Maria and Zoe are taking Biology 105 but are in different classes. Maria's class has an average of 78% with a standard deviation of 5% on the midterm, whereas Zoe's class has an average of 83% with a standard deviation of 12%. Assume that scores in both classes follow a normal distribution.
A) Convert Maria's midterm score of 84 to a standard z score.
(A) 0.083 (B) 0.5 (C) 0.2 (D) 1.2 (E) 6
There is a part B to this problem, but if someone could explain how to they came up with answer to part A then I can do the rest!!! Thank you
A) Convert Maria's midterm score of 84 to a standard z score.
(A) 0.083 (B) 0.5 (C) 0.2 (D) 1.2 (E) 6
There is a part B to this problem, but if someone could explain how to they came up with answer to part A then I can do the rest!!! Thank you
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use the formula
z=(x-avg)/sd
z=(84-78)/5
z=1.2
this means that maria's score is 1.2 standard deviations above the class average.
additional, but important:
look up 1.2 in z score table to get .8849. this means that maria scored better than 88.49% of her classmates.
z=(x-avg)/sd
z=(84-78)/5
z=1.2
this means that maria's score is 1.2 standard deviations above the class average.
additional, but important:
look up 1.2 in z score table to get .8849. this means that maria scored better than 88.49% of her classmates.