POLS 4150 – Practice problems for calculating areas under the normal curve
In a large class, the average final exam score was 75 with a standard deviation of
8.
Assume
that the scores are distributed normally.
Let X = exam score
1)
What proportion of students scored above a 65?
Z = (6575) / 8 = 1.25
Area below 1.25 = 10.56%
100%  10.56% =
89.44%
2)
What proportion of students scored between an 81 and an 89?
Z = (8175) / 8 = .75.
Area above .75 = 22.66%
Z = (8975) / 8 = 1.75.
Area above 1.75 = 4.01%
22.66%  4.01%
= 18.65%
3)
What proportion of students scored below a 61?
Z = (6175) / 8 = 1.75
Area below 1.75 =
4.01%
4)
What proportion of students scored between a 61 and an 85?
Z = (6175) / 8 = 1. 75.
Area below 1.75 = 4.01%
Z = (8575) / 8 = 1.25.
Area above 1.25 = 10.56%
100%  (46.99% + 39.44%)
= 85.43
5)
What proportion of students scored above a 91?
Z = (9175) / 8 = 2.0
Area beyond Z =
2.28%
6)
What proportion of students scored between 55 and 65?
Z = (5575) / 8 = 2.5.
Area below 2.5 = 0.62%
Z = (6575) / 8 = 1.25.
Area below 1.25 = 10.56%
10.56%  0.62%
= 9.94%
7)
What proportion of students scored below an 89?
Z = (8975) / 8 = 1.75
Area above 1.75 = 4.01%
100%  4.01% =
95.99%
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 Fall '10
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 Normal Distribution

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