QM2610 Midterm Exam A ecUse this statement to answers questions 36-41 The popula
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QM2610 Midterm Exam A ecUse this statement to answers questions 36-41 The population mean is $19,228 with a standard deviation of $7,696 represents the median income for all ountries. For questions 3S-40, base your answers on the population mean and the standard deviation (given above) and a normal distribution. What is the probability that a country's annual median income (AMI) is greater than Germany's AMI, which is $21,240.76 (10)? 36. 37. What is the probability that a country's annual median income (AMI) is greater than Chile's AMI, which is $7,850.73 (10)? 38. What is the probability that a country's annual median income (AMIl) falls between Germany's AMI and Chile's AMI (10)? 39. What is the probability that a country's annual median income (AMI) is greater than country with the lowest AMI in the sample (Mexico at $4,689.17) (10)? 40 What is the probability that a country's annual median income (AMI) is less than country with thoe lowest AMI in the sample (Mexico at $4,689.17) (5)? 41. At what value of AMI would 20% of the countries fall above (10)?Explanation / Answer
as we know for normal distirbution z score =(X-mean)/std deviaiton
hence
36) probability that income is greater then 21240.76 =P(X>21240.76)=1-P(X<21240.76)
=1-P(Z<(21240.76-19228)/7696)=1-P(Z<0.2615)=1-0.6032 =0.3968
37)
probability that income is greater then 7850.73 =P(X>7850.73)=1-P(X<7850.73)
=1-P(Z<(7850.73-19228)/7696)=1-P(Z<-1.4783)=1-0.0697=0.9303
38) probability that income falls between 7850.73 and 21240.76 =0.9303-0.3968 =0.5335
39)
probability that income is greater then 4689.17 =P(X>4689.17)=1-P(X<4689.17)
=1-P(Z<(4689.17-19228)/7696)=1-P(Z<-1.8891)=1-0.0294=0.9706
40)
probability that income is less then 4689.17 =1-0.9706 =0.0294
41)for top 20% ; z=0.8416
hence corresponding value =mean +z*std deviation =19228+0.8416*7696=25705.12
please revert for any clarification
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