Test completion times are normally distributed with a mean of 58 minutes and a standard deviation of 12 minutes. By what time have 90% of students finished?

Step-by-step solution

1. 1

   Recognise this as an inverse-normal problem. You are given a probability and asked for the value — the opposite of the usual "find the probability" direction. In symbols, find $x$ with

   $P(X \le x) = 0.90, \qquad X \sim N(\mu = 58,\ \sigma = 12)$

2. 2

   Look up the $z$-score for the 90th percentile. From a standard normal table or calculator,

   $z_{0.90} \approx 1.2816$

   (Many textbooks round this to $1.28$; the difference shifts the final answer by about $0.02$ minutes.)

3. 3

   Convert the $z$-score back to the original units. Rearranging $z = \frac{x-\mu}{\sigma}$ gives

   $x = \mu + z\sigma$

4. 4

   Substitute and evaluate.

   $x = 58 + 1.2816 \times 12 = 58 + 15.38 = 73.38 \text{ minutes}$

5. 5

   Sanity-check the direction. The 90th percentile must lie above the mean, and it does — about $1.28$ standard deviations above. If your answer comes out below $58$, you have used $-z$ by mistake.