# Calculating Standard Deviation,What Standard Deviation tells us

Standard Deviation in Math is a measure of the spread regarding a set of data.

Specifically, what standard deviation tells us is how far away the values of a data set are from the average/mean value.
The smaller the Standard Deviation value, then the less spread out from the average a set/group of data is.

Standard Deviation is a measure that can often be useful. As sometimes an average value can be a good fit for a set/group of data values, but there are also situations when an average value may not be a good fit.

We can observe this below in &nbsp2&nbsp examples of working out the average/mean of a set of numbers.

### Calculating the Mean/Average Examples

(1.1)&nbsp

List of &nbsp7&nbsp numbers:

5 , 7 , 3 , 5 , 6 , 4 , 5

Mean/average &nbsp= &nbsp \bf{\frac{5 \space + \space 7 \space + \space 3 \space + \space 5 \space + \space 6 \space + \space 4 \space + \space 5}{7}} &nbsp = &nbsp \bf{\frac{35}{7}} &nbsp = &nbsp 5

The average for the list of numbers is &nbsp5,&nbsp and it turns out that all values in the list are close to this value.

(1.2)&nbsp

List of &nbsp7&nbsp numbers:

3 , 10 , 12 , 5 , 18 , 6 , 2

Mean/average &nbsp = &nbsp \bf{\frac{3 \space + \space 10 \space + \space 12 \space + \space 5 \space + \space 18 \space + \space 6 \space + \space 2}{7}} &nbsp = &nbsp \bf{\frac{56}{7}} &nbsp = &nbsp 8

The average of the list of numbers here is &nbsp8,&nbsp however there are some values that are quite a distance away from that value, especially &nbsp18.

## Calculating Standard Deviation

The Standard Deviation of a set of data values, is a vale that helps to give a measure of how spread out the values in that set are from the mean/average. This is what standard deviation tells us.

The Standard Deviation of a set of data values, is in fact the square root of the variance of the data values.

So in order to establish the standard deviation of a data set, we first need to know the variance number.

Variance is given by the formula:

σ &nbspis known as the delta symbol, and it is the common symbol used as notation for Standard Deviation.

### Calculating Standard Deviation Example

(2.1)&nbsp

If we look at the same list of &nbsp7&nbsp numbers from example &nbsp(1.1):

5 , 7 , 3 , 5 , 6 , 4 , 5 &nbsp = &nbsp \lbrace \space x_1 \space , \space x_2 \space , \space x_3 \space , \space x_4 \space , \space x_5 \space , \space x_6 \space , \space x_7 \space \rbrace

n = 6 &nbsp , &nbsp μ = 5

\sum_{i=1}^7 \space(x_i \space – \space \mu)^2

= &nbsp ( x_1 5 )2 &nbsp+&nbsp ( x_2 5 )2 &nbsp+&nbsp ( x_3 5 )2 &nbsp+&nbsp ( x_4 5 )2 &nbsp+&nbsp ( x_5 5 )2 &nbsp+&nbsp ( x_6 5 )2 &nbsp+&nbsp ( x_7 5 )2

= &nbsp ( 5 5 )2 + ( 7 5 )2 + ( 3 5 )2 + ( 5 5 )2 + ( 6 5 )2 + ( 4 5 )2 + ( 5 5 )2

= &nbsp 0 + 4 + 4 + 0 + 1 + 1 + 0 &nbsp = &nbsp 10

σ &nbsp = &nbsp \bf{\sqrt{\frac{10}{7}}} &nbsp = &nbsp 1.195

### What Standard Deviation tells us,Using the Standard Deviation Value

Now considering the Standard Deviation is an indicator of how far away a set of values are from the mean/average.

Let’s see how we can make use of it with the list of &nbsp7&nbsp numbers.

For &nbsp 5 , 7 , 3 , 5 , 6 , 4 , 5, &nbsp &nbsp the mean is &nbsp5.

While the Standard Deviation is &nbsp1.29.

We can establish how far way &nbsp1&nbsp Standard Deviation is from the mean/average in both a positive direction and a negative direction also.

51.195 &nbsp=&nbsp 3.805 &nbsp&nbsp , &nbsp&nbsp 5 + 1.195 &nbsp=&nbsp 6.195

What this means is that we should expect to see the majority of the values in the list of 7 to be between &nbsp3.805&nbsp and &nbsp6.195.

From looking at the list, this does turn out to be the case.

With only &nbsp3&nbsp and &nbsp7&nbsp being out side of this range.

## Alternative Formula for Standard Deviation

The variance does have a slightly simpler and quicker formula that can be used. This leads us to an alternative formula Standard Deviation that can also be used, which is:

Again &nbspμ&nbsp is the mean/average, and &nbspn&nbsp represents the number of values.

So the calculations required for example &nbsp(2.1)&nbsp from above using this formula are:

σ &nbsp=&nbsp \bf{\sqrt{{\frac{5^2 \space + \space 7^2 \space + \space 3^2 \space + \space 5^2 \space + \space 6^2 \space + \space 4^2 \space + \space 5^2}{7}} \space – \space 5^2}} &nbsp = &nbsp \bf{\sqrt{\frac{185}{7} \space – \space 25}} &nbsp = &nbsp 1.195

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