Change in single observation of data set and how it will affect sample variance Q/ Consider the sample of size n=10, where the sample mean is 6 and the sample variance is 12.
Consider a second sample that is exactly the same as the first except that there is an additional observation, that takes on the value of 6.
What is the sample variance for this larger sample with n=11 observations? 
any help would be appreciated :) Thanks! 
 A: For our particular numbers, there is a very quick way. But it depends on a numerical "accident." So we do the problem a slow way, and then a quick way.
Slow way: In general, let $y_1,y_2,\dots,y_n$ be the observations. Then the sample variance is 
$$\frac{1}{n-1} \sum_1^n (y_i-\bar{y})^2,$$
where $\bar{y}$ is the sample mean. 
Note that 
$$\sum (y_i-\bar{y})^2=\sum y_i^2-2\bar{y}\sum y_i+n\bar{y}^2=\sum y_i^2-n\bar{y}^2.$$
Thus the sample variance is 
$$\frac{1}{n-1}\left(\sum y_i^2-n\bar{y}^2\right).\tag{1}$$
With our sample of $10$, the sample variance is
$$\frac{1}{9}\left(\sum_1^{10} y_i^2-(9)(36)\right).$$
This is $12$, and therefore $\sum_1^{10}y_i^2=432$.
Now take $n=11$, and let our new observation be $y_{11}$. We know the old sample mean, so we know the old sample sum. Thus we know the new sample sum, and therefore the new sample mean. By good luck, the new observation is the old sample mean, so the new sample mean is also $6$. 
Since $y_{11}=6$, we get $\sum_1^{11}y_i^2=468$. Now we can use formula (1). Our new sample variance is
$$\frac{1}{10}\left(468-(10)(36)\right).$$
Quick way: Go back to the fundamental formula
$$\frac{1}{n-1} \sum_1^n (y_i-\bar{y})^2$$
for the sample variance. The new observation is exactly the mean of the old ones. So the new mean is also $6$. Also,  if we call the new observation $y_{11}$, we have $y_{11}-\bar{y}=0$. Thus $\sum(y_i-\bar{y})^2$ has not changed, we are just dividing by $10$ rather than $9$. so the new sample variance is $\frac{9}{10}\cdot 12$.
