Question

in a survey of a group of men, the heights in the 20 - 29 age group were normally distributed, with a mean of 69.7 inches and a standard deviation of 4.0 inches. a study participant is randomly selected. complete parts (a) through (d) below.
(a) find the probability that a study participant has a height that is less than 68 inches.
the probability that the study participant selected at random is less than 68 inches tall is 0.3352 (round to four decimal places as needed.)
(b) find the probability that a study participant has a height that is between 68 and 72 inches.
the probability that the study participant selected at random is between 68 and 72 inches tall is 0.3825 (round to four decimal places as needed.)
(c) find the probability that a study participant has a height that is more than 72 inches.
the probability that the study participant selected at random is more than 72 inches tall is (round to four decimal places as needed.)

Explanation:

Step1: Recall the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. Here, $\mu = 69.7$ inches and $\sigma=4.0$ inches.

Step2: Calculate the z - score for $x = 72$ inches

$z_1=\frac{72 - 69.7}{4.0}=\frac{2.3}{4.0}=0.575$

Step3: Use the standard normal distribution table

We know that the total area under the standard - normal curve is 1. The probability that a value is less than 72 inches is $P(Z<0.575)$. From the standard - normal table, $P(Z < 0.575)\approx0.7177$.

Step4: Calculate the probability that $x>72$ inches

The probability that a value is more than 72 inches is $P(X > 72)=1 - P(X\leq72)$. Since $P(X\leq72)$ corresponds to $P(Z\leq0.575)\approx0.7177$, then $P(X > 72)=1 - 0.7177 = 0.2823$

Answer:

$0.2823$