Question

adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. find the z-score of a man 70.2 inches tall. (to 2 decimal places)
this means that this mans height is standard deviations select an answer the mean.
please remember that although the z-score tells how far many standard deviations away from the mean a given data value is, z-scores have no units.
question help: message instructor
submit question jump to answer

Explanation:

Step1: Recall z - score formula

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the data point, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Here, $x = 70.2$, $\mu=69.0$, and $\sigma = 2.8$.

Step2: Substitute values into the formula

Substitute the given values into the formula: $z=\frac{70.2 - 69.0}{2.8}$
First, calculate the numerator: $70.2-69.0 = 1.2$
Then, divide by the standard deviation: $z=\frac{1.2}{2.8}\approx0.43$ (rounded to two decimal places)

Answer:

The z - score is approximately 0.43. This means that this man's height is 0.43 standard deviations above the mean.