Thread: easy mean problem

let $\displaystyle x$ be a variable assuming the values $\displaystyle 0,1,2,3.....k$ and let $\displaystyle f_{0},f_{1},f_{2},f_{3},......,f_{k} $be the corresponding cumulative frequency of the greater than type. Prove that the arithmetic mean of x is, $\displaystyle \bar{x} = \frac{1}{N}(\sum_{i=1}^{k} F_{i})$, where N is the total frequency.

Thanks in advance!

Hey earthboy.

For this problem, you might want to consider the definition of the mean in terms of the summation and what the frequency corresponds to in terms of the actual observation.

Recall that the frequency for a variable says how many times it occurs so the observations in terms of the frequencies are F_i*i which gives the total number of observations.

The mean is simply adding up all the observations and dividing by the total number.

So write the frequencies in terms of the observations and relate that to the definition of the sample mean.

Originally Posted by chiro

Hey earthboy.

For this problem, you might want to consider the definition of the mean in terms of the summation and what the frequency corresponds to in terms of the actual observation.

Recall that the frequency for a variable says how many times it occurs so the observations in terms of the frequencies are F_i*i which gives the total number of observations.

The mean is simply adding up all the observations and dividing by the total number.

So write the frequencies in terms of the observations and relate that to the definition of the sample mean.

Do you mean this:
So the mean $\displaystyle \bar{x}=\frac{0.(f_{0}-f_{1})+1.(f_{1}-f_{2})+2(f_{2}-f_{3})+....+k-1(f_{k-1}-f_{k})+kf_{k}}{N}$ where $\displaystyle N=f_{0}$

expanding the numerator, all the terms would cancel out, leaving$\displaystyle \bar{x}=\frac{f_{1}+f_{2}+f_{3}......+f_{k}}{N}= \frac{1}{N} \sum_1^{k} f_{i}$


Thanks very much!

You will have to use the fact that CF_(i+1) - CF_i = f_(i+1) where CF is the cumulative frequency and f is the normal frequency.