# Will the mean equal the standard deviation

In my problem, there are three classes that have each taken a test and in these classes, there are 99 students. Class A had 1 student who scored a 1, 1 students who scored a 99, and 97 students who scored a 50. In class B 49 students scored a 1, 49 students scored a 99 and 1 student scored a 50. In class C 1 student scored a 1, 1 student scored a 2, 1 student scored a 3 and so on until this class reached the 1 student who scored 99. The mean of all of these classes is 495 points and the range is equal as well. Does this mean the standard deviation is equal, too?

• Pretty sure the mean is 50 (not 495). The sum is 4950 in each case. Sep 16, 2015 at 4:03
• Try computing the standard deviations and seeing if they are equal. Hint: For the first one (using population standard deviation) it's $\sqrt{\frac{(49)^2 + (-49)^2 + 97(0)^2}{99}}$ Sep 16, 2015 at 4:07
• @DanielR.Collins: Evidently this is for Class A (only): You need to divide by $n - 1 = 98$ for the standard deviation of a sample. Sep 16, 2015 at 7:13
• Perhaps also useful: $\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6.$ Sep 16, 2015 at 7:28

As you say, means and ranges (not shown below) are equal, for each of the three classes separately. (Also, for the aggregate of all three classes.) But the standard deviations are $not$ equal. The standard deviation is a 'more sensitive' measure of variability than is the range. Here are computations from R statistical software:

 a = c(1, 99, rep(50,97))            # quick ways
b = c(50, rep(1,49), rep(99,49))    #  to represent
c = 1:99                            #   your data

mean(a);  mean(b);  mean(c);  mean(c(a,b,c))
## 50
## 50
## 50
## 50
sd(a);  sd(b);  sd(c);  sd(c(a,b,c))
## 7               # least variation
## 49              # most variation
## 28.72281
## 32.92857


Using the hints in the Comments, you should be able to find the standard deviations using a calculator without much effort.