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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(height is between 61 and 79 inches)
[tex]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\%[/tex]
[tex]P(61 \leq x \leq 79) = 99.73\%[/tex]
0.9973 is the percentage of American men whose height is between 61 and 79 inches.
