Question

adult american males have heights that have a normal distribution with mean 70 inches and standard deviation 3 inches. what height separates the tallest 5% of people from the shortest 95%?
74.93 inches
65.06 inches
67.48 inches
60 inches

Explanation:

Step1: Find the z - score

We want to find the z - score corresponding to the 95th percentile. Looking up in the standard normal distribution table (z - table), the z - score $z$ such that $P(Z\leq z)=0.95$ is approximately $z = 1.645$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the original normal distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know $\mu = 70$, $\sigma=3$, and $z = 1.645$. Rearranging the formula for $x$ gives $x=\mu+z\sigma$.

Step3: Calculate the height

Substitute $\mu = 70$, $z = 1.645$, and $\sigma = 3$ into the formula $x=\mu+z\sigma$. So $x=70 + 1.645\times3=70+4.935 = 74.935\approx74.93$ inches.

Answer:

74.93 inches