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A sample of n=400 observations is drawn from a population with mean μ=1,000 and σ=400. Find the following probabilities:
P(x<1050)is:
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- The mean was found to be u = 70 so now we need to find the standard deviation o. The standard deviation is calculated as follows where n is the sample size, 350, and p is the probability of a success, 0.2. Find the standard deviation. V np(1 – p) O = V 350( )(1 – 0.2)The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is o = $2,400. (a) What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 20, 60, 100, and 300? (Round your answers to four decimal places.) sample size n = 20 sample sizen = 60 sample sizen = 100 sample size n = 300 (b) What is the advantage of a larger sample size when attempting to estimate the population mean? A larger sample has a standard error that is closer to the population standard deviation. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean. A larger sample increases the probability that the…A large consumer goods company ran a television advertisement for one of its soap products. On the basis of a survey that was conducted, probabilities were assigned to the following events. B = individual purchased the product S = individual recalls seeing the advertisement BNS = individual purchased the product and recalls seeing the advertisement The probabilities assigned were P(B) = 0.20, P(S) = 0.40, and P(BnS) = 0.12. %3D a. What is the probability of an individual's purchasing the product given that the individual recalls seeing the advertisement (to 1 decimal)? Does seeing the advertisement increase the probability that the individual will purchase the product? - Select your answer- As a decision maker, would you recommend continuing the advertisement (assuming that the cost is reasonable)? - Select your answer - b. Assume that individuals who do not purchase the company's soap product buy from its competitors. What would be your estimate of the company's market share (to the…
- The wildlife department has been feeding a special food to rainbow trout fingerlings in a pond. Based on a large number of observations, the distribution of trout weights is normally distributed with a mean of402.7grams and a standard deviation of12.8grams. What is the probability that the mean weight for a sample of 42 trout exceeds405.5grams? a.0.4338 b.1.0 c.0.5 d.0.0778A health expert evaluates the sleeping patterns of adults. Each week she randomly selects 40 adults and calculates their average sleep time. Over many weeks, she finds that 5% of average sleep time is less than 3 hours and 5% of average sleep time is more than 3.4 hours. What are the mean and standard deviation (in hours) of sleep time for the population? (Round "Mean" to 1 decimal places and "standard deviation" to 3 decimal places.) Mean Standard deviationAccording to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers. What is the probability that the sampling error would be more than 20 minutes?
- 4.33 124 For a population of 1000 items, μ = 50 and o= 10. What is the mean and standard error of the theoretical sampling distribution of the mean for sample sizes of (a) 25 and (b) 81? Ans. (a) x = 50 units and ox= 2 (b) μx = 50 units and ox = 1.07 XX1. +1₁ f+h f th 10 if thGiven a sample with a mean of 57 and a standard deviation of 14, find x such that Pr(X > x) = 0.715. Round your answer to two decimal places, e.g. 22.96.A large consumer goods company ran a television advertisement for one of its soap products. On the basis of a survey that was conducted, probabilities were assigned to the following events. B = individual purchased the product S = individual recalls seeing the advertisement BnS = individual purchased the product and recalls seeing the advertisement The probabilities assigned were P(B) = 0.20, P(S) = 0.40, and P(BS) = 0.12. a. What is the probability of an individual's purchasing the product given that the individual recalls seeing the advertisement (to 1 decimal)? Does seeing the advertisement increase the probability that the individual will purchase the product? - Select your answer - As a decision maker, would you recommend continuing the advertisement (assuming that the cost is reasonable)? - Select your answer - + b. Assume that individuals who do not purchase the company's soap product buy from its competitors. What would be your estimate of the company's market share (to the…
- Recall that a bank manager has developed a new system to reduce the time customers spend waiting to be served by tellers during peak business hours. The mean walting time during peak business hours under the current system is roughly 9 to 10 minutes. The bank manager hopes that the new system will have a mean waiting time that is less than six minutes. The mean of the sample of 104 bank customer waiting times is 5.48. If we let u denote the mean of all possible bank customer waiting times using the new system and assume that o equals 2.47: (a) Calculate 95 percent and 99 percent confidence intervals for u. (Round your answers to 3 decimal places.) 95 percent confidence intervals for u is 99 percent confidence intervals for u is (b) Using the 95 percent confidence interval, can the bank manager be 95 percent confident that u is less than six minutes? Explain. 95 percent interval is (c) Using the 99 percent confidence interval, can the bank manager be 99 percent confident that u is less…The heights of NBA players are normally distributed, with an average height of 6'7" (i.e. 79 inches), and a standard deviation of 3.5". You take a random sample of 8 players, and calculate their average height. What is the probability that this sample average is between 78 and 81 inches?this population, and the mean of each sample is determined. The mean, ux,is The standard error of the mean, Oy, is (Round your answer to 2 decimal places). QUESTION 4 The average number of miles driven each day by a group of drivers is 20.7 miles with a standard deviation of 6.5 miles. If you randomly select 50 drivers, what is the probability that the average number of miles driven each day is between 19.4 and 22.5 miles? Round your answer to 4 decimal places. Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers 5,337 MAY 24 MacBook Air B口 F3 888 F2 F4 1880 %23 24 4 3 T