Question

QUESTION 13 5 points Save Answer Information for Questions 12 and 13: The 2012 American Academy of Physician Assistants (AAPA) survey of the population of 84,000 physicians' assistants who were working full time in 2011 reported a mean annual income of $92,460 and a standard deviation of $11,000. Suppose the AAPA collected data on the salaries of 100 randomly selected Physicians' Assistants. Find the probability that the sample mean (which we call x-bar) would fall within $1500 of the population mean. (That is find P($90,960 < x-bar < $93,960). O a. 1.00 O b. 0.827 O c. 0.977 O d. 0.108 O e. 0.023

Answer 1

Question 19 Hese el= $92460 (mean) 6 - $ 11000 C Standard deviation n = 100 Then INN (92460 , 11000 100) Capproximately) we have to find P( 8 goggo Lac $9960) 0 / 90960-92460, ã-l [ 93960 - 92460) 11000/ Too 6 1 1000/tico = P(- 1.36 42 <1:36) - P(Z <1-36) - PC25-1:36) = P(Z <1.36) - [i-PCZC 1.36) = 2pcz .36) - 1 By using standard Roomal table P( 90960 LãL 93960) = 2 coigis!) - || = 0.827 Loption b_correct. Ans: b. 0.827