Question

In a health club, research shows that on average, patrons spendan average of 42.5 minutes on the treadmill, with a standarddeviation of 4.8 minutes. It is assumed that this is a normallydistributed variable. Find the probability that randomly selectedindividual would spent between 30 and 40 minutes on thetreadmill.

Answer 1

Given that :

The avearge(mean) time patrons spend on the treadmill is

$\mu=42.5$

The standard deviation is

$\sigma=4.8$

We now need to calculate the Z- score for 30 minutes and 40 minutes

$\begin{gathered} \text{Let Z}_1\text{ represent the Z-score for 30 minutes} \\ Z_2\text{ represent the Z-score for 40 minutes} \end{gathered}$$\begin{gathered} Z_1=\frac{X-\mu}{\sigma} \\ Z_1=\frac{30-42.5}{4.8}=\frac{-12.5}{4.8} \\ Z_1=-2.604 \end{gathered}$

Similarly,

$\begin{gathered} Z_2=\frac{40-42.5}{4.8}=\frac{-2.5}{4.8} \\ Z_2=-0.521 \end{gathered}$

Therefore,

[tex]\begin{gathered} Pr(Z_1

Therefore, the probability that a randomly selected individual would spend between 30 minutes and 40 minutes on the treadmill is 0.2965 (to four decimal places)