For the following question, round the associated z-score to the nearest hundredth (second decimal value). If test scores for an exam were normally distributed with a mean of 75 and a standard deviation of 5, find the probability that a randomly selected student scored less than 82. Please type your solution (as a percentage) in the text entry box provided.

To find the probability that a randomly selected student scored less than 82, we first need to find the z-score associated with a score of 82 using the z-score formula:

z = \frac{x - \mu}{\sigma}

where:
x = score (82),
\mu = mean (75),
\sigma = standard deviation (5).

Plugging in the values, we get:

z=\frac{82 - 75}{5}=\frac{7}{5}=1.40

So, the z-score for a score of 82 is 1.40.

Next, we look up the probability associated with a z-score of 1.4 using a standard normal distribution table or a calculator.

The probability of scoring less than 82 is the probability to the left of a z-score of 1.4. From the z-table, this probability is approximately 0.9192 or 91.92%.

Therefore, the probability that a randomly selected student scored less than 82 is approximately \boxed{91.92\%} .