a) from the given table Anna's and Megan's averages for all languages are
Anna's average = (83+83)/2 = 83
Megan's average = (77+95)/2 = 86
Here Megan's all languages average is 86 which is greater than 84.
So Megan should qualify the language honors
b) Anna's standard deviation s1 = 0
And Megan's standard deviation s2 = 12.73
Anna has no variation in language scores .
Anna's coefficient of variation =( 0/83)*100 = 0
Megan's coefficient of variation (12.73/86)*100 = 14.80
Low coefficient of variation means good performance .
Here Anna has low coefficient of variation so Anna's overall performance is good.
11. Below is some data about two students who took the same exams as weliras the...
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