Saturday, October 30, 2010

Median, Average and Standard Deviation

Mathematicians often use the terms "median," "average" and "standard deviation." They are important for data analysis. I will provide examples to illustrate how these can be applied.

Imagine that seven students write an exam and obtain the following scores: 60%, 65%, 65%, 75%, 80%, 90% and 95%. The total number of exams is seven. The median is the number which separates the higher half from the lower half. This is known as the central value. In this case, it is the fourth highest/lowest score. There are three lower scores and three higher scores from the central value. The median is precisely in the middle. In this case, the median is 75%.

With an odd number it is easy to find the median. With an even number, however, the method is a little different. Imagine that instead of seven exams we have eight. The scores are: 60%, 65%, 65%, 75%, 80%, 90%, 95% and 100%. To calculate the median it is necessary to determine the average of the two scores in the middle. In this case we have two scores below the top three and two scores above the top three. They are 75% and 80%. To calculate the average of these two scores, we add them together and then divide by two. 75+80=155. 155/2=77.5. 77.5 is the median of the eight exam scores.

The average in a data set is the value obtained by adding up all the values in a set and dividing them by the number of values. For example, if we have the scores 60%, 65%, 65%, 75%, 80%, 90%, 95% and 100%, the total is 630. We now divide this number by eight, the total number of scores. 630/8=78.75. Thus, 78.75 is the average of the eight exam scores.

Standard deviation measures average variation from the average value of a data set. Low standard deviation indicates that values tend to be close to the average and high standard deviation indicates that they are spread over a wide range. To calculate standard deviation, we first determine the average value of our data. Next we compute the difference of each value from the average and square the result of each. Then we divide the this number by the total number of values minus one and calculate the square root.

Consider the following scores on an quiz out of ten: 5, 6, 7, 8, 9, 9, 10, 10. To find the average, we need the sum of the eight scores. This is 64. We now divide 64 by 8. 64/8=8. The average value is 8.

Next we compute the difference of each value from the average. Here are the results:

5-8=-3
6-8=-2
7-8=-1
8-8=0
9-8=1
9-8=1
10-8=2
10-8=2

Negative values are not a problem for calculating standard deviation. Now we square these values. -3x-3=9, -2x-2=4, -1x-1 =1, 0x0=0, 1x1=1, 1x1=1, 2x2=4 and 2x2=4.

The next step is to calculate the sum of these values. 9+4+1+0+1+1+4+4=24. The sum is 24.

We must now divide by the number of values minus one. We have eight values so we divide by seven. 24/7=3.42.

The final step is to determine the square root of 3.42. The square root of 3.42 is approximately 1.85. This means that the average variation from the value of 8 is 1.85. In other words, most exam scores deviate 1.85 points from the average value of 8 which is from 6.15 to 9.85. Of the eight exam scores, five fall within this range.

Mathematicians often calculate the median, average and standard deviation of a data set. They are very useful in the interpretation of data. For this reason, it is a good idea to become familiar with their many applications.

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