SOLUTION: In the following problem, I have found some of the answers but I'm stuck on the last part. Here is what I have tried. According to a study done by UCB students, the height for

Question 1148632: In the following problem, I have found some of the answers but I'm stuck on the last part. Here is what I have tried.
According to a study done by UCB students, the height for Martian adult males is normally distributed with an average of 64 inches and a standard deviation of 2.3 inches. Suppose one Martian adult male is randomly chosen. Let X = height of the individual. Round all answers to 4 decimal places where possible.
My answers:
What is the distribution of X? X ~ N
mean is 64, standard deviation is 2.3
Find the probability that the person is between 63.1 and 66.9 inches?
0.5485 (which is correct according to my online class)
The middle 40% of Martian heights lie between what two numbers?
62.7948 and 65.2052 (which are INCORRECT)....
how can I figure this out? THANKS!!!!!! (I do not have a TI calculator and I don't want one!)

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The middle 40% are where the probability is 0.2000 to the left and to the right of the midline. Want the z values for both of them, which are additive inverses of each other.
this is z=-0.5244 and +0.5244
z=(x-mean)/sd
0.524=(x-64)/2.3
1.2061=x-64
x=65.2061 and also 62.7939
You did it properly--it is a matter of rounding, and 4 decimal places is not appropriate given the numbers they gave you. I would probably try doing it with rounding to 5 decimal places throughout and then round at the end. Normally, the z is rounded to two decimal places, rarely three, and the probability to four.

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