According to a certain organization, adults worked an average of 1,823 hours last year. Assume the population standard deviation is 350 hours and that a random sample of 50 adults was selected. Complete parts a through e below. a. Calculate the standard error of the mean. 0'; = 49.50 (Round to two decimal places as needed.) b. What is the probability that the sample mean will be more than 1,840 hours? P6013340) = 0.3656' (Round to four decimal places as needed.) 0. What is the probability that the sample mean will be between 1,800 and 1,810 hours? P (1,800 5 )1 51,810) = 0.0753' (Round to four decimal places as needed.) d. Would a sample mean of 1,848 hours support the claim made by the organization? Select the correct choice below and ll in the answer box within your choice. (Round to four decimal places as needed.) ' A- The probability P ()1 21,848) = 0.3068'. This supports the claim. 5- The probability P ()7: 21,848) = . This contradicts the claim. 9. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 1,823 hours. 765 hours 5 i 5 hours (Round to one decimal place as needed.)