SOLUTION: A set of final examination grades in an introductory statistics course was founded to be normally distributed with a mean of 73 and a standard deviation of 8. a. what is the pro

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Question 385745: A set of final examination grades in an introductory statistics course was founded to be normally distributed with a mean of 73 and a standard deviation of 8.
a. what is the probability of getting at most a grade of 91 on the exam?
b. what percentage of students scored between 65 and 85?
c. what percentage of students scored between81 and 89?
d. what is the final exam grade in only 5% of the students taking the test scored higher?
e. If the professor "curves"( gives A`s to the top 10% of the class regardless of the score), are you better off with a grade of 81on this exam or a grade of 68 on a different exam where the mean is 62 and the standard deviation is 3? show statistically and explain
Include a normal distribution curve

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A set of final examination grades in an introductory statistics course was founded to be normally distributed with a mean of 73 and a standard deviation of 8.
a. what is the probability of getting at most a grade of 91 on the exam?
---
z(91) = (91-73)/8 = 2.25
P(x <= 91) = P(z < 2.25) = 0.9878
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b. what percentage of students scored between 65 and 85?
P(65 < x < 85) = 0.7745
----------------------------------
c. what percentage of students scored between81 and 89?
P(81< x <89) = 0.1359
------------------------
d. what is the final exam grade if only 5% of the students taking the test scored higher?
Find the z-value with a right tail of 5%: z = 1.645
Find the corresponding exam score:
x = zs+u
x = 1.645*8+73
x = 86.16
=============
e. If the professor "curves"( gives A`s to the top 10% of the class regardless of the score), are you better off with a grade of 81 on this exam or a grade of 68 on a different exam where the mean is 62 and the standard deviation is 3? show statistically and explain
Include a normal distribution curve
I'll leave that to you.
cheers,
Stan H.

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