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#1 2006-07-09 12:20:17

babyangel1972
Member
Registered: 2005-12-12
Posts: 18

Any help is appreciated

Imagine the exam scores of a class of 10 students.  Make three very different distributions that all have a mean of 80.    ??  help

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#2 2006-07-09 20:27:42

Patrick
Real Member
Registered: 2006-02-24
Posts: 1,005

Re: Any help is appreciated

I'm not going to solve this for you because it's fairly easy, but I'll tell you how to solve it.

First off - what is the (arithmetic) mean? Basicly it is when you take a number of observations(in this case exam scores), add them up and divide by the total amount of numbers.
Instead of 10 students, imaginge 3 students - yet the mean remains the same. So, what do we know?

There are 3 students.
The mean score is 80.

Where do we go from here? Well, the obvious solution is:


That's only one solution though, you asked for 3. The way to to find more is to find out what the sum of the scores is. You do this by multiplying the number of observations(3 students) and the mean (80). 3*80=240, so what you have to do now is just to find 3 numbers that adds up to 240:


I don't know if any of this will make sense to you, but I'll write it anyway. It's just another way of showing what the mean is. If you understand this, however, you prolly know your mean. Anyway tongue Here it goes:

First off, the x with a bar over it's head(x bar) is the mean of the set X with n elements(in your case n is 10 and "x bar" is 80)


If you consider my example with 60+80+100, it would look like this:



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