Question 1169484: According to a study done by students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
I need help with the following:
1) find the probability of an Asian male is over 71 inches tall
2) find the middle 40% of heights fall between what two values?
3) Write the probability statement.
P(x1 < X < x2) =???
Thank you!
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
or (71-66)/2.5
so the first is the probability z>2 which=0.0228
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middle 40% is between the 30th and 70th percentiles.
from the table or calculator invnorm(.30,0,1) and (.70,0,1) or between-0.5244 nd +0.5244
-0.5244=(x-66)/2.5
x=64.69
and at the upper end 67.31
(64.69, 67.31) inches
P(64.69 inches
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