The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of 20 minutes.
What proportion of members spend less than 55 minutes at the gym? Use the standard normal distribution tables of your Wegner textbook. Give your answer to the nearest percentage, e.g. 54% or 28% (remember to include the % sign!!!).
(Based on Wegner, Exercise 19, p149, 3rd ed / Exercise 8.17, p208, 2nd ed)
The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of 20 minutes.
What proportion of members spend less than 55 minutes at the gym?
Z = (score-mean)/SD
Find table in the back of your statistics text (Wegner?) labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Multiply by 100.
75
To find the proportion of members who spend less than 55 minutes at the gym, we can use the standard normal distribution.
1. Start by calculating the z-score for 55 minutes using the formula:
z = (x - μ) / σ
where x is the value you want to find the proportion for, μ is the mean, and σ is the standard deviation.
z = (55 - 80) / 20
z = -1.25
2. Look up the z-score of -1.25 in the standard normal distribution table. In this case, a z-score of -1.25 corresponds to a cumulative probability of 0.1056.
3. The proportion of members spending less than 55 minutes at the gym is equal to the cumulative probability. Convert the probability to a percentage:
Proportion = 0.1056 * 100 = 10.56%
Therefore, approximately 10.56% of members spend less than 55 minutes at the gym.