Question

The heights of American men have an approximately bell-shaped distribution with a mean of 70 inches (5' 10") and standard deviation of 3 inches. What is the percentage of American men whose height is between 61 and 79 inches (5' 1" and 6' 7")?

Answer 1

Answer:

0.9973  is the percentage of American men whose height is between 61 and 79 inches.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70 inches

Standard Deviation, σ = 3 inches

We are given that the distribution of heights of American men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(height is between 61 and 79 inches)

$P(61 \leq x \leq 79) = P(\displaystyle\frac{61 - 70}{3} \leq z \leq \displaystyle\frac{79-70}{3}) = P(-3 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -3)\\= 0.9987 - 0.0014 = 0.9973= 99.73\%$

$P(61 \leq x \leq 79) = 99.73\%$

0.9973  is the percentage of American men whose height is between 61 and 79 inches.