# Standard deviation - use z-score to determine how the student placed comparing with the class average?

A student took a math exam and scored 77. If the class exam scores were mound-shaped with a mean score of 70 and standard deviation of 16, use z-score to determine how the student placed comparing with the class average i.e. is the student's mark an outlier or not?

So my attempt: I used this equation $$n = \mu + \sigma z$$

where

• $$n$$ is the data point (77)
• $$\mu$$ is the mean (70)
• $$\sigma$$ is the standard deviation (16)
• and $$z$$ is the z-score (unknown)

To solve for the unknown:

\begin{aligned} 77 &= 70 + 16z \\ 7 &= 16z \\ 0.4375 &= z \\ \end{aligned}

This value is equivalent to 0.6700 on the z table. I just don't know what to do with this information and what steps I would take next to figure out the question.

The $$Z$$-score is the number of standard deviations away from the mean. The data point is within one standard deviation so clearly it is not an outlier.
What the table tells you is that a $$Z$$-score of 0.4375 means that 67% of people would have scored less than him, or alternatively:
$$\mathrm{P}(Z\leq 0.4375) = \mathrm{P}(n\leq 77) = 0.67$$
Where $$\mathrm{P}(xyz)$$ is the probability that $$xyz$$ is true.