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Question
the mean of a distribution is 16.2 with a standard deviation of 2.5. which value has the most extreme z - score? 16.2 24 9 10
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 data - point. Given $\mu = 16.2$ and $\sigma=2.5$.
Step2: Calculate z - score for $x = 16.2$
$z_1=\frac{16.2 - 16.2}{2.5}=\frac{0}{2.5}=0$
Step3: Calculate z - score for $x = 24$
$z_2=\frac{24 - 16.2}{2.5}=\frac{7.8}{2.5}=3.12$
Step4: Calculate z - score for $x = 9$
$z_3=\frac{9 - 16.2}{2.5}=\frac{-7.2}{2.5}=-2.88$
Step5: Calculate z - score for $x = 10$
$z_4=\frac{10 - 16.2}{2.5}=\frac{-6.2}{2.5}=-2.48$
Step6: Compare the absolute values of z - scores
$|z_1| = 0$, $|z_2|=3.12$, $|z_3| = 2.88$, $|z_4| = 2.48$. Since $3.12>2.88>2.48>0$, the value with the most extreme z - score is the one with the largest absolute z - score.
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