Question

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?

Answer 1

Answer:

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