Answer to Question #152010 in Statistics and Probability for JG

Question

Based on a sample of 450 AUS students, we found the average score is 165, with a standard deviation of 6.
a) At least how many percent of students scored between 150 to 180.
b) Assuming the histogram of GPA has a right skewed distribution, calculate the range of GPA that will cover at least 71% of the students?
c) Assuming the histogram of GPA follows the normal distribution, approximately how many students score between 147 to 159?

Answer 1

Let $X=$ the score of the student: $X\sim N(\mu, \sigma^2)$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)$

Given $\mu=165, \sigma=6$

a)

$$P(150<X<180)=P(X<180)-P(X\leq150)$$

$$=P(Z<\dfrac{180-165}{6})-P(Z\leq\dfrac{150-165}{6})$$

$$=P(Z<2.5)-P(Z\leq-2.5)$$

$$\approx0.993790-0.006210\approx0.9876$$

$98.76\%$

b)

$$P(X\geq x)=1-P(Z<\dfrac{x-165}{6})=0.71$$

$$P(Z<\dfrac{x-165}{6})=0.29$$

$$\dfrac{x-165}{6}\approx-0.553385$$

$$x=165-6(0.553385)=162$$

71% of the students will receive the score no less than $162$

c)

$$P(147<X<159)=P(X<159)-P(X\leq147)$$

$$=P(Z<\dfrac{159-165}{6})-P(Z\leq\dfrac{147-165}{6})$$

$$=P(Z<-1)-P(Z\leq-3)$$

$$\approx0.158655-0.001350\approx0.1573$$

$15.73\%$