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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.**
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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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2022-01-10T20:43:52+00:00
2022-01-10T20:43:52+00:00 1 Answer
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## Answers ( )

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) =

0.9927b)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) =

0.6585b)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) =

0.0781c)P ( X > a ) = 0.05 , a?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