Question

According to a certain organization, adults worked an average of 1,792 hours last year. Assume the population standard deviation is 410 hours and that a random sample of 50 adults was selected. Complete parts a through e below. O = 57.98 (Round to two decimal places as needed.) b. What is the probability that the sample mean will be more than 1,820 hours? P(X>1,820) = .3146 (Round to four decimal places as needed.) c. What is the probability that the sample mean will be between 1,740 and 1,780 hours? P(1,740 sās 1,780) = .2331 (Round to four decimal places as needed.) d. Would a sample mean of 1,821 hours support the claim made by the organization? Select the correct choice below and fill in the answer box within your choice. (Round to four decimal places as needed.) O A. The probability P(X2 1,821) This contradicts the claim. B. The probability P(X2 1,821) This supports the claim. Click to select and enter your answer(s) and then click Check Answer. ? 1 part remaining Clear All Final Check

Answer 1

P(Mean > 1821)

$= P(z > \frac{1821-1792}{410/\sqrt{50}})$

= P(z > 0.50)

= 0.3085

Hence,

Option B; 0.3085