Question

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.

Answer 1

Answer:

z-score is 2.16

98.46% of men are SHORTER than 6 feet 3 inches

Step-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 = $\mu = 69.3 inches$

Standard deviation = $\sigma = 2.64 inches$

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  = 72+3=75 inches

Formula : $Z=\frac{x-\mu}{\sigma}$

$Z=\frac{75-69.3}{2.64}$

$Z=2.159$

So, his z-score is 2.16

No to find percentage of men are SHORTER than 6 feet 3 inches

We are supposed to find P(x< 6 feet 3 inches)

z-score is 2.16

Refer the z table

P(x< 6 feet 3 inches) =0.9846

So, 98.46% of men are SHORTER than 6 feet 3 inches