilon>0$: $$P(|X-\mu|\geq\sigma\epsilon)\leq \frac{1}{\epsilon^2}$$ Since we want the probability that the mean is within $5$ to $75$ (i.e., between $70$ and $80$) to be $1 - \alpha:= 0.9375$, then the probability that the mean is not within $5$ of $75$ (i.e., that $|\overline{X}-\mu|\geq 5$) must be $\alpha=1-0.9375=0.0625$ (or less than $\alpha$). In this case, because of Fisher's Lemma, Chebyshev's inequality reads: $$P\left(|\overline{X}-\mu|\geq\frac{\sigma}{\sqrt{n}}\epsilon\right)\leq \frac{1}{\epsilon^2}=\alpha$$ We set it equal to $\alpha$ because of the reasons I stated earlier. Solving for epsilon, $\epsilon$ must be $\epsilon = 1/\sqrt{\alpha}=4$

Finally, notice that we want $n$ such that $\frac{\sigma}{\sqrt{n}}\epsilon=5$, since, in that case, Chebyshev's inequality would read as we want. Hence, solving for $n$, we obtain that $$n=\frac{\sigma^2\epsilon^2}{5^2}=16,$$ (or greater than 16, of course). I hope it helps.