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=(X-\mu)/(\sigma) \\ Z_1=(30-42.5)/(4.8)=(-12.5)/(4.8) \\ Z_1=-2.604 \end{gathered}`

Similarly,

`\begin{gathered} Z_2=(40-42.5)/(4.8)=(-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)