Question

Exam grades in my statics class tend to be normally distributed. I've been keeping long term records, and the mean grade is a 67, with a standard deviation of 13 (on a of 100 point exam).

9. What is the probability that a student gets a grade lower than a 90?

10. What fraction of the class would be expected to get a grade between 60 and 80?

11. What is the probability that one of my statics sections with 30 students has an average midterm grade greater than 70?

12. The final grade in my class consists of 3 equally weighted exams. Out of a total of 300 points, what is the mean overall grade?

13. Continuing question 12, what is the standard deviation of the overall grade?

14. Continuing questions 12 & 13, out of the 300 points possible, what is the probability that a person scores greater than 240?

Answer 1

Solution:

We are given:

$\mu=67, \sigma =13$

9. We are required to find:

P(Z <90)

Using the z-score formula, we have:

90 - 67 P(Z <90) = P(Z < 13

= P(Z < 1.77)

Now using the standard normal table, we have:

Par <90) = P(Z < 1.77) = 0.9616

10. We are required to find:

Using the z-score formula, we have:

$P(60<x<80)=P \left(\frac{60-67}{13}<z<\frac{80-67}{13} \right )$

  $=P(-0.54<z<1)$

$=P(z<1)-P(z<-0.54)$

Now using the standard normal table, we have:

$\boldsymbol{P(60<x<80)=P(z<1)-P(z<-0.54)=0.8413-0.2946=0.5467}$

11. We are required to find:

$P(\bar{x}>70)$

Using the z-score formula, we have:

$P(\bar{x}>70)=P \left(z> \frac{70-67}{\frac{13}{\sqrt{30}}} \right )$

$=P(z>1.26)$

Now using the standard normal table, we have:

$\boldsymbol{P( \bar{x}>70)=P(z>1.26)=0.1038 }$

12. We know that:

$X \sim N(67,13)$

Therefore, the mean of overall grade is:

$\boldsymbol{Mean= 3 \times67=201 }$

13. The standard deviation of the overall grade is:

$\boldsymbol{Standard \ deviation = \sqrt{3} \times 13=22.5167}$

14. We are required to find:

$P \left( \sum x >240 \right )$

Using the z-score formula, we have:

$P \left( \sum x >240 \right )=P \left(z> \frac{240-201}{22.5167} \right )$

$=P \left(z> 1.73\right )$

Now using the z-score formula, we have:

$P \left( \sumx > 240 \right )=P \left(z> 1.73\right )=0.0418$