he heights of the 228 men who filled out Survey 1 last semester follow the norma

Question

he heights of the 228 men who filled out Survey 1 last semester follow the normal curve fairly closely with an average of about 70.5" and a SD of about 2.5 ". What percent of the students are over 71.5 "? Do the problem in steps First convert 71.5 " to a z score.= 0.40

In the problem above you were given the height and asked to find the percentile. Now, you'll be going in the opposite direction-- you'll be given the percentile (placement on the normal curve) and asked to find the height. In both cases the first step is to find the z-score. If someone is in the 92 th percentile in height (i.e., is taller than 92 % of the other males in the class), let's work on finding his height. First we'll find the z score. (Do NOT look up the z score for Area= 92 % on the table b/c the Areas given on the table are MIDDLE areas. 92 th percentile means 92 % to the LEFT, and 8 % in the right tail.)

If someone is in the 8 th/st percentile in height (i.e., is taller than only 8 % of the other males in the class), let's find his height. First find the Z score. Tries 0/5

Now change the z score to a height. Round to the nearest tenth. "

Explanation / Answer

from normal value table for 92th percentile ; z score =1.405

thefore corresponding height for 92th percentile =mean+z*Std deviaiton=70.5+1.405*2.5=74.0

here from normal value table for 8th percentile ; z score =-1.405

thefore corresponding height for 8th percentile =mean+z*Std deviaiton=70.5-1.405*2.5=67.0

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