find the standard deviation to the nearest tenth for a data set with the following values 122,139,189

WHAT IS STANDARD DEVIATION?
One way of thinking about the standard deviation is to think of it at "the average distance from the average." While this is not the most precise mathematical definition it might help you remember what this is about and how to find it.

THE ARITHMETIC MEAN
If we want the "average distance from the average" the first thing we need to find is the "average." While there are various averages, the one that jumps to mind when you hear the word is the one we want to use. Mathematically we call it the "arithmetic mean" and we find it by adding the data points we are given (here the numbers 122, 139 and 189) and dividing by however many there are (here 3). The arithmetic mean is denoted by an x with a line over it.
The arithmetic mean here is [tex] \frac{122+139+189}{3}=150 [/tex]

DISTANCE FROM THE MEAN
Next we find how far from the mean each data point is. To do this we subtract.
122-150 = -28
139 - 150 = -11
189 - 150 = 39

DISTANCE?
Technically, distance is non-negative. We do not say a tree is -20 feet high or that the distance from here to the store is -37 miles. So the numbers above (some of which are positive and some of which are negative) have to be cleaned up somehow. The way statisticians have decide to do this is by squaring each number. We get:
(-28)(-28) = 784
(-11)(-11)=121
(39)(39)=1521
Let's treat these as the distances for now.

THE "AVERAGE DISTANCE"
The average (arithmetic mean) of the "distances" above is:
[tex] \frac{784+121+1521}{3} = \frac{2426}{3}=808.6666 [/tex]

VARIANCE
The number we found above (808.6666) is called the variance. This is actually an approximation ... in reality the 6s at the end there go on forever.

THE STANDARD DEVIATION
You might have realized that 808.6666 is pretty far away from the numbers we originally found as (-28, -11, 39) and if we ignore the negatives we could say the distances are 28, 11 and 39 BUT we squared these numbers which did get rid of negatives but also made the numbers much bigger. So to "fix" this we take the square root of the variance. The square root of the variance is the standard deviation.

The standard deviation here is approximately [tex] \sqrt{808.6666}=24.437[/tex]

Since you were asked to round to the nearest tenth (one decimal place) the answer is 24.4