What is the process for finding the standard deviation after I have my data? 

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violy | High School Teacher | (Level 1) Associate Educator

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If we are finding the sample standard deviation, we will apply the formula:

 `s = sqrt((x - barx)^2/(n-1))`

where `barx` is the sample mean` `

If we are finding the population standard deviation, we will apply the formula:

`sigma = sqrt((x - mu)^2/(n-1))`

where `mu` is the population mean

When we already have a data to find the standard deviation of, we will first find the mean.
The formula for the mean is:

`barx = (sumx)/(n)`  

Then, we will subtract each number by the mean, then square the result. Then, take the sum of the results that we got. Divide the result by "n-1" where n is the sample size, and finally take the square root.

Here is a similar example: 
Suppose you are asked to find the standard deviation for a sample data of:
132     144     125       141      134       137 

1. Find the mean.
Apply the formula for the mean: `barx = (sumx)/(n)`

`barx` `=` `(132 + 144 + 125 + 141 + 134 + 137)/(6) = (813)/(6) = 135.5`

2. Subtract each number by the mean.


` x - barx` 

`132- 135.5 = -3.5`

`144 - 135.5 = 8.5`

`125 - 135.5 = -10.5`

`141 - 135.5 = 5.5`

`135 - 135.5 = -1.5`

`137 - 135.5 = 1.5`

3. Square the results.

`(x - barx)^2`

`(-3.5)^2 = 12.25`
`(8.5)^2 = 72.25 `
`(-10.5)^2 = 110.25`
`(5.5)^2 = 30.25`
`(-1.5)^2 = 2.25`
`(1.5) = 2.25`

4. Take the sum of the results.

`sum(x - barx)^2 = 12.25 + 72.25 + 110.25 + 30.25 + 2.25 + 2.25 = 229.50`

5. Divide by `n - 1` .

`(sum(x - barx)^2)/(n-1) = (229.50)/(6-1) = (229.50)/(5) = 45.9`

6. Take the square root of the result.

`s = sqrt((sum(x - barx)^2)/(n-1)) = sqrt(45.9) = 6.77`  

Therefore, the sample standard deviation is `s = 6.77` .

That is it!

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