Hi How are you?
I did a question, I get it right, but the other two TRUE sentences, I didn`t understand
-
The arithmetic mean of a frequency distribution is equal to the sum of the class frequency times the midpoint of the frequency class all divided by the number of observations. It doesn’t make any sense to me, maybe is my english. I am not a native english speaker. 2) The arithmetic mean is the only measure of central tendency where the sum of the deviations of each observation from the mean is always zero. ( I think this is right if is normal distributed but it gets harder to understand, when I think in nonnormal )
2/ This is essentially your definition of the arithmetic mean. Suppose your population is X1, X2,…, Xn and the mean E(X). The deviation of each obs from the mean is: X1 - E(X), X2 - E(X), …, Xn - E(X).
Then the sum of the deviations of each observation from the mean is X1 + X2 + … + Xn - n*E(X), which, by definition, equals 0.
1)The arithmetic mean of a frequency distribution. For example, there are 3 classes, class1: 8~10 the frequencies are 4; class2: 11~15, the frequencies are 5; class3: 16~18 the frequencies are 6. Thus, in this case, for class1: the class frequency times the midpoint means 4*[(10+8)/2]=36, class2: 5*[(11+15)/2]=65; class3: 6*[(16+18)/2]=102. so the sum these data= 36+65+102=203. Then divided by 4+5+6=15, you can get the answer
- the mean is E(x), the diviation is Xn-E(x). Then, the sum of diviations equal to [X1-E(x)]+[X2-E(x)]+…+[Xn-E(x)]. rearrange this equation, you can get X1+X2+…+Xn-n*E(x)=0. since the E(x) can be difined as (X1+X2+…+Xn)/n
Sorry for a mistake in the 1) you should divided by 4+5+6=15
tks very much for the answer