Question

3. Professor Brown has been teaching the same university psychology course for many years, and she has been keeping a record of all the grades given to the students. Suppose that her student’s grades follow a normal distribution, with an average of 70 and a standard deviation of 12.

(a) Suppose you will be taking her class next semester. What is the probability that you will get a grade between 70 and 85?

(b) What minimum grade do you need in order to qualify for Professor Brown’s Honour Roll, which is reserved for the top 8% of students?

(c) Suppose this professor will have a group of 25 students next semester. What is the probability that her class average will not be between 65 and 75?

Answer 1

We have a normal distribution, $\mu=70, \sigma=12.$

(a) Let's convert it to the standard normal distribution,

$z=\cfrac{x-\mu}{\sigma};$

$z_1=\cfrac{70-70}{12}=0;\\ z_2=\cfrac{85-70}{12}=1.25;\\ P(70<X<85)=P(0<Z<1.25)=\\ =P(Z<1.25)-P(Z<0)=$

$=0.8944-0.5=0.3944$ ​(from z-table).

(b) 92^th percentile $\approx$ 1.405

We look for 0.9200 inside the z-table.

Although 0.9200 does not appear, both 0.9192 and 0.9207 do, corresponding to z = 1.40 and 1.41, respectively.

Since 0.9200 is approximately halfway between the two probabilities that do appear, we will use 1.405 as the 92th percentile.

The upper 8% of the normal curve:

$P(Z\leq1.405)\approx0.92, \\P(Z\geq1.405)\approx1-0.92=0.08.$

$z=\cfrac{x-\mu}{\sigma}; x=z\sigma+\mu=1.405\cdot12+70=86.86.$

So (if the grade is an integer number) the minimum grade for the top 8% of students is 87.

(c) We have a normal distribution, $\mu=70, \sigma=12, n=25.$

Let's convert it to the standard normal distribution,

$z=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}};$

$z_1=\cfrac{65-70}{12/\sqrt{25}}=-2.08,\\ z_2=\cfrac{75-70}{12/\sqrt{25}}=2.08,$

$1-P(65<\bar{X}<75)=\\1-P(-2.08<\bar{Z}<2.08)=\\ =1-(P(\bar{Z}<2.08)-P(\bar{Z}<-2.08))=$

$=1-(0.9812-0.0188)=1-0.9624=0.0376.$ ​(from z-table)