Question

In a health club, research shows that on average, patrons spend an average of 42.5 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 35 and 48 minutes on the treadmill.

Answer 1

Using the normal distribution, there is a 0.8155 = 81.55% probability that a randomly selected individual would spent between 35 and 48 minutes on the treadmill.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation for this problem are given, respectively, by:

$\mu = 42.5, \sigma = 4.8$

The probability that a randomly selected individual would spent between 35 and 48 minutes on the treadmill is the p-value of Z when X = 48 subtracted by the p-value of Z when X = 35, hence:

X = 48:

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

$Z = \frac{48 - 42.5}{4.8}$

Z = 1.15

Z = 1.15 has a p-value of 0.8749.

X = 35:

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

$Z = \frac{35 - 42.5}{4.8}$

Z = -1.56

Z = -1.56 has a p-value of 0.0594.

0.8749 - 0.0594 = 0.8155.

0.8155 = 81.55% probability that a randomly selected individual would spent between 35 and 48 minutes on the treadmill.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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