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.
+128407 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%
