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$M$ is the possible mean (average) score of a class of $n$ students and $E$ is the expected score of a student. Can someone please tell me the relationship between these two?

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More context and definitions are needed here but I guess here what is meant by mean is the empirical mean or average. (just the sum of the $n$ scores divided by the size of the class. This is a random variable because the sample is supposedly random). The expected sore is the true mean or the theoretical mean and is a constant. The empirical mean is an estimator of the expectation (expected score) and by the law of large numbers, as the sample size goes to infinity, it should converge to the expected value (expected score).


Let $X_1,...,X_n$ be the scores. (I am assuming here they are independent and identically distributed. The identical distribution means that $V[X_i]$ is constant for all $i$. Let's write it $V[X]$. The independence implies that all $cov[X_i,X_j]=0$ for all distint pairs $i, j$. Therefore, the variance of the sample mean is given by

$$\begin{array}{lll} V \left[ \frac{1}{n} \sum_{i = 1}^n X_i \right] & = & \frac{1}{n^2} V \left[ \sum_{i = 1}^n X_i \right]\\ & = & \frac{1}{n^2} \sum_{i = 1}^n V \left[ X_i \right]\\ & = & \frac{1}{n^2} n V \left[ X \right] \end{array}$$

This is $V[X]/n$ as mentioned in my comment.

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what about the expected value of the mean and it's variance. i guess the expected value of the mean is the expected score but what about the variance? –  user31280 Dec 5 '12 at 6:28
    
The expected value of the empirical mean is the mean (expected score in your example). The (theoretical or true) variance of the empirical mean is equal to the (theoretical or true) variance of one variable (a single score in your example) divided by $n$. At the limit, that variance goes to zero. –  Learner Dec 5 '12 at 6:31
    
sorry to bother but can you show a demonstration, please? –  user31280 Dec 5 '12 at 6:33
    
See also my answer here (about interchaing variance ans sum signs ) math.stackexchange.com/a/249660/48763 –  Learner Dec 5 '12 at 6:41
    
Thank you very much! –  user31280 Dec 5 '12 at 6:42

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