Question

In: Statistics and Probability

A certain gym selected a random sample of 10 customers and monitored the number of times...

A certain gym selected a random sample of 10 customers and monitored the number of times each customer used the workout facility in a​ one-month period. The data are shown in the table below. The​ gym's managers are considering a promotion in which they reward frequent users with a small gift. They have decided that they will only give gifts to those customers whose number of visits in a​ one-month period is 1 standard deviation above the mean. Find the minimum number of visits required to receive a gift.

19,16,16,15,18,22,19,17,14,13

The minimum number of visits required to receive a gift is:

​(Type a whole​ number.)

Solutions

Expert Solution

Mean Calculation

Mean = 16.9

Standard Deviation Calculation

In this case, Mean = 16.9

Standard Deviation SD = 2.6854

As gifts will be given to customers whose number of visits in a​ one-month period is 1 standard deviation above the mean.

Minimum number of visits = Mean + SD = 16.9 + 2.6854 = 19.5854

As no. of visits cannot be in decimal places.

Minimum number of visits = 20

The minimum number of visits required to receive a gift is: 20


Related Solutions

A simple random sample of 12 reaction times of helicopter pilots is selected. The reaction times...
A simple random sample of 12 reaction times of helicopter pilots is selected. The reaction times have a normal distribution. The sample mean is 1.67 sec with a standard deviation of 0.16 sec. Construct a 90% confidence interval for the population standard deviation. 1. 0.11 < σ < 0.23 2. 0.01 < σ < 0.05 3. 0.12 < σ < 0.25 4. 0.13 < σ < 0.22
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Also, the sample is from a normal population. Note that σ is unknown. 21 22 22 17 21 17 23 20 20 24 9 22 16 21 22 21 The sample mean is 19.875 and the sample standard deviation is 3.65. Which of the following represents the 80 percent confidence interval for µ? Select one: a. [13.75, 25.25] b. [18.65, 21.10] c. [19.55, 20.425] d. [18.8,...
13. A television manufacturer takes a random sample of 30 customers and determines the times (after...
13. A television manufacturer takes a random sample of 30 customers and determines the times (after purchasing) until the televisions need servicing have a mean of 6 years and a standard deviation of 1.2 years. a) Determine the 95% confidence interval for the mean and explain whether it is reasonable for the company to claim the mean time until servicing is needed is 6.4 years. b) Repeat a), but use a 90% interval. c) Suppose instead a larger sample of...
1. A simple random sample of 50 customers is selected from an account receivable portfolio and...
1. A simple random sample of 50 customers is selected from an account receivable portfolio and the sample mean account balance is $1000. The population standard deviation σ is known to be $200. (12 marks) a. Construct a 95% confidence interval for the mean account balance of the population. b. What is the margin of error for estimating the mean account balance at the 95% confidence level? c. Construct a 95% confidence interval for the mean account balance of the...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 26 16...
A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 26 16 25 18 17 24 18 23 14 20 10 18 19 13 17 16 Click here for the Excel Data File Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population. (Round your standard deviation answer to 4 decimal places and t-value to 3 decimal places. Round your answers to 3 decimal places.)    The 90% confidence...
The annual incomes of a random sample of 10 individuals in a certain city are given...
The annual incomes of a random sample of 10 individuals in a certain city are given below. Construct at 95% confidence interval for the average yearly income of individuals in this city. 65100, 52400, 32000, 33300, 72200, 39700, 39400, 42800, 54300, 68200,
The grades of a sample of 10 applicants, selected at random from a large population, are...
The grades of a sample of 10 applicants, selected at random from a large population, are 71, 86, 75, 63, 92, 70, 81, 59, 80, and 90. Compute the sample variance. Can we infer at the 90% confidence that the population variance is significantly less than 100? (i.e. Perform Hypothesis testing)
A simple random sample of 30 households was selected from a particular neighborhood. The number of...
A simple random sample of 30 households was selected from a particular neighborhood. The number of cars for each household is shown below. Construct the 98% confidence interval estimate of the population mean. 2 0 1 2 3 2 1 0 1 4 1 3 2 0 1 1 2 3 1 2 1 0 0 5 0 2 2 1 0 2 A) 1.0 cars < μ < 2.0 cars B) 0.9 cars < μ < 2.1 cars C)...
The distribution of wait times for customers at a certain department of motor vehicles in a...
The distribution of wait times for customers at a certain department of motor vehicles in a large city is skewed to the right with mean 23 minutes and standard deviation 11 minutes. A random sample of 50 customer wait times will be selected. Let x¯W represent the sample mean wait time, in minutes. Which of the following is the best interpretation of P(x¯W>25)≈0.10 ? For a random sample of 50 customer wait times, the probability that the total wait time...
A random sample of 100 employees from 5 different companies was randomly selected, and the number...
A random sample of 100 employees from 5 different companies was randomly selected, and the number who take public transportation to work was recorded. The results are listed below. Perform a homogeneity of proportions test to test the claim that the proportion who take public transportation to work is the same in all 5 companies. Use α= 0.01. 1 2 3 4 5 Use Public Trans. 18 25 12 33 22 Don't Use Public Trans. 82 75 88 67 78...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT