A production manager would like to know the precision of the operator producing the product. The product thickness is assumed to be normally distributed, with standard deviation of 2 mm. How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from the sample mean by 0.25 mm.
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- Suppose that the hemoglobin levels among healthy females are normally distributed with a mean of 14. Research shows exactly 95% of healthy females have hemoglobin levels below 15.9. What is the standard deviation of the distribution of hemoglobin levels in healthy females?carry computations to four decimal places and round your answer to two.Mason earned a score of 226 on Exam A that had a mean of 250 and a standard deviation of 40. He is about to take Exam B that has a mean of 550 and a standard deviation of 25. How well must Mason score on Exam B in order to do equivalently well as he did on Exam A? Assume that scores on each exam are normally distributed.Sam and Peter are classmates. In an examination, Sam gets 75 marks and Peter gets 89 marks. It is known that the standard normal score of Sam's marks is 0, z(Sam) = 0 and the standard normal score of Peter's marks is 1.75, z(Peter) = 1.75. Find the standard deviation of the examination marks.
- A notable indicator of a baby's health is the weight gained in the first year of the baby's life. Assume that the population of all such weight gains for baby boys is approximately normally distributed. A study claimed that the mean of this population is 4.93 kg. As a practicing pediatrician, you want to test this claim. So, you select a random sample of 16 baby boys, and you record the weight each gained in their first year. Follow the steps below to construct a 99% confidence interval for the population mean of all the weight gains for baby boys in their first year. Then state whether the confidence interval you construct contradicts the study's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Sample standard Number of baby boys Sample mean deviation Take Sample 16 5.496 1.873 Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for…For data that is not normally distributed we can't use z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula is p = 1 - 7.2 where p is the minimum percentage of scores that fall within k standard deviations on both sides of the mean. Use this formula to answer the following questions. a) If you have scores and you don't know if they are normally distributed, find the minimum percentage of scores that fall within 2.1 standard deviations on both sides of the mean? b) If you have scores that are normally distributed, find the percentage of scores that fall within 2.1 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 96 percent of the scores? Note: To answer part c you will need to solve the equation for k.A fitness center bought a new exercise machine called the Mountain Climber. They decided to keep track of how many people used the machine over a 3-hour period. Here X is the number of people who used the machine. X 0 1 2 3 4 PX 0.2 0.2 0.1 0.4 0.1 Find the mean,variance and standard deviation
- "Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for two types of treated fabrics: Permafresh 12 13 15 11 10 16 14 Hylite 14 19 20 16 A manufacturer wants to know how large is the difference in mean wrinkle recovery angle. Give a 98% confidence interval for the difference in mean wrinkle recovery angle: [three decimal accuracy] [three decimal accuracy]To compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 42 feet. Assume the population standard deviation is 4.7 feet. The mean braking distance for Make B is 45 feet. Assume the population standard deviation is 4.4 feet. At a = 0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) rari rz (b) Find the critical value(s) and identify the rejection region(s). The critical value(s) is/are (Round to three decimal places as needed. Use a comma to separate answers as needed.)Please show the complete solution
- Solve the problem. The systolic blood pressures of the patients at a hospital are normally distributed with a mean of 138 mmHg and a standard deviation of 12.8 mmHg. Find the two blood pressures having these properties: the mean is midway between them and 90 % of all blood pressures are between them. (Hint: the area between the two values is 0.9. Since they are symmetrical on either side of the mean, that leaves an area of 0.1 divided by 2 = 0.05 in the tails of the distribution.) ο οο 126.5 mm Hg, 149.5 mm Hg 116.9 mm Hg, 159.1 mm Hg 117.9 mm Hg, 160.1 mm Hg O.121.6 mm Hg, 154.4 mm HgSolve this for me. Thank you.If we meet these conditions, the sampling distribution of the mean will have a normal shape and ... The mean of the sampling distribution will be u, (i.e., the same as the population mean) and the standard deviation of the sampling distribution will be (that's the population standard deviation divided by the square root of the sample size). In the AP Stats Guy video, he talks about the number of text messages his students send during class. Suppose the average number of text messages his students send during class is u = 30 text messages. If we take samples of say, n = 36 students at a time, we would expect the mean of the sampling distribution we create to be the same as the population mean, 30 text messages. If we can further say that standard deviation of the number of text messages is 12 text messages, by how much would we expect the sample means to vary? (hint, use the formula above) text messages
