ShopOnline advertises that 95% of its online orders ship within two .2/ working days. You select...
ShopOnline advertises that 95% of its online orders ship within two .2/ working days. You select a random sample of 100 orders received over the past month to audit. The audit reveals that 91 of these orders were shipped on time. Assuming that this is a normal distribution. What is the probability that at least 90% of the orders in the sample will be shipped on time?
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ShopOnline advertises that 95% of its online orders ship within
two .2/ working days. You select...
Cassandra's Cogs advertises on its website that 90% of customer orders are recived within three working...
Cassandra's Cogs advertises on its website that 90% of customer orders are recived within three working days. They perfromed an adult from a random sample of 100 of the 2,500 orders that month and it shows 85 orders were recieved on time. Part A: Can we use a normal approximation? Explain Part B: If Cassandra's Cogs customers recieve 90% of their orders within three working days, where is the probability that the proportion in the random sample of 100 orders...
8.1.5 Question Help Determine the 95% confidence interval estimate for the population mean of a normal...
8.1.5 Question Help Determine the 95% confidence interval estimate for the population mean of a normal distribution given n = 100, o = 133, and x = 1,500 The 95% confidence interval for the population mean is from to (Round to two decimal places as needed. Use ascending order.) 8.1.14-T Question Help As a follow-up to a report on gas consumption, a consumer group conducted a study of SUV owners to estimate the mean mileage for their vehicles. A simple...
An insurance company has business objectives of reducing the amount of time it takes to approve...
An insurance company has business objectives of reducing the amount of time it takes to approve applications for life insurance. During a period of one month, you collect a random sample of 27 approved policies and the total processing time, in days, stored in insurance are: 73 19 16 64 28 28 31 90 60 56 31 56 22 18 45 48 17 17 17 91 92 63 50 51 69 16 17 A. On the past, the mean processing...
you are working with a team of vascular surgeons who seeks to compare the 2. Assu...
you are working with a team of vascular surgeons who seeks to compare the 2. Assu me that patient outcome of ruptured aneurysm their patient that, on average, 6% of patients with he previous five year's data, you determine that 4 of the team's aneurysm patients ruptured out of 15 a. Calçulate and report the binomial distribution. in their clinic with national averages. You determine t demographics are similar to several similar studies in the literature who showed aneurysms rupture...
This course is actually Quantitative Methods and Analysis In Unit 2, you have learned about three...
This course is actually Quantitative Methods and Analysis In Unit 2, you have learned about three different types of distributions: Normal, binomial, and Poisson. You can take data that you collect and plot it out onto graphs to see a visual representation of the data. By simply looking at data on a graph, you can tell a lot about how related your observed data are and if they fit into a normal distribution. For this submission, you will be given...
During a period of one month, a random sample of 27 approved life insurance policies is...
During a period of one month, a random sample of 27 approved life insurance policies is selected, and the total processing time, in days, is recorded as shown in the data table below. Complete parts (a) through (d). In the past, the mean processing time was 46 days. At the 0.05 level of significance, is there evidence that the mean processing time has changed from 46 days? State the null and alternative hypotheses. Determine the test statistic. The test statistic...
2)According to a 2009 Reader's Digest article, people throw away about 12% of what they buy...
2)According to a 2009 Reader's Digest article, people throw away about 12% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 68 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.18? Note: You should keep at least 4 decimal places for any intermediate calculations you make to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate...
sbrino19 In a continuous probability distribution, the probability that z will ake on n exact value a is equal to the height of the curve at that value b. is calculated using the probability den...
sbrino19 In a continuous probability distribution, the probability that z will ake on n exact value a is equal to the height of the curve at that value b. is calculated using the probability density c. is always greater than 0. d. is always equal to O. e. None of these is correct. 2. Using the standard normal table, the total area between:--132ada.Ois a. 0.0494 b. 0.4656 c. 0.1554 d. 0.4162 e. 0.1005 3. The z-score representing the 10th percentile...
ask back to top Background You are a manager in the audit division at Miller Yates...
ask back to top Background You are a manager in the audit division at Miller Yates Howarth (MYH), an accounting firm with offices throughout the major regional centres of NSW and Queensland. Although a medium sized firm by national standards, MYH is the second largest regional accounting firm in Australia. Most of MYH’s audit clients are in the agriculture, mining, manufacturing and property industries. All those industries are currently under pressure, either from a downturn in commodity prices or fierce...
A. B. The 95% confidence interval is _____ <μd< _____. (Round to two decimal places as...
A.
B.
The 95% confidence interval is _____ <μd< _____.
(Round to two decimal places as needed.)
What does the confidence interval suggest about farmer Joe's
claim that type 1 seed is better than type 2 seed?
A. Because the confidence interval only includes positive values
and does not include zero, there
is not sufficient evidence to support farmer Joe's claim.
B. Because the confidence interval includes zero, there is
sufficient evidence to support farmer Joe's claim.
C. Because the...
8.1.5 Question Help Determine the 95% confidence interval estimate for the population mean of a normal distribution given n = 100, o = 133, and x = 1,500 The 95% confidence interval for the population mean is from to (Round to two decimal places as needed. Use ascending order.) 8.1.14-T Question Help As a follow-up to a report on gas consumption, a consumer group conducted a study of SUV owners to estimate the mean mileage for their vehicles. A simple...
you are working with a team of vascular surgeons who seeks to compare the 2. Assu me that patient outcome of ruptured aneurysm their patient that, on average, 6% of patients with he previous five year's data, you determine that 4 of the team's aneurysm patients ruptured out of 15 a. Calçulate and report the binomial distribution. in their clinic with national averages. You determine t demographics are similar to several similar studies in the literature who showed aneurysms rupture...
sbrino19 In a continuous probability distribution, the probability that z will ake on n exact value a is equal to the height of the curve at that value b. is calculated using the probability density c. is always greater than 0. d. is always equal to O. e. None of these is correct. 2. Using the standard normal table, the total area between:--132ada.Ois a. 0.0494 b. 0.4656 c. 0.1554 d. 0.4162 e. 0.1005 3. The z-score representing the 10th percentile...
A.
B.
The 95% confidence interval is _____ <μd< _____.
(Round to two decimal places as needed.)
What does the confidence interval suggest about farmer Joe's
claim that type 1 seed is better than type 2 seed?
A. Because the confidence interval only includes positive values
and does not include zero, there
is not sufficient evidence to support farmer Joe's claim.
B. Because the confidence interval includes zero, there is
sufficient evidence to support farmer Joe's claim.
C. Because the...
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