Find the probability and interpret the results. If convenient, use technology to find the probability. The population mean annual salary for environmental compliance specialists is about $66,000. A random sample of 44 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $62,500? Assume 0' = $6,500. The probability that the mean salary of the sample is less than $62,500 is (Round to four decimal places as needed.) Interpret the results. Choose the correct answer below. Only 2% of samples of 44 specialists will have a mean salary less than $62,500. This is an extremely unusual event. Only 0.02% of samples of 44 specialists will have a mean salaw less than $62,500. This is an extremely unusual event. About 2% of samples of 44 specialists will have a mean salary less than $62,500. This is not an unusual event. 9.0.65.5 About 0.02% of samples of 44 specialists will have a mean salary less than $62,500. This is not an unusual event. Find the critical value zo necessary to form a confidence interval at the level of confidence shown below. C = 0.96 . . . Z= (Round to two decimal places as needed.)Find the margin of error for the given values of c, o, and n. c=0.95, o=3.8, n = 81 a Click the icon to view a table of common critical values. E = D (Round to three decimal places as needed.) You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 50 home theater systems has a mean price of $139.00. Assume the population standard deviation is $15.20. . . . Construct a 90% confidence interval for the population mean. The 90% confidence interval is (,). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is (,). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%. O B. With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%. O C. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%.Find the critical value to for the confidence level c = 0.98 and sample size n = 7. Click the icon to view the t-distribution table. . . . ts (Round to the nearest thousandth as needed.)Find the margin of error for the given values of c, s, and n. c=0.95.s=5, n=26 3 Click the icon to View the tdistribution table. The margin of error is El. (Round to one decimal place as needed.) Construct the indicated confidence interval for the population mean u using the t-distribution. Assume the population is normally distributed. C = 0.98, x = 13.2, s =0.67, n= 14 . . . (Round to one decimal place as needed.)