Question

In a certain​ study, women's heights are found to be approximately normally distributed with a mean of 61 inches and a standard deviation of 2.5 inches.
a. What would be the​ z-score for a woman who is 5 feet 3 inches ​tall?

b. What percentage of women is she taller​ than?

Answer 1

Answer:

a. Z=0.8

b. 78.81%

Step-by-step explanation:

The z-score for a woman measuring 'X' inches is given by:

$Z=\frac{X - \mu}{\sigma}$

Where μ is the distribution mean and σ is the standard deviation.

Since each feet equals 12 inches, the woman's height is:

$X= 5*12 +3 = 63 \ inches$

a. The z-score is:

$Z=\frac{63 - 61}{2.5}\\Z=0.8$

b. The percentage of women she is taller than can be found by checking for the corresponding percentile in a z-score table. A z-score of 0.8 is equivalent to the 78.81-th percentile. Therefore, she is taller than 78.81% of women.