MATH SOLVE

3 months ago

Q:
# The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.53 inches. a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? z = 2.16 b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

Accepted Solution

A:

Answer:z-score is 2.1698.46% of men are SHORTER than 6 feet 3 inchesStep-by-step explanation:The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches.Mean = [tex]\mu = 69.3 inches[/tex]Standard deviation = [tex]\sigma = 2.64 inches[/tex]a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)x = 6 feet 3 inches 1 feet =12 inches 6 feet = 12*6 = 72 inches So, x = 6 feet 3 inches Β = 72+3=75 inchesFormula : [tex]Z=\frac{x-\mu}{\sigma}[/tex][tex]Z=\frac{75-69.3}{2.64}[/tex][tex]Z=2.159[/tex]So, his z-score is 2.16No to find percentage of men are SHORTER than 6 feet 3 inchesWe are supposed to find P(x< 6 feet 3 inches) z-score is 2.16Refer the z table P(x< 6 feet 3 inches) =0.9846So, 98.46% of men are SHORTER than 6 feet 3 inches