Question 1123379: A survey found that women's heights are normally distributed with mean 63.2 in. and standard deviation of2.7 in. The survey also found that men's heights are normally distributed with a mean 68.1 in. and standard deviation of2.8. Complete parts a through c below.
a) 98.37% Woman (z score was hard to find on both)
b) 99.87% Men
c) if the height requirements are changed to exclude only the tallest 5% of? the men and the shortest 5% of women, what are the new height requirements?
how do you set up and work the problem to get the answer of at least 58.7 in and at least 72.1in round to one place as needed
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 98.37% of women are at a z-score of 2.14, and 2.14*2.7 is 5.78, so the percentile is at 68.98 or 69.0 inches
99.87% of men are at a z-score of 3.015, and 3.015*2.8 is 8.44, so the percentile is at 68.1+8.44 or 76.5 inches
The tallest 5% of men are a z-score <1.645
z*sd=4.61 in
it would be men below 68.1+4.6 or 72.7 in.
The shortest 5% of women are at a z-score of -1.645, and -1.645*2.7 is 4.4, so it would exclude those more than 4.4 inches shorter than 63.2 or shorter than 58.8 inches.
The basic formula is z=(x-mean)/sd, and solve for the missing variable.
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