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Question
adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. find the z-scor 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.
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.
Step2: Identify values
We are given that $\mu = 69.0$ inches, $\sigma=2.8$ inches, and $x = 70.2$ inches.
Step3: Substitute values into formula
Substitute $x = 70.2$, $\mu=69.0$, and $\sigma = 2.8$ 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 denominator: $z=\frac{1.2}{2.8}\approx0.43$ (rounded to two decimal places)
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The z - score is approximately 0.43. This means that this man's height is 0.43 standard deviations above the mean.