Let's assume there are 3 people in Room A and another 3 people in Room B. The ages of the people in Room A are 25, 25, and 25. The ages of the people in Room B are: 10, 25, and 40. The mean of the ages in both rooms is 25. Does that mean that both room have populations that are equivalent? Is it fair to say that the number 25 is truly an indication of how old the people in the two rooms are? Well, not exactly... There is NO dispersion in the ages of the people in Room A (since everybody is the same age...), but there is large dispersion in Room B.



The standard deviation of the ages in Room A is zero; the standard deviation of the ages in Room B is 12.25. So, when we say that for Room B the mean is 25 and the standard deviation is 12.25, we get a better picture of the age distribution in the room. Standard deviation is a measure of the "spread" in the data.



Also, how did I get 12.25? How was it calculated?

Standard Deviation = SQRT [ {sqr(10-25) + sqr (25-25) + sqr (40-25)}/3 ]

To calculate the standard deviation, you need to follow these steps:

1. Find the mean (average) of the data set.
In this case, the mean is already given as 25.

2. Subtract the mean from each data point and square the result.
For Room B, the data points are 10, 25, and 40. Subtracting the mean from each gives -15, 0, and 15. Squaring these values gives 225, 0, and 225.

3. Find the mean of the squared differences from step 2.
Add up the squared differences (225 + 0 + 225 = 450) and divide by the number of data points (3 in this case). The mean of the squared differences is 150.

4. Take the square root of the mean from step 3.
The square root of 150 is approximately 12.25.

So, the standard deviation of the ages in Room B is 12.25. This indicates that the ages in Room B are more dispersed or spread out compared to the ages in Room A.