Question

SECTION2- Sampling Distributions 6. The amount of time that you have to wait before seeing the...

SECTION2- Sampling Distributions

6. The amount of time that you have to wait before seeing the doctor in the doctor's office is normally distributed with a mean of 15.2 minutes and a standard deviation of 15.2 minutes. If you take a random sample of 35 patients, what is the probability that the average wait time is greater than 20 minutes? (Hint: Round the probability value to 2 decimal places.) Show your calculations.

  • A. 0.16
  • B. 0.09
  • C. 0.03
  • D. 0.28

7. A basketball player shoots 60 free throws a day and historically makes 75 percent of them. What is the probability that he will make at most 80 percent tomorrow? Show your calculations.

  • A. 0.1056
  • B. 0.1867
  • C. 0.8944
  • D. 0.8133

8. If a sample of size 100 is taken from a population whose standard deviation is equal to 100, then the standard error of the mean is equal to: Show your calculations.

  • A. 10
  • B. 10,000
  • C. 100
  • D. 1,000

SECTION 3 - Confidence 1

9. The following sample was taken from a normally distributed population: 20, 27, 15, 20, 16, 22, and 13. A 95% confidence interval for the population mean using this sample is: Show your calculations.

  • A. 19 ± 4.402
  • B. 19 ± 5.603
  • C. 19 ± 2.201
  • D. 19 ± 8.804

10. In a recent survey of 600 adults, 16.4% indicated that they had fallen asleep in front of the television in the past month. Which of the following intervals represents a 98% confidence interval for the population proportion? Show your calculations.

  • A. 0.137 to 0.192
  • B. 0.140 to 0.189
  • C. 0.129 to 0.199
  • D. 0.143 to 0.186

Homework Answers

Answer #1

P(z<Z) table :

section 2:

6.

SD of sample = SD / (n^0.5)

= 15.2 / 35^0.5

= 2.57

ANSWER : C. 0.03

(please UPVOTE)

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A hospital was concerned about reducing its wait time. A targeted wait time goal of 25...
A hospital was concerned about reducing its wait time. A targeted wait time goal of 25 minutes was set. After implementing an improvement framework and? process, a sample of 364 patients showed the mean wait time was 23.06 ?minutes, with a standard deviation of 16.4 minutes. Complete parts? (a) and? (b) below. a. If you test the null hypothesis at the 0.10 level of? significance, is there evidence that the population mean wait time is less than 25 ?minutes? State...
38. If a population is not normally distributed, the distribution of the sample means for a...
38. If a population is not normally distributed, the distribution of the sample means for a given sample size n will A. take the same shape as the population                        B. approach a normal distribution as n increases C. be positively skewed D. be negatively skewed E. none of the above 39. A random sample of 66 observations was taken from a large population. The population proportion is 12%. The probability that the sample proportion will be more than 17% is...
assume that the amount of time (x), in minutes that a person must wait for a...
assume that the amount of time (x), in minutes that a person must wait for a bus is uniformly distributed between 0 & 20 min. a) find the mathematical expression for the probability distribution and draw a diagram. assume that the waiting time is randomly selected from the above interval b) find the probability that a eprson wait elss than 15 min. c) find the probability that a person waits between 5-10 min. d) find the probability the waiting time...
A manager of a cafeteria wants to estimate the average time customers wait before being served....
A manager of a cafeteria wants to estimate the average time customers wait before being served. A sample of 51 customers has an average waiting time of 8.4 minutes with a standard deviation of 3.5 minutes. a) With 90% confidence, what can the manager conclude about the possible size of his error in using 8.4 minutes to estimate the true average waiting time? b) Find a 90% confidence interval for the true average customer waiting time. c) Repeat part (a)...
Queuing theory is the study waiting. Many banks are interested in the mean amount of time...
Queuing theory is the study waiting. Many banks are interested in the mean amount of time their customers spend waiting on line. Chase bank says their average wait time is 3 minutes with a population standard deviation of 1.25 minutes. To prove this the bank takes a random sample of 1250 people, and finds that their average time waiting on line is 3.73 minutes. A) construct a 99% confidence interval estimate for the population mean waiting time at Fleet. B)...
A) The amount of time it takes students to complete a quiz is uniformly distributed between...
A) The amount of time it takes students to complete a quiz is uniformly distributed between 35 to 65 minutes. One student is selected at random. Find the following events: a. The probability density function (pdf) of amount of time taken for the quiz. b. The probability that the student requires more than 50 minutes to complete the quiz.(1 mark) c. The probability that completed quiz time is between 45 and 55 minutes. (1 mark) d. The probability that the...
The amount of time a bank teller spends with each customer has a population​ mean, muμ​,...
The amount of time a bank teller spends with each customer has a population​ mean, muμ​, of 2.902.90 minutes and a standard​ deviation, sigmaσ​, of 0.500.50 minute. Complete parts​ (a) through​ (d). a. If you select a random sample of 1616 ​customers, what is the probability that the mean time spent per customer is at least 2.8 ​minutes? . 7881.7881 ​(Round to four decimal places as​ needed.)b. If you select a random sample of 1616 ​customers, there is an 84​%...
An on-going study requires you to make a determination on the average amount of time it...
An on-going study requires you to make a determination on the average amount of time it takes for athletes to complete running a 10k marathon. Data was taken from 50 college student-athletes. It was found that the average time to completion was 43.22 minutes and a standard deviation of 4.16 minutes. -- Show work where applicable A) Determine the observational unit B) Identify the sample and the size of the sample C) Using correct symbols, identify the observed statistic d)...
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 20 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 6 minutes. Complete parts​ (a) through​ (d) below. a. If you select a random sample of 16 ​sessions, what is the probability that the sample mean is between 19.5 and 20.5 ​minutes? ​(Round to three decimal places as​...
Suppose X1,...,X20 are a random sample of size 20 from N(100,100) population. What are the distributions...
Suppose X1,...,X20 are a random sample of size 20 from N(100,100) population. What are the distributions of the following quantities? (a) sample mean: X(bar) = (1/20) (X1 + . . . + X20); (b) a scaled sample variance: (19/100)S^2, where S^2 = (1/19) *SIGMA from i=1 to 20* (Xi - X(bar))^2; (c) standardized mean: (X(bar) - 100) / (10/sqrt(20)); (d) studentized mean: (X(bar) - 100) / (S/sqrt(20))
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT