Q. 8.13 - A First Course in Probability

Question

Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give two exams, one to a class of size 25 and the other to a class of size 64.

(a) Approximate the probability that the average test score in the class of size 25 exceeds 80.

(b) Repeat part (a) for the class of size 64.

(c) Approximate the probability that the average test score in the larger class exceeds that of the other class by more than 2.2 points.

(d) Approximate the probability that the average test score in the smaller class exceeds that of the other class.

by more than 2.2 points.

Step-by-Step Solution

Answer

The summary of the solution is as follows

$$\begin{gathered} \mathrm{a}.) P(\overline{X}>80)\approx 0.0162 \\ \mathrm{b}.) P(\overline{Y}>80)\approx 0.0003 \\ \mathrm{c}.) P(\overline{Y}-\overline{X}>2.2)\approx 0.2514 \\ \mathrm{d}.) P(\overline{Y}-\overline{X}<2.2)\approx 0.7485 \end{gathered}$$

Given Information

$𝜇=74, 𝜎=14, n=25 \mathrm{and} m=64$

where symbols have their usual meanings.

Part (a)

$$\begin{gathered} P(\overline{X}>80)=P\left(\frac{\overline{X}-74}{14/5}>\frac{15}{7}\right) \\ \approx P(Z>2.14)\approx 0.0162 \end{gathered}$$

Part (b)

$$\begin{gathered} P(\overline{Y}>80)=P\left(\frac{\overline{Y}-74}{14/8}>\frac{24}{7}\right) \\ \approx P(Z>3.43)\approx 0.0003 \end{gathered}$$

Part (c)

We have 

$SD(\overline{Y}-\overline{X})=\sqrt{196/64+196/25}\approx 3.33$

Therefore

$$\begin{gathered} P(\overline{Y}-\overline{X}>2.2)=P\left(\frac{\overline{Y}-\overline{X}}{3.3}>\frac{2.2}{3.3}\right) \\ \approx P(Z>0.6667)\approx 0.2514 \end{gathered}$$

Part (d)

Using part (c) we get

$P(\overline{Y}-\overline{X}<2.2)\approx 0.7485$