We used the equation for standard deviation at the beginning of this course as an example of a formula. In this activity, we’re going to learn what the equation does and where it comes from.

If we take a data set:
(x = 3, 4, 5)
We are able to determine that the mean is 4.

Standard deviation lets us know how far the other data points lie from the mean.

Let’s unpack standard deviation in the following steps:

1. Distance from mean = ((x – bar{x}))
2. Calculate the average of the mean difference: (frac{Sigma(x – bar{x})}{n})

Let’s apply our variable to see how this works:

Remember that our standard deviation formula is: (sigma = sqrt{frac{Sigma((x – bar{x})^2)}{n}})

(sigma = sqrt{frac{Sigma((x – bar{x})^2)}{n}})

(therefore sigma =sqrt{frac{(3-4)^2 + (4-4)^2 + (5-4)^2}{3}})
(therefore sigma =sqrt{frac{(-1)^2 + (0)^2 + (1)^2}{3}})
(therefore sigma = sqrt{frac{1 + 0 + 1}{3}})
(therefore sigma = sqrt{frac{2}{3}})
(therefore sigma = 0,81)

This answer of 0.81 gives us a good view of the average distance of our data from the mean. In this example, it is quite accurate because there is 1 data point on each side of the mean (each one within 1 data point of the mean, and 0.81 lies in this range – so it is correct).

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Why do you think it important for us to be able to calculate the standard deviation as a more accurate measure than just using the mean?

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