# Exam scores in a MATH 1030 class is approximately normally distributed with mean 87 and standard deviation 5.8. Round answers..?

Exam scores in a MATH 1030 class is approximately normally distributed with mean 87 and standard deviation 5.8. Round answers to the nearest tenth of a percent.

a) What percentage of scores will be less than 92?

b) What percentage of scores will be more than 85?

c) What percentage of scores will be between 83 and 89?

### 3 Answers

- 6 months agoFavorite Answer
Find the z-scores for each:

a) (92 - 87) / 5.8

5 / 5.8

25/29

0.86206896551724137931034482758621

z = 0.86, p = 0.8051

80.51%

b)

(85 - 87) / 5.8 =>

-2/5.8 =>

-10/29 =>

-0.34482758620689655172413793103448

1 - z(-0.34) =>

1 - 0.3669 =>

0.6331

63.31%

c)

(83 - 87) / 5.8 =>

-4/5.8 =>

-20/29

(89 - 87) / 5.8 =>

2/5.8 =>

10/29

z(10/29) - z(-20/29) =>

z(0.344) - z(-0.690) =>

0.6331 - 0.2451 =>

0.3880

38.8%

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- Anonymous6 months ago
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- az_lenderLv 76 months ago
(a) z = (92 - 87)/5.8 = 0.86207;

normal distribution table says 80.6% of scores are less than this.

(b) z = (85 - 87)/5.8 = -0.34483;

normal distribution table says 13.5% of scores are between z=0 and z=-0.345;

therefore 63.5% of scores are higher than z = -0.345.

(c) z1 = (83 - 87)/5.8 = -0.68966 and

z2 = (89 - 87)/5.8 = +0.34483;

normal distribution table says 13.5% of scores are between z = 0 and z2,

while 25.5% of scores are between z1 and z=0;

therefore 49.0% of scores are between z1 and z2.

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