MATH SOLVE

3 months ago

Q:
# The heights of women aged 20 to 29 are approximately normal with mean 64 inches and standard deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches. a) What is the z-scores for a woman 6 feet tall? b) What is the z-scores for a man 6 feet tall? c) What information do the z-scores give that the actual heights do not? d) What percent of men are shorter than 5 feet, 5 inches tall? e) What percent of women are over 6 feet, 1 inch tall? f) Find the interquartile range for the height of women aged 20-29.

Accepted Solution

A:

Answer:Step-by-step explanation:[tex]Z=\frac{x-mu}{\sigma }[/tex]X - Heights of women aged 20 to 29: X is N(64, 2.7)Y - Heights of men aged 20 to 29: X is N(69.3, 2.8)a) X=6'=72"Z=[tex]\frac{72-64}{2.7} =2.963[/tex]b) Y = 72"z=[tex]\frac{72-69.3}{2.8} =0.965[/tex]c) 72' women can be taken as very tall but men moderately taller than averaged) Y<5'5" =65"Z<[tex]\frac{65-69.3}{2.8} =-1.536[/tex]e) X>73"Z>[tex]\frac{73-64}{2.7} =3.33[/tex]Almost 0% can be takenf) IQR =Q3-Q1Z scores =