Best tips for calculating standard deviation?

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The standard deviation is a statistical number calculated to give the specific bounds of data clusters below and above the mean of an ideal population within a normal curve. In other words, a calculated standard deviation gives the limits of the data indicated by three equally spaced lines on either side of the midline of a bell curve. Most procedures for calculating standard deviation without statistical programs or statistical calculators are referred to as “one-step” or “two-step” procedures, referring to the number of times each number must be noted and manipulated as part of the solution overall. Despite having to go through each number a second time, “two-step” methods for calculating standard deviation are easier to explain without referring to or understanding the statistical formula actually calculated. The best tips for calculating standard deviation include working with small amounts of data when first learning the process, using an example problem a student might encounter in real life, writing down all the arithmetic and calculations to double-check for errors and understand how your individual calculations lead to your final answer.

To establish a reasonable example problem, consider calculating the standard deviation on a list of 10 exam grades: 99, 78, 89, 71, 92, 88, 59, 68, 83, and 81.
The calculation is done using a formula known as the Welford method:
s = √ (1/n-1)(∑(x – µ)2
The variables in this equation are as follows:
s = standard deviation
√ = square root of the entire calculation
n = the number of data points, for example 10 test grades
∑ = summation symbol indicating that all the results calculated to follow must be added with simple calculations
x = each of the different data, for the example of test grades: 99, 78, 89, etc.
µ = the mean, or mean, of all your data; for example all 10 test marks added up and divided by 10
(x – µ)2 = squaring the result of the equation or multiplying the result by itself

Now, as you solve for certain variables, plug them into the equation.
The first step is the simplest. The denominator, n-1, of the fraction 1/n-1 can be easily solved. With n equal to 10 test marks, the denominator will clearly be 10 – 1 or 9.
The next step is to get the mean, or mean, of all the test grades by adding them up and dividing by the number of grades. The result should be µ = 80.8. This will be the middle, or mean, line that bisects the standard curve graph into two bilateral halves.

Then, subtract the mean – µ = 80.8 – from each of the 10 test degrees and square each of these deviations in a second pass through the data. As,
99 – 80.8 = 18.2331.2478 – 80.8 = -2.87.8489 – 80.8 = 8.267.2471 – 80.8 = -9.896.0492 – 80.8 = 11.2125.4488 – 80.8 = 7.251.8459 – 80.8 = -21.8475.2468 – 80.8 = – 12.8163.8483 – 80.8 = 2.24.8481 – 80.8 = 0.20.04

Add all of these calculations to reach the sum of the data represented by . Basic arithmetic now shows that = 1,323.6
now it needs to be multiplied by 1/9 as the denominator of this fraction was established in the first step of calculating the standard deviation. This results in a product of 147.07.

Finally, calculating the standard deviation requires the square root of this product to be 12.13.
So, for our example exam problem with 10 test scores ranging from 59 to 99, the average test score was 80.8. Calculating the standard deviation for our example problem yielded a value of 12.13. Based on the expected distribution of a normal curve, we could estimate that 68% of the grades would be within one standard deviation of the mean (68.67 to 92.93), 95% of the grades would be within two standard deviations of the mean (56.54 to 105.06) and 99.5% of the grades would be within three standard deviations of the mean.

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