Among American women aged 20-29 years, 10% are less than 60.8 inches tall,

Question

80% are between 60.8 inches and 67.6 inches tall, and 10% are more than 67.6 inches tall. Assuming that the height distribution can adequately be approximated by a normal curve, find the mean and standard deviation of the distribution.

I am trying to substitute equations into the z-score formula but I am getting stuck.

Answer 1

Mean = (60.8 + 67.6)/2 = 64.2

So in this case, from 64.2 to 67.6, the area of the region is 40% because it is 80% fro 60.8 to 67.6.

According to Z- calculator on calculator.net the Z = 1.28157 at 67.6 and Z = -1.28157 at 60.8. You can also look at the table if you prefer.

The formula you can use to find the standard deviation is:

Z = (x - μ) / σ

Where σ = standard deviation

Z = z-score

x = raw score

μ = mean

1.28157 = (67.6-64.2) / σ

σ = (67.6-64.2) / 1.28157

σ ≈ 2.65 inches