Question

In: Statistics and Probability

A local gym has found that each member, on average, spends 73 minutes at the gym...

A local gym has found that each member, on average, spends 73 minutes at the gym per week, with a standard deviation of 11 minutes. Assume the amount of time that members spend at the gym has a normal distribution.

a. What is the probability that a randomly selected customer spends less than 85 minutes at the gym?

b. Suppose the gym surveys a random sample of 25 members about the amount of time they spend at the gym each week. What are the expected value and standard deviation (standard error) of the sample mean of the time spent at the gym?

c. If 25 members are randomly selected, what is the probability that the average time spent at the gym exceeds 75 minutes?

Solutions

Expert Solution

hii... Although I am trying to provide the detailed answer but if you have any doubt please ask by comment. your rating encourage us to provide the detailed and effective answers so please like the answer. thanks..


Related Solutions

Suppose that a typical customer at a local restaurant spends 20 minutes in the store. It...
Suppose that a typical customer at a local restaurant spends 20 minutes in the store. It is estimated that the average number of customers in the restaurant is 4. Assume that the service time in this restaurant is random with some known distribution. How many customers does the restaurant serve during a typical hour that it is open? Suppose the restaurant wants to decrease the average number of customers in the restaurant at any time. However, assume it cannot control...
The average time a subscriber of a local newspaper The News spends reading it is 24...
The average time a subscriber of a local newspaper The News spends reading it is 24 minutes. Assume that the times are normally distributed and the standard deviation is 10 minutes. (a) Compute the probability that a subscriber spends more than 30 minutes reading the paper. (b) Find the range of reading times for the middle 90% of the subscribers. [Hint: First, find zc such that P(−zc < z < zc) = 0.90.] (c) The average time was calculated from...
A local business promotes oil changes that last on the average of 45 minutes with a...
A local business promotes oil changes that last on the average of 45 minutes with a standard deviation of 10 minutes.  If the process of changing oil follows a normal distribution, answer the following: What is the probability an oil change takes less than 40 minutes? What is the probability that the oil change takes between 35 and 55 minutes? What is the probability that an oil change takes less than 49 minutes. What is the probability that an oil change...
James has a $100 monthly exercise budget, which he spends at Tarheel Gym on yoga classes...
James has a $100 monthly exercise budget, which he spends at Tarheel Gym on yoga classes (Y) and spin classes (S). The price of yoga class is $10, and the price of a spin class is $5. Draw his budget constraint for each of the following scenarios. Use a separate graph for each part and label all axes, intercepts and the coordinates of any kink points. Treat each case separately and graph yoga classes on the x-axis and spin classes...
A survey indicates that for each trip to the supermarket, a shopper spends an average of...
A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the Z scores for the 2 Variables 35 and 65. Z 65 =    Z 35 = Probability that the shopper will be in the store for between...
A local pizza place claims that they average a delivery time of 7.32 minutes. To test...
A local pizza place claims that they average a delivery time of 7.32 minutes. To test this claim, you order 11 pizzas over the next month at random times on random days of the week. You calculate the average delivery time and sample standard deviation from the 11 delivery times (minutes), and with the sample mean and sample standard deviation of the time (minutes), you create a 95% confidence interval of (7.648, 9.992). (delivery time is normally distributed). What is...
6. Last year it was found that on average it took students 20 minutes to fill...
6. Last year it was found that on average it took students 20 minutes to fill out the forms required for graduation. This year the department has changed the form and asked graduating student to report how much time it took them to complete the forms. Of the students 22 replied with their time, the average time that they reported was 18.5 minutes, and the sample standard deviation was 5.2. Can we conclude that the new forms take less time...
pls give the process of function!thank you. 1.​Revenue Canada spends, on the average, 30 minutes per...
pls give the process of function!thank you. 1.​Revenue Canada spends, on the average, 30 minutes per income tax return it decides to audit. These audit times are known to be normally distributed with a standard deviation of 10 minutes. a.​Suppose a tax return is selected at random for audit. What is the probability that it will take less than 24 minutes to complete the audit? b.​In less than how many minutes that 90% of all these returns can each be...
A survey indicates that for each trip to the Target store, a shopper spends an average...
A survey indicates that for each trip to the Target store, a shopper spends an average of 25 minutes with a standard deviation of 14 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 25 and 50 minutes. Probability that the shopper will be in the store for between...
A telco, using a sample of 85 accounts, found out that the average talk minutes per...
A telco, using a sample of 85 accounts, found out that the average talk minutes per home user every month is 510 minutes. Lets assume that the standard deviation is 46 minutes. Create a 90%, 95% και 99% confidence interval for the average talk minutes.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT