Question

In a health club, research shows that on average, patrons spend an average of 46.2 minutes
on the treadmill, with a standard deviation of 4.8 minutes. It is assumed that this is a normally
distributed variable. Find the probability that randomly selected individual would spent
between 30 and 40 minutes on the treadmill.

Answer 1

As the problem states, the variable has a normal distribution, and thus, concepts of this distribution are used to solve this question.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In a health club, research shows that on average, patrons spend an average of 46.2 minutes  on the treadmill, with a standard deviation of 4.8 minutes.

This means that $\mu = 46.2, \sigma = 4.8$

Find the probability that randomly selected individual would spent  between 30 and 40 minutes on the treadmill.

This is the p-value of Z when X = 40 subtracted by the p-value of Z when X = 30.

X = 40

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 46.2}{4.8}$

$Z = -1.29$

$Z = -1.29$ has a p-value of 0.0985

X = 30

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30 - 46.2}{4.8}$

$Z = -3.38$

$Z = -3.38$ has a p-value of 0.0004.

0.0985 - 0.0004 = 0.0981

Thus, 0.0981 = 9.81% probability that randomly selected individual would spent  between 30 and 40 minutes on the treadmill.

For another example of the normal distribution, you can check https://brainly.com/question/15181104.