Compute the critical value za/2 that corresponds to a 81% level of confidence. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Zu/2=[ (Round to two decimal places as needed.)
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- Until your a For a recent 10k run, the finishers are normally distributed with mean 63 minutes and standard deviation 14 minutes. Complete parts (a) through (d) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. ... a. Determine the percentage of finishers with times between 45 and 75 minutes. Approximately 70.50 % of finishers had times between 45 and 75 minutes. (Round to two decimal places as needed.) b. Determine the percentage of finishers with times less than 80 minutes. torre Que Approximately % of finishers had times less than 80 minutes. (Round to two decimal places as needed.) ✔Question 1 (1 Question 5 (0 Name: Due: Last Worked: Current Score: Attempts:The working life of a certain electrical equipment is normally distributed with a mean of 180 days and a standard deviation of 4 days. For each of the following questions, construct a normal distribution curve and provide the answer. About what percent of the products last between 176 and 184 days? About what percent of the products last between 180 and 184 days? For each of the following questions, use the standard normal table and provide the answer. About what percent of the products last 172 or less days? About what percent of the products last 184 or more days?A vending machine is designed to dispense a mean of 7.8 oz of coffee into an 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.5 oz and the amount is normally distributed, determine the percent of times the machine will dispense more than 7.1 oz. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. *** % of the time the machine will dispense more than 7.1 oz. (Type an integer or decimal rounded to two decimal places as needed.) 4 Inco
- Compute the critical value z,a/2. that corresponds to a 86% level of confidence. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Za/2 = (Round to two decimal places as needed.)Salaries of 32 college graduates who took a statistics course in college have a mean, x, of $62,000. Assuming a standard deviation, 6, of $17,889, construct a 95% confidence interval for estimating the population mean µ. Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table.A random sample of 100 automobile owners in a region shows that an automobile is driven on average 25,500 kilometers per year with a standard deviation of 3500 kilometers. Assume the distribution of measurements to be approximately normal. Construct a 90% prediction interval for the kilometers traveled annually by an automobile owner in the region. Click here to view page 1 of the standard normal distribution table. of the standard normal distribution table. Click here to view page 2 Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. The prediction interval is < X < (Round to the nearest integer as needed.) -CFind the z-scores that separates the middle 85 % of the distribution from the area in the tail of the standard normal distribution.First z-score = [ Note: If it did not work, try switching your answer around]Second z-score =In a survey, 39% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected 160 pet owners and discovered that 56 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.05 level of significance. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Because npo (1-po) =U V10, the sample size is V 5% of the population size, and the sample the requirements for testing the hypothesis satisfied, (Round to one decimal place as needed.) What are the null and alternative hypotheses? V versus H,: (Type integers or decimals. Do not round.) Но Determine the test statistic, zo Zo = |(Round to two decimal places as needed.) Determine the critical value(s). Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as…In a survey, 34% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 210 pet owners and discovered that 70 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.05 level of significance. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Because npo (1- Po): V 10, the sample size is V 5% of the population size, and the sample the requirements for testing the hypothesis satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? versus H: Ho: (Type integers or decimals. Do not round.) Determine the test statistic, Zo Z0 = (Round to two decimal places as needed.) Determine the critical value(s). Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as…In a recent poll, 45% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 160 randomly selected students on your campus and find that 79 of them would prefer a boy. Complete parts (a) and (b) below. Click here to view the standard normal distribution table (page 1). LOADING... Click here to view the standard normal distribution table (page 2). LOADING... (a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 160 students, at least 79 would prefer a boy, assuming the true percentage is 45%. The probability that at least 79 students would prefer a boy is 3636. (Round to four decimal places as needed.)Salaries of 49 college graduates who took a statistics course in college have a mean, x, of $63,500. Assuming a standard deviation, o, of $14,619, construct a 90% confidence interval for estimating the population mean u. Click here to view at distribution table. 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