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For a class of 43 students, the mean grade is 67 and the standard deviation is 7, how many students, should receive a C?
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Hi bodie001;
I am interpreting a C aa between 60 - 69 inclusive.
Someday, probably thousands of years from now when they teach statistics correctly, everyone will know that these types of problems are nothing more than the area under a curve. The present method with the stupid use of tables masks that fact. The result, know one realizes that z-scores is just a numerical integration problem. This is how I solve them, but I will also do it the way your teacher requires.
This what the Normal distribution really is, notice the parameters of mean and SD.
After plugging in your mean and SD.
So .4538 * 43 = 19.51 students are expected to get a C
Now using tables: You see that 60-67 is 1 standard deviation to the the left of the mean, by the table that is .3413 Now z = 2 / 7 (represents the 67 to 69) to the right of the mean which is about .28:
Now go here: http://www.statsoft.com/textbook/distribution-tables/#z
Find the z table.
Find .28 = .1103 and add it to your .3413 = .4516, not quite as accurate as you can do with integration but the table value of .28 is only an approximation of 2/7.
.4516 * 43 = 19.42 students are expected to get a C. Again only an approximation.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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