Question

suppose that the scores on a certain test are normal with mean 120 and standard deviation 30. (a) what proportion of scores are over 100? (b) find the 57th percentile, i.e. the best score in the low 57% gro

Answer 1

Part a: The proportion of scores are over 100 is 25.46%.

Part b: Best score below 57th percentile is 125.4.

What is the definition of normal distribution?

• Although all symmetrical distributions are normal, not all normal distributions are symmetrical.
• Many naturally occurring phenomena resemble the normal distribution.
• However, most pricing distributions in finance are not perfectly normal.

For the given question;

• mean score μ = 120
• standard deviation σ = 30

Part a: For sample mean x = 100.

z = (x - μ)/σ

z = (100 - 120)/30

z = -0.66

Use negative z table

P(x = 100) = 0.2546

P(x = 100) = 25.46%

Thus, the proportion of scores are over 100 is 25.46%.

Part b: Best score below 57th percentile.

p = 0.57

Se z score from z score table for  p= 0.57

z = 0.18

0.18 = (x - 120)/30

x = 125.4

Thus, the best score below 57% is 125.4.

To know more about the normal distribution, here

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