Question

At a certain school, the distribution of sleep (in hours) is normally distributed with a mean of 7.5 hours and a standard deviation of 0.75 hours. what amount of sleep is at the 97.5th percentile of this distribution?

Answer 1

To find the amount of sleep at the 97.5th percentile, use the z-score formula. The amount of sleep at the 97.5th percentile is approximately 9.09 hours.

Step-by-step explanation:

In order to find the amount of sleep at the 97.5th percentile, we can use the z-score formula. The z-score formula is given as: z = (x - μ) / σ, where z is the z-score, x is the value we want to find, μ is the mean, and σ is the standard deviation.

First, we need to find the z-score corresponding to the 97.5th percentile. Using a standard normal distribution table or a calculator, we find that the z-score is approximately 1.96.

Now, we can rearrange the z-score formula to solve for x: x = z * σ + μ.

Substituting the values, we have x = 1.96 * 0.75 + 7.5.

Evaluating this expression, we find that the amount of sleep at the 97.5th percentile is approximately 9.09 hours.