SOLUTION: Maria and Zoe are taking Biology 101 but are in different classes. Maria’s class has an average of 78% with a standard deviation of 5% on the midterm while Zoe’s class has an avera
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-> SOLUTION: Maria and Zoe are taking Biology 101 but are in different classes. Maria’s class has an average of 78% with a standard deviation of 5% on the midterm while Zoe’s class has an avera
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Question 937336: Maria and Zoe are taking Biology 101 but are in different classes. Maria’s class has an average of 78% with a standard deviation of 5% on the midterm while Zoe’s class has an average of 83% with a standard deviation of 12%. Assume that scores in both classes follow a normal distribution. Now suppose that Maria scored an 84 on the midterm exam and Zoe scored an 89.
Who did better relative to their class?
How do I convert raw score to percentage when I do not know the total number of test questions. how do I solve this problem? Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! assume 100 point tests
Maria: p(x = .84) = p(z < .06/.05) = normalcdf(-100,1.2) **** did better
Zoe: p(x = .89) = p(z < .06/.12) = normalcdf(-100, .5)
For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
area to the left of z = 1.2 > area to the left of z = .5
Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right
