Chapter 06, Section 6.4, Problem 044A machine at Keats Corporation fills 64-ounce detergent jugs. The machine can be adjusted to pour, on average, any amount of detergent into these jugs. However, the machine does not pourexactly the same amount of detergent into each jug; it varies from jug to jug. It is known that the net amount of detergent poured into each jug has a normal distribution with a standarddeviation of 0.33 ounce. The quality control inspector wants to adjust the machine such that at least 95% of the jugs have more than 64 ounces of detergent. What should the mean amount ofdetergent poured by this machine into these jugs be?Round your answer to 2 decimal places.ounces

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Answered: Chapter 06, Section 6.4, Problem 044 A…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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b) What is the probability that is less than 2.7 grams? 0.3085 0.6915 0.9982 0.0018
Question 6 A manufacturer produces, packages, and sells packs of their product targeted to weigh 50 grams. A quality control manager working for the company was concerned that the variation in the actual weights of the targeted 50-gram packs was larger than acceptable. That is, he was concerned that some packs weighed significantly less than 50-grams and some weighed significantly more than 50 grams. In an attempt to estimate the standard deviation of the weights of all of the 50-gram packs the manufacturer makes, he took a random sample of 21 packs off of the factory line. The random sample yielded the sample mean 49.86 grams and the standard deviation of 0.655 grams. Use the random sample to derive a 99% confidence interval for the variance (standard deviation) of the actual weights of the packs. Note: if the normality plot is not provided you may assume that the actual weights of the packs are normally distributed. i. Procedure: Select an answer Select an answer One mean Z procedure…
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…
Downloadable eTextbook ment FULL SCREEN PRINTER VERSION 1 BACK NEXT Chapter 09, Section 9.2, Problem 025a The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 24 such batteries and found that the mean life for this sample is 43.35 months. The lives of all such batteries have a normal distribution with the population standard deviation of 4.5 months. Find the p-value for the test of hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months. Round your answer to four decimal places. Will you reject the null hypothesis at a = 0.025? the tolerance is +/-2% Question Attempts: 0 of 2 used SAVE FOR LATER SUBMIT ANSWER powered by MapleNet Policy 2000-2020 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons, Inc. Version 4.24.20.1 tv hulu W
Question A - A company producing steel nails receives an order for 5000 nails of length 50 mm. Due to externalfactors such as temperature and pressure, the lengths of the nails produced by the machine is normallydistributed with a standard deviation of 1.6 mm. Acceptable lengths (lengths that any customer willconsent to buy) must have a tolerance of 2 mm. If the machine is set to produce nails with a meanlength of 5.0 cm, determine the percentage of nails that are acceptable. A. 0.7888 / B. 0.8877 / C. 0.9921 / D. 0.7878 Further to the information provided in question A above, we are informed that nails produced thatare shorter than the minimum acceptable length have to be scrapped. Consider that 4.8cm is still theminimum threshold for acceptable nail length. If the cost to produce one nail is 45 cents, how muchdoes the company incur in Rand losses due to scrapping? A. R247.70 / B. R347.91 / C. R237.60 / D. R471.99 Further considering information given in questions A and question B, if…
Given the mean and standard deviation of data set Y as 20 and 5, how can I find the mean and standard deviation of Y + 6? Question refers to chapter 16 problem 25 of Stats, Modeling the World, AP Edition, part B. Thanks so much!!
Chapter 06, Section 6.3, Problem 035 The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6.0 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.79 ounces and a standard deviation of 0.14 ounces. What percentage of the pucks produced this factory cannot be used by the National Hockey League? Round your answer to two decimal places. The percentage of pucks that cannot be used is %.
Section 9.2 Question #7 Listed in the data table are IQ scores for a random sample of subjects with medium lead levels in their blood. Also listed are statistics from a study done of IQ scores for a random sample of subjects with high lead levels. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. View the data table of IQ scores. Medium Lead Level High Lead Level 72 n2 = 11 96 92 _ x2 = 91.815 85 90 s2 = 10.365 97 83 92 104 111 91 a. Use a 0.05 significance level to test the claim that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead levels. What are the null and alternative hypotheses? Assume that population 1 consists of subjects with medium lead levels and population…
Section 9.2 Question #8 Listed below are time intervals (min) between eruptions of a geyser. Assume that the "recent" times are within the past few years, the "past" times are from around 20 years ago, and that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Does it appear that the mean time interval has changed? Is the conclusion affected by whether the significance level is 0.10 or 0.01? Recent 77 91 90 78 57 100 61 87 69 89 83 83 55 81 73 101 61 Past 90 89 93 94 64 85 85 92 88 91 89 91 Let μ1 be the recent times and let μ2 be the past times. What are the null and alternative hypotheses? A. H0: μ1 = μ2 H1: μ1 ≠ μ2 B. H0: μ1 < μ2 H1: μ1 = μ2 C. H0: μ1 = μ2 H1: μ1 > μ2 D. H0: μ1 ≠ μ2…
A certain virus affects 0.7% of the population. A test used to detect the virus in a person is positive 87% of the time if the person has the virus (true positive) and 14% of the time if the person does not have the virus (false positive). Fill out the remainder of the following table and use it to answer the two questions below based on a total sample of 100,000 people. Virus No Virus TotalPositive Test Negative Test Total 100,000a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest hundredth of a percent and do not include a percent sign. % b) Find the probability that a person does not have the virus given that they test negative. Round your answer to the nearest hundredth of a percent and do not include a percent sign. %
This cumulative review problem uses material from Chapters 3, 5, and 10. Recall that the Poisson distribution deals with rare events. Death from the kick of a horse is a rare event, even in the Prussian army. The following data are a classic example of a Poisson application to rare events. The data represent the number of deaths from the kick of a horse per army corps per year for 10 Prussian army corps over a period of time. Let x represent the number of deaths and f the frequency of x deaths. x 0 1 2 3 or more f 111 61 24 4 (a) First, we fit the data to a Poisson distribution.The Poission distribution states P(x) = e−λλx x! , where λ ≈ x (sample mean of x values). From our study of weighted averages, we get the following. x = Σxf Σf Verify that x = 0.605. Hint: For the category 3 or more, use 3. x = (b) Now we have P(x) = e−0.605(0.605)x x! for x = 0, 1, 2, 3 . Find P(0), P(1), P(2), and P(3 ≤ x). Round to three places…

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