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A large statistics class took an exam. The scores followed a normal curve with mean 78 and standard deviation 10. Students who scored 90 or above got an A on the exam. Those who scored between 80 and 90 got a B, between 70 and 80 a C, between 60 and 70 a D, and below 60 an F. a. Estimate the percentage of students who earned an A on the exam. b. Estimate the percentage of students who earned a B on the exam. c. Estimate the percentage of students who earned a C on the exam. d. Estimate the percentage of students who earned a D on the exam. e. Estimate the percentage of students who earned an F on the exam. Please show as much work as you can and don\'t round. Thank you. Solution http://www.math.umt.edu/elias/NormalDistributionTable.jpg given is the normal distribution table = 78 = 10 1) for students scoring A : we find the area under the curve of standard normal distribution using Z value Z for x=90 : (90-78)/10 = 1.2 (z)=(1.2)=.8849 so the required area is 1-(z)= 1=0.8849= .1151 hence the percentage will be 11.51% 2) for students scoring B : we find the area under the curve of standard normal distribution using Z value Z for x=80 : (80-78)/10 = 0.2 (z)=(0.2)=.5793 % students scoring above B = (1-.5793)*100=42.07% % students scoring B = % students scoring above B -% students scoring A = 42.07-11.51 = 30.56% 3) for students scoring C : we find the area under the curve of standard normal distribution using Z value Z for x=70 : (70-78)/10 = -0.8 (z)=(-0.8)=.2119 % students scoring above C = (1-.2119)*100=78.81% % students scoring C = % students scoring above C -% students scoring above B = 78.81-42.07 = 36.74% 4) for students scoring D : we find the area under the curve of standard normal distribution using Z value Z for x=60 : (60-78)/10 = -1.8 (z)=(-1.8)=.0359 % students scoring above D = (1-.0359)*100=96.41% % students scoring D = % students scoring above D -% students scoring above C = 96.41-78.81 = 17.60% 5) for students scoring F : we find the area under the curve of standard normal distribution using Z value Z for x=60 : (60-78)/10 = -1.8 (z)=(-1.8)=.0359 % students scoring F =100- % students scoring above D % students scoring D = % students scoring above D -% students scoring above C = = 100 - 96.41 = 3.59% Verification: Sum of all 5 percentages = 100.

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- 1. A large statistics class took an exam. The scores followed a normal curve with mean 78 and standard deviation 10. Students who scored 90 or above got an A on the exam. Those who scored between 80 and 90 got a B, between 70 and 80 a C, between 60 and 70 a D, and below 60 an F. a. Estimate the percentage of students who earned an A on the exam. b. Estimate the percentage of students who earned a B on the exam. c. Estimate the percentage of students who earned a C on the exam. d. Estimate the percentage of students who earned a D on the exam. e. Estimate the percentage of students who earned an F on the exam. Please show as much work as you can and don't round. Thank you. Solution http://www.math.umt.edu/elias/NormalDistributionTable.jpg given is the normal distribution table = 78 = 10 1) for students scoring A : we find the area under the curve of standard normal distribution using Z value Z for x=90 : (90-78)/10 = 1.2 (z)=(1.2)=.8849 so the required area is 1-(z)= 1=0.8849= .1151 hence the percentage will be 11.51% 2) for students scoring B : we find the area under the curve of standard normal distribution using Z value Z for x=80 : (80-78)/10 = 0.2 (z)=(0.2)=.5793 % students scoring above B = (1-.5793)*100=42.07% % students scoring B = % students scoring above B -% students scoring A = 42.07-11.51
- 2. = 30.56% 3) for students scoring C : we find the area under the curve of standard normal distribution using Z value Z for x=70 : (70-78)/10 = -0.8 (z)=(-0.8)=.2119 % students scoring above C = (1-.2119)*100=78.81% % students scoring C = % students scoring above C -% students scoring above B = 78.81-42.07 = 36.74% 4) for students scoring D : we find the area under the curve of standard normal distribution using Z value Z for x=60 : (60-78)/10 = -1.8 (z)=(-1.8)=.0359 % students scoring above D = (1-.0359)*100=96.41% % students scoring D = % students scoring above D -% students scoring above C = 96.41-78.81 = 17.60% 5) for students scoring F : we find the area under the curve of standard normal distribution using Z value Z for x=60 : (60-78)/10 = -1.8 (z)=(-1.8)=.0359 % students scoring F =100- % students scoring above D % students scoring D = % students scoring above D -% students scoring above C = = 100 - 96.41 = 3.59% Verification: Sum of all 5 percentages = 100