The scores on an aptitude test are normally distributed with a mean of 18 and a standard deviation of 6.

a. What score would a student need to be in the top 50%?

b.What score would a student need to be in the 10th percentile?

c.What score would a student need to be in the 99th percentile?

d.What score would a student need to be in the top 25% of students?

I don't understand how I do probability, percentages, and percentiles. Please help and show how I can get the answer, I really need to understand this. Thank you.

Guest Feb 11, 2020

#1**+1 **

We heed to use z scores here

The formula is :

score - mean

___________ = z score

std deviation

a. This one is easy.....to be in the top 50%......since the mean is 18.....then you need a scure greater than or equal to 18 to be in the top 50%

b. We need to use a z score table ( here) : http://www.z-table.com/ and find the z score closest to .10......this is the z score -1.28

So we can solve this

x - 18

_______ = -1.28 multiply both sides by 6

6

x - 18 = -7.68 add 18 to both sides

x = 10.32 = score we need to be in the 10th percentile

c. To be in the 99th percentile......we can find the z score closest to .99.....using the table we find that this is about 2.33

So we can solve this

x - 18

_____ = 2.33 nultiply both sides by 6

6

x -18 = 13.98 add 18 to both sides

x = 31.98 = the 99th percentile score

d. To be in the top 25% of the students we need to find the z score closest to .75.....using the table we have the z score of about .68

So

x - 18

______ = .68 multiply both sides by 6

x - 18 = 4.08 add 18 to both sides

x = 22.08 = we need at least this score to be in the top 25%

CPhill Feb 11, 2020