Question

adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. find the height of a man with a z-score of 0.6429 (to 4 decimal places)

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $z$ is the z - score, $x$ is the value we want to find (the height), $\mu$ is the mean, and $\sigma$ is the standard deviation. We need to solve this formula for $x$. Rearranging the formula gives $x = \mu+z\times\sigma$.

Step2: Identify the values

We know that $\mu = 69.0$ inches, $\sigma=2.8$ inches, and $z = 0.6429$.

Step3: Substitute the values into the formula

Substitute $\mu = 69.0$, $\sigma = 2.8$, and $z=0.6429$ into the formula $x=\mu + z\times\sigma$. So $x=69.0+0.6429\times2.8$.
First, calculate $0.6429\times2.8$: $0.6429\times2.8 = 1.79912$.
Then, add this to the mean: $x=69.0 + 1.79912=70.79912$.

Step4: Round to 4 decimal places

Rounding $70.79912$ to 4 decimal places gives $70.7991$.

Answer:

$70.7991$