Question

According to a survey by the National Center for Health Statistics, the heights of adult men in the U.S. are normally distributed with a mean of 68 inches and a standard deviation of 2.75 inches. What is the probability that adult men are at most 59 inches or at least 74 inches

Answer 1

Answer:

0.0151 is the probability  that adult men are at most 59 inches or at least 74 inches.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 68 inches

Standard Deviation, σ = 2.75 inches

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

Formula:

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

P(height between 59 and 74)

$P(59 \leq x \leq 74) = P(\displaystyle\frac{59 - 68}{2.75} \leq z \leq \displaystyle\frac{74-68}{2.75}) = P(-3.2727 \leq z \leq 2.1818)\\\\= P(z \leq 2.1818) - P(z < -3.2727)\\= 0.9854 - 0.0005=98.49\%$

$P(x\leq59 \cup x\geq 74)=1 - P(59\leq x \leq 74) \\=1 - 0.9849\\=0.0151\\=1.51\%$

0.0151 is the probability  that adult men are at most 59 inches or at least 74 inches.