Question

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule.
(a) What proportion of the students scored at least 19 points on this test, rounded to five decimal places?

(b) What is the 85 percentile of the distribution of test scores, rounded to three decimal places?

Answer 1

6.68% of the students scored at least 19 points on this test while the 85 percentile has a test score of 24.08 points

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ \mu=mean.\sigma=standard\ deviation, x=raw\ score\\\\Given\ that\ mean(\mu)=22,standard \ deviation (\sigma)=2\\\\For\ x>19:\\\\z=\frac{22-19}{2}=1.5$

P(x > 19) = P(z > 1.5) = 1 - P(z < 1.5) = 1 - 0.9332 = 0.0668

6.68% of the students scored at least 19 points on this test

b) 85 percentile corresponds to a z score of 1.04

$z=\frac{x-\mu}{\sigma}\\\\1.04 =\frac{x-22}{2} \\\\2.08=x-22\\\\x=24.08$

The 85 percentile has a test score of 24.08

Find out more at: brainly.com/question/25012216