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.

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?

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Abigail 2 weeks 2022-01-10T20:43:52+00:00 1 Answer 0 views 0

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    2022-01-10T20:45:37+00:00

    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              

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