The following questions must be answered with appropiately** drawn bell-shaped curves, and labeled and shaded respectively.Waiting times to see a doctor at a clinic are normally distributed with a mean of 15 minutes and a standard deviation of 7.4 minutes. a) Find the probability that a randomly selected patient had to wait more than 30 minutes.**______________--b) If 2500 patients are randomly selected, about how many of them do you expect to have to wait less than 30 minutes?, (no bell-curve is needed for this part).______________**The amounts that Statistics students study a week are normally distributed with a mean of 7 hours, and a standard deviation of 2.46 hours. What percentage of Statistics students study between 5 and 10 hours a week?___________________**When you apply for a job at the ABC Company, you must take a basic skills test. The applicants with a score in the top 1% will be given an interview. The population of test scores are normally distributed with a mean of 83.2 and a standard deviation of 4.25. What is the minimum score needed for an applicant to get an interview?____________In referring to the last question just answered, What score on the skills test is at the third quartile?**____________________A random sample of 12 American women revealed that the number of pairs of shoes they own are as follows: 11, 10, 7, 12,13, 9, 11, 8,14,22,20, 15. Assume a normal population.a) What do you estimate the mean number of pairs of shoes owned by all American women to be?_____________b) Construct a 95% confidence interval for the mean number of a pair of shoes owned by all American women.____________< M < _____________c) Interpret the interval. ______________________________________________________________________________________________________________________________________________ A random sample of 85 adults reveals that the mean age of their oldest living grandparent is 82.4 years. Assuming that o-- (standard deviation) = 5.23 years, construct a 95% confidence interval for the mean age of the oldest living grandparent._____________< M < _____________Fill in each blank with the correct symbol.a) The point estimate for M is __________ b) ___________is the point estimate for o--(standard deviation).Given the interval 54 < M < 98, find the following:a) _ __________ b) E___________ c) width of interval_____________ x
Already Tried: So far it's been kahn academy, and my math textbook, and about 7 various books of statistical reference books.All the books looked good when I viewed them in the store. But when I got them home (3 were bought off Amazon), they were all vague with sensibly explained nononsense approach, as if they were written for an engineer or economist studying statistics.
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Here are the answers ...
(1)
(a) 0.0213
(b) 2447
(2) 68.06%
(3)
(a) 93.09
(b) 86.07
(4)
(a) 12.67
(b) 9.77 < M < 15.57
(c) If samples of 12 women are drawn repataedly from the population and confidence intervals constructed, such intervals will contain the true mean 95% of the time.
(5)
81.29 < M < 83.51
(a) 82.4
(b) 5.23
(a) 76
(b) 22
(c) 44
I hope that helps. Let me know if you need any clarifications on my answers. I will be happy to provide them. Else, please give a rating for my answer.
Experience: Excellent Tutor with a long teaching experience at different levels.
answers: a.) Find P(x > 30) z = (x - mu) / sd = (30 - 15) / 7.4 = 2.0270 then P( x > 30) = P( z > 2.0270) Using table of normal distribution: P( z > 2.0270) = 0.0213 or 2.13% b.) P( z < 2.0270) = 1 - 0.0213 = 0.9787 then the number of those who will wait less than 30 mins is = 2500*0.9787 = 2446.75 or approx 2447 **The amounts that Statistics students study a week are normally distributed with a mean of 7 hours, and a standard deviation of 2.46 hours. What percentage of Statistics students study between 5 and 10 hours a week?___________________ Find P( 5 < x < 10) z5 = (x - mu) / sd = (5 - 7) / 2.46 = -0.8130 and z10 = (x - mu) / sd = (10 - 7) / 2.46 = 1.2195 then P(5 < x < 10) = P(-0.8130< z < 1.2195) Using table of normal distribution: P(-0.8130< z 1.2195) =0.6806 Using table of normal distribution: Z = 2.3263 then using raw score formula (from z score) raw score = mean + sd * z score = 83.2 + 4.25*2.3263 = 93.087 or approx 93 In referring to the last question just answered, What score on the skills test is at the third quartile?**____________________ Find Z such that P(z > Z) 0.25 Using table of normal distribution: Z = 0.6745 then using raw score formula (from z score) raw score = mean + sd * z score = 83.2 + 4.25*0.6745 = 86.067 or approx 86 A random sample of 12 American women revealed that the number of pairs of shoes they own are as follows: 11, 10, 7, 12,13, 9, 11, 8,14,22,20, 15. Assume a normal population. a) What do you estimate the mean number of pairs of shoes owned by all American women to be?_____________ answer: estimate = sample mean = 12.6667 b) Construct a 95% confidence interval for the mean number of a pair of shoes owned by all American women. ____________< M < _____________ 95% CI for mean = sample mean +/- T * [s / sqrt(n)] = 12.6667 +/-2.201*[4.5593726 / sqrt(12)] = 12.6667 +/- 2.8969 = 9.7698 < M < 15.5636 c) Interpret the interval. answer: there is 95% chance that the true mean number of pairs of shoes owned by all American women to be within the interval 9.7698 < M < 15.5636 A random sample of 85 adults reveals that the mean age of their oldest living grandparent is 82.4 years. Assuming that o-- (standard deviation) = 5.23 years, construct a 95% confidence interval for the mean age of the oldest living grandparent. _____________< M < _____________ 95% CI for mean = sample mean +/- Z * [o- / sqrt(n)] = 82.4 +/- 1.96*[5.23 / sqrt(85)] = 82.4 +/- 1.1118 = 81.29 < M < 83.51 Fill in each blank with the correct symbol. _ a) The point estimate for M is __x________ b) _____s______is the point estimate for o-- (standard deviation). Given the interval 54 < M < 98, find the following: a) _ __________ b) E___________ c) width of interval_____________ x answers: a.) mean = (98 + 54) / 2 = 76 b.) error = 98 - 76 = 22 c.) width of the interval = 2*22 = 44TRIGEOSCAL41079.8441467593
answers: a.) Find P(x > 30) z = (x - mu) / sd = (30 - 15) / 7.4 = 2.0270 then P( x > 30) = P( z > 2.0270) Using table of normal distribution: P( z > 2.0270) = 0.0213 or 2.13% b.) P( z < 2.0270) = 1 - 0.0213 = 0.9787 then the number of those who will wait less than 30 mins is = 2500*0.9787 = 2446.75 or approx 2447 **The amounts that Statistics students study a week are normally distributed with a mean of 7 hours, and a standard deviation of 2.46 hours. What percentage of Statistics students study between 5 and 10 hours a week?___________________ answerFind P( 5 < x < 10) z5 = (x - mu) / sd = (5 - 7) / 2.46 = -0.8130 and z10 = (x - mu) / sd = (10 - 7) / 2.46 = 1.2195 then P(5 < x < 10) = P(-0.8130< z < 1.2195) Using table of normal distribution: P(-0.8130 Z) = 0.01Using table of normal distribution: Z = 2.3263 then using raw score formula (from z score) raw score = mean + sd * z score = 83.2 + 4.25*2.3263 = 93.087 or approx 93 In referring to the last question just answered, What score on the skills test is at the third quartile?**____________________ Find Z such that P(z > Z)= 0.25 Using table of normal distribution: Z = 0.6745 then using raw score formula (from z score) raw score = mean + sd * z score = 83.2 + 4.25*0.6745 = 86.067 or approx 86 A random sample of 12 American women revealed that the number of pairs of shoes they own are as follows: 11, 10, 7, 12,13, 9, 11, 8,14,22,20, 15. Assume a normal population. a) What do you estimate the mean number of pairs of shoes owned by all American women to be?_____________ answer: estimate = sample mean = 12.6667 b) Construct a 95% confidence interval for the mean number of a pair of shoes owned by all American women. ____________< M < _____________ 95% CI for mean = sample mean +/- T * [s / sqrt(n)] = 12.6667 +/-2.201*[4.5593726 / sqrt(12)] = 12.6667 +/- 2.8969 = 9.7698 < M < 15.5636 c) Interpret the interval. answer: there is 95% chance that the true mean number of pairs of shoes owned by all American women to be within the interval 9.7698 < M < 15.5636 A random sample of 85 adults reveals that the mean age of their oldest living grandparent is 82.4 years. Assuming that o-- (standard deviation) = 5.23 years, construct a 95% confidence interval for the mean age of the oldest living grandparent. _____________< M < _____________ Answer95% CI for mean = sample mean +/- Z * [o- / sqrt(n)] = 82.4 +/- 1.96*[5.23 / sqrt(85)] = 82.4 +/- 1.1118 = 81.29 < M < 83.51 Fill in each blank with the correct symbol. _ a) The point estimate for M is __x________ b) _____s______is the point estimate for o-- (standard deviation). Given the interval 54 < M < 98, find the following: a) _ __________ b) E___________ c) width of interval_____________ x answers: a.) mean = (98 + 54) / 2 = 76 b.) error = 98 - 76 = 22 c.) width of the interval = 2*22 = 44
Hi
I had already posted the correct answers to you and the other expert has over-posted. Please review my answers and let me know if you have any queries. Else, please rate my answer.
Guru2009.
I've already left feedback, and the customer care line told me you would receive the credit for assisting me.
Yes, I have received.Thanks.
When it comes to the bonus to leave the experts, will I find a place at the answers page to do that? I want to leave you a bonus which is reflective on this time plus the last time. You have been so good at helping me with the difficulties of Statistical problems. Really, I am a reader and I seek out books when I don't understand the full process of how things work. But some of the books I have look very good in the bookstores when I view them. However, when I get home, sometimes I realize my purchase of books may not have been the best help (per an example, my professor "literally" told me, "You can get rid of that book (Principles of Statistics), because it goes into Calculus, and it's like a theory book, rather than a book showing you how to do problems. You don't need any of the advanced stuff offered in that book!" I responded to him, "It kind of gives new meaning to Deep Water Horizon!" You remember that was the tanker that spilled oil in the Gulf Coastal region.
I can always help you with your Statistics problems.There should be a button which you can click to add a bonus. I am not sure where it will be, but it should be on the Ratings page.By the way, I have asked you for a clarification on one of the questions in your latest post. Please see that posting and provide the clarification there.Thanks.
Please check your e-mail for the Q6 data I sent 10 minutes ago.
We can't exchange emails on this site. Please post the data in that posting itself as a reply.Thanks.
Data for Q6 is as follows:
x(age) / 18 24 36 40 58
________________________________
y(TV hours)3.9 2.6 2.0 2.7 1.8
Ok, thanks. I will be posting the solutions in that posting,