Question

From past experience, a professor knows that the test score of a student taking her final examination is a random variable with mean $75$.

a. Give an upper bound for the probability that a student's test score will exceed $85$.

Suppose, in addition, that the professor knows that the variance of a student's test score is equal to equal to $25$.

b. What can be said about the probability that a student will score between $65$ and $85$.

c. How many students would have to take the examination to ensure with probability at least $0.9$ that the class average would be within $5$ of $75$?

Do not use the central limit theorem.

Answer 1

Let $X$ be a random variable representing marks scored.

$X~N\left(\mu =75,{\sigma }^2=5^2\right)$

Part (a):

$\begin{array}{rcl}P\left(X>85\right)& =& P\left(\frac{X-75}{5}>\frac{85-75}{5}\right)\\ & \Rightarrow & P\left(X>85\right)=P\left(z>2\right)\\ & \Rightarrow & P\left(X>85\right)=0.023\end{array}$

Part (b):

$\begin{array}{rcl}P\left(65<X<85\right)& =& P\left(X<85\right)-P\left(X<65\right)\\ & \Rightarrow & P\left(65<X<85\right)=P\left(\frac{X-75}{5}<\frac{85-75}{5}\right)-P\left(\frac{X-75}{5}<\frac{65-75}{5}\right)\\ & \Rightarrow & P\left(65<X<85\right)=P\left(z<2\right)-P\left(z<-2\right)\\ & \Rightarrow & P\left(65<X<85\right)=P\left(z<2\right)-\left[1-P\left(z<2\right)\right]\\ & \Rightarrow & P\left(65<X<85\right)=2P\left(z<2\right)-1\\ & \Rightarrow & P\left(65<X<85\right)=2\left(0.977\right)-1\\ & \Rightarrow & P\left(65<X<85\right)=0.954\end{array}$

Part (c):

For the average score distribution is,

$\overline{X}~N\left(\mu =75,{\sigma }^2={\left(\frac{5}{\sqrt{n}}\right)}^2\right)$

Professor want,

$\begin{array}{rcl}P\left(70<\overline{X}<80\right)& =& 0.9\\ & \Rightarrow & P\left(\overline{X}<80\right)-P\left(\overline{X}<70\right)=0.9\\ & \Rightarrow & P\left(\frac{X-75}{{\displaystyle \frac{5}{\sqrt{n}}}}<\frac{80-75}{{\displaystyle \frac{5}{\sqrt{n}}}}\right)-P\left(\frac{X-75}{{\displaystyle \frac{5}{\sqrt{n}}}}<\frac{70-75}{{\displaystyle \frac{5}{\sqrt{n}}}}\right)=0.9\\ & \Rightarrow & P\left(z<\sqrt{n}\right)-P\left(z<-\sqrt{n}\right)=0.9\\ & \Rightarrow & P\left(z<\sqrt{n}\right)-\left[1-P\left(z<\sqrt{n}\right)\right]=0.9\\ & \Rightarrow & 2P\left(z<\sqrt{n}\right)-1=0.9\\ \therefore P\left(z<\sqrt{n}\right)& =& 0.95\end{array}$

Using standard normal tables,

$\begin{array}{rcl}P\left(z<1.6449\right)& =& 0.95\\ & \Rightarrow & \sqrt{n}\ge 1.6449\\ & \Rightarrow & n\ge 1.{6449}^2\\ & \Rightarrow & n\ge 2.7056\end{array}$

Therefore, at least $3$ students would have to take the examination for at least $0.9$ probability that average will be within $5$ of $75$.