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 30 and 40 minutes on the treadmill.


0,30
0.70
0.40
Less than 1%

Answer 1

0.30

Explanation:

To find the probability that a randomly selected individual would spend between 30 and 40 minutes on the treadmill, we need to calculate the z-scores corresponding to these values and then use the z-table or a statistical calculator to find the probability.

First, we calculate the z-scores using the formula:

z = (x - μ) / σ

where x is the value (in this case, 30 and 40), μ is the mean (42.5), and σ is the standard deviation (4.8).

For x = 30:

z = (30 - 42.5) / 4.8 ≈ -2.604

For x = 40:

z = (40 - 42.5) / 4.8 ≈ -0.521

Next, we look up the probabilities associated with these z-scores in the z-table or use a statistical calculator.

From the z-table or calculator, the probability corresponding to z = -2.604 is approximately 0.0047, and the probability corresponding to z = -0.521 is approximately 0.3015.

To find the probability between 30 and 40 minutes, we subtract the probability associated with z = -2.604 from the probability associated with z = -0.521:

P(30 ≤ x ≤ 40) = P(z = -0.521) - P(z = -2.604)

≈ 0.3015 - 0.0047

≈ 0.2968

Therefore, the probability that a randomly selected individual would spend between 30 and 40 minutes on the treadmill is approximately 0.2968, which is equivalent to 29.68%. Rounding up we will get 0.30.

Hope this helps!