+0  
 
0
77
1
avatar

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. 

 Feb 11, 2020
 #1
avatar+109740 
+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%

 

 

 

cool cool cool

    

 Feb 11, 2020

52 Online Users

avatar
avatar
avatar
avatar
avatar
avatar