A set of final examination grades in a calculus course wasfound to be normally distributed with a mean of 69 and a standarddeviation of 9.

a. what is the probality of getting a grade of 91or less on this exam?

b. What percentage of students scored between 65 and89?

c. What percentage of students scored between 81 and89?

d. Only 5% of the students taking the test scored higherthan what grade?

a) P (X < 91) = 0.9927

b) P (65 < X < 91) = 0.6585

c) P (81 < X < 89) = 0.0781

d) X = 83.8

Step-by-step explanation:

Given:

- Mean of the distribution u = 69

- standard deviation sigma = 9

Find:

a. what is the probability of getting a grade of 91 or less on this exam?

b. What percentage of students scored between 65 and 89?

c. What percentage of students scored between 81 and 89?

d. Only 5% of the students taking the test scored higher than what grade?

Solution:

- We will declare a random variable X denoting the score that a student gets on a final exam. So,

X ~ N (69, 9)

- After defining our variable X follows a normal distribution. We can compute the probabilities as follows:

a) P (X < 91) ?

- Compute the Z-score value as follows:

Z = (91 - 69) / 9 = 2.4444

- Now use the Z-score tables and look for z = 2.444:

P (X < 91) = P (Z < 2.4444) = 0.9927

b) P (65 < X < 89) ?

- Compute the Z-score values as follows:

Z = (89 - 69) / 9 = 2.2.222

Z = (65 - 69) / 9 = - 0.4444

- Now use the Z-score tables and look for z = 2.222 and Z = - 0.4444:

P (65 < X < 89) = P (-0.444< Z < 2.2222) = 0.6585

b) P (81 < X < 89) ?

- Compute the Z-score values as follows:

Z = (89 - 69) / 9 = 2.2.222

Z = (81 - 69) / 9 = 1.3333

- Now use the Z-score tables and look for z = 2.222 and Z = 1.333:

P (81 < X < 89) = P (1.333< Z < 2.2222) = 0.0781

c) P (X > a) = 0.05, a?

- Compute the Z-score values as follows:

Z = (a - 69) / 9 = q

- Now use the Z-score tables and look for z value that corresponds to:

P (X > a) = P (Z > q) = 0.05

- The corresponding Z-value is: q = 1.6444

Hence,

Z = (a - 69) / 9 = 1.644

a = 83.8