SOLUTION: The mean score on the the exam was 63%. The Standard deviation was 12%. Find the smallest interval about the mean that contains at least 82% of the scores.

Algebra ->  Permutations -> SOLUTION: The mean score on the the exam was 63%. The Standard deviation was 12%. Find the smallest interval about the mean that contains at least 82% of the scores.       Log On


   

Question 662286: The mean score on the the exam was 63%. The Standard deviation was 12%. Find the smallest interval about the mean that contains at least 82% of the scores.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

μ = 63% and  σ = 12%  

Find the smallest interval about the mean that contains at least 82% of the scores

(100-82)/2 = 18/2 = 9% on either side

z = 1.341

 1.341 = (x-63)/12

 1.341*12 + 63 = 79

 79-63 = 13

 82% of the scores will be between 50%-79%