Math 225N Week 5 Assignment_2020 {Graded A} | Central Limit Theorem for Means – Chamberlain College of Nursing

Math 225N Week 5 Assignment (2020) : Central Limit Theorem for Means – Chamberlain College of Nursing 1. Question A family of statisticians is trying to decide if they can afford for their child to play youth basebal ...

Math 225N Week 5 Assignment (2020) : Central Limit Theorem for Means – Chamberlain College of Nursing 1. Question A family of statisticians is trying to decide if they can afford for their child to play youth baseball. The cost of joining a team is normally distributed with a mean of $750 and a standard deviation of $185 . If a sample of 40 teams is selected at random from the population, select the expected mean and standard deviation of the sampling distribution below. Correct answer: σx¯=$29.25 μx¯=$750 The standard deviation of the sampling distribution σx¯=σn−−√=$18540−−√=$29.25 When the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=$750 . Question A cupcake baker is planning a supplies order and needs to know how much flour he needs. He knows that his recipes use an average of 100 grams of flour, normally distributed, with a population standard deviation of 15 grams. If he is consulting a sample size of 30 recipes, select the mean and standard deviation of the sampling distribution to help him order his supplies from the options below. σx¯=2.74 grams μx¯=100 gramsThe standard deviation of the sampling distribution is σx¯=σn−−√=1530−−√=2.74 grams Likewise, when the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=100 grams. Question A head librarian for a large city is looking at the overdue fees per user system-wide to determine if the library should extend its lending period. The average library user has $19.67 in fees, with a standard deviation of $7.02 . The data is normally distributed and a sample of 72 library users is selected at random from the population. Select the expected mean and standard deviation of the sampling distribution from the options below. Correct answer: σx¯=$0.83 μx¯=$19.67 The standard deviation of the sampling distribution is σx¯=σn−−√=$7.0272−−√=$0.83 When the distribution is normal, the mean of the sampling distribution is equal to the mean of the population μx¯=μ=$19.67 . 2. Question A well known social media company is looking to expand their online presence by creating another platform. They know that they current average 2,500,000 users each day, with a standard deviation of 625,000 users. If they randomly sample 50 days to analyze the use of their existing technology, identify each of the following, rounding to the nearest whole number if necessary:We are given population mean μ=2,500,000 and population standard deviation σ=625,000 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=50 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=2,500,000 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=625,00050−−√≈88,388 3. Question A bank is reviewing its risk management policies with regards to mortgages. To minimize the risk of lending, the bank wants to compare the typical mortgage owed by their clients against other homebuyers. The average mortgage owed by Americans is $306,500 , with a standard deviation of $24,500 . Suppose a random sample of 150 Americans is selected. Identify each of the following, rounding your answers to the nearest cent when appropriate: • 1306500$306500$306500 • 224500$24500$24500 • 3150$150$150 • 4306500$306500$306500 • $2000.42 We are given population mean μ=$306,500 and population standard deviation σ=$24,500 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=150 .By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=$306,500 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=$24,500150−−−√=$2,000.42 4. Question The average time it takes a certain brand of ibuprofen to start working is 25 minutes, with a standard deviation of 13 minutes, distributed normally. A pharmacist randomly samples 20 pills from this brand, because she is researching different brands in order to find the quickest acting ibuprofen to recommend to her customers. Identify the following to help her make her recommendations, rounding to the nearest hundredth if necessary: We are given that the population mean is μ=25 minutes and that the population standard deviation σ=13 minutes, distributed normally. We want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=20 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=25 minutes To find the standard deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=1320−−√=2.91 minutes 5. QuestionMajor league baseball recruiters are analyzing college players as potential draft choices. In a survey of college baseball players, the recruiters found that they hit an average of 13 home runs per season, with a standard deviation of 5 . Suppose a random sample of 45 baseball players is selected. Identify each of the following and remember to round to the nearest whole number: We are given population mean μ=13 and population standard deviation σ=5 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=45 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=13 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=545−−√≈1- - - - - - - - - - - - -Continued