document.write( "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.\r
\n" ); document.write( "\n" ); document.write( " 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.\r
\n" ); document.write( "\n" ); document.write( "My answers:
\n" ); document.write( "What is the distribution of X? X ~ N
\n" ); document.write( " mean is 64, standard deviation is 2.3
\n" ); document.write( "Find the probability that the person is between 63.1 and 66.9 inches?
\n" ); document.write( " 0.5485 (which is correct according to my online class)
\n" ); document.write( "The middle 40% of Martian heights lie between what two numbers?
\n" ); document.write( " 62.7948 and 65.2052 (which are INCORRECT)....
\n" ); document.write( " how can I figure this out? THANKS!!!!!! (I do not have a TI calculator and I don't want one!)
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Algebra.Com's Answer #770057 by Boreal(15235)\"\" \"About 
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.
\n" ); document.write( "this is z=-0.5244 and +0.5244
\n" ); document.write( "z=(x-mean)/sd
\n" ); document.write( "0.524=(x-64)/2.3
\n" ); document.write( "1.2061=x-64
\n" ); document.write( "x=65.2061 and also 62.7939\r
\n" ); document.write( "\n" ); document.write( "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.\r
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