SOLUTION: The lengths of marbles are normally distributed with a mean of 3.9m and a standard deviation of 1.8m. Find an interval centered on the mean that contains 60% of the values. Ple

Algebra ->  Probability-and-statistics -> SOLUTION: The lengths of marbles are normally distributed with a mean of 3.9m and a standard deviation of 1.8m. Find an interval centered on the mean that contains 60% of the values. Ple      Log On


   

Question 959013: The lengths of marbles are normally distributed with a mean of 3.9m and a standard deviation of 1.8m.
Find an interval centered on the mean that contains 60% of the values.
Please help, TEST in 20minutes - confused - Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Find the z score corresponding to a probability of 0.2 and 0.8 because the difference is 0.6 or the 60% you're looking for.
When P=0.2
z%5B1%5D=-0.84162
When P=0.8
z%5B2%5D=0.84162
That the z-scores are additive inverses shouldn't be a surprise to you, hopefully.
So then,
%28x-mu%29%2Fsigma=0.84162
%28x-3.9%29%2F%281.8%29=0.84162
x-3.9=1.514918
x%5B2%5D=5.41
DELTA=5.41-3.9=1.51
So x%5B1%5D=3.9-1.51=2.39
(2.39,5.41)