Question

Owners of an exercise gym believe that a Normal model is useful in projecting the number of clients who will exercise in their gym each

week. They use a mean of 800 clients and a standard deviation of 90 clients.
The owner believes they need at least 700 clients per week to be profitable. What percent of weeks does this model project the gym to
be profitable?
87%
16%
13%
84%

Answer 1

The percentage of weeks the exercise gym will be profitable, based on the Normal distribution model with a mean of 800 clients and a standard deviation of 90 clients, is approximately 87%. Therefore correct option is A

Step-by-step explanation:

The student's question is related to using a Normal distribution to determine the percentage of time an exercise gym will be profitable based on weekly client attendance. The gym is projected to be profitable when there are at least 700 clients per week.

Given a mean (μ) of 800 clients and a standard deviation (σ) of 90 clients, we need to find the probability that the number of clients will be greater than or equal to 700. To find this, we calculate the z-score for 700 clients.

Z = (X - μ) / σ

Z = (700 - 800) / 90

Z = -100 / 90

Z = -1.11 (approximately)

Using the standard Normal distribution table or calculator, we can find the area to the right of the z-score of -1.11, which will give us the probability that the gym will have 700 or more clients. This area corresponds to the percentage of weeks the gym will be profitable.

Since the Normal distribution is symmetric about the mean, the area to the left of z = 1.11 will be the same as the area to the right of z = -1.11.

Consulting a standard Normal distribution table or using a calculator with the z-score of 1.11 will give us approximately 0.8665 or 86.65%, which means the percentage of weeks the gym will be profitable is about 87%.

Answer 2

87%

Step-by-step explanation:

The number of clients who will exercise at a gym each week (X) is normally distributed with a mean (μ) of 800 clients and a standard deviation (σ) of 90 clients:

`\rm X \sim N(\mu,\sigma^2)\implies \boxed{\rm X \sim N(800,90^2)}`

The owner believes they need at least 700 clients per week to be profitable. Therefore, to determine the percentage of weeks that the gym is projected to be profitable, we need to find P(X ≥ 700).

Calculator input for "normal cumulative distribution function (cdf)":

• Lower bound: x = 700
• Upper bound: x = 1500
• σ = 90
• μ = 800

This gives the probability as:

`\rm P(X \geq 700)=0.8667397371`

Therefore, the percent of weeks that the model projects that the gym will be profitable is 87% (rounded to the nearest percent).