The scores for the Algebra 2 CFE are normally distributed with a mean score of 43 and

a standard deviation of 3.8. If you scored 47 on the test, what percentage of test takers

scored lower than you? Do not round your answer. [percent]

85.375%

Step-by-step explanation:

We would be using the z-score formula which is

z = (x-μ) / σ, where

x is the raw score,

μ is the population mean, and

σ is the population standard deviation.

In the question, we are given:

x is the raw score = 47

μ is the population mean = Mean score = 43

σ is the population standard deviation = 3.8

z = (x-μ) / σ

z = (47 - 43) / 3.8

z = 4/3.8

z = 1.0526315789

The z score = 1.0526315789

Checking my z score on my normal distribution table,

The percentage of test takers that scored lower than you =

[1 - P (x>Z) ] * 100

= (1 - 0.14625) * 100

= 0.85375 * 100

= 85.375%

Therefore, the percentage of test takers that scored lower than you is 85.375%