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# Variance and Standard Deviation Help

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By — McGraw-Hill Professional
Updated on Sep 13, 2011

## Variance and Standard Deviation

Let's further analyze the results of the hypothetical 10-question quiz given to the class of 130 students. Again, in verbal form, here are the results:

• 0 correct answers: 0 students
• 1 correct answer: 4 students
• 2 correct answers: 7 students
• 3 correct answers: 10 students
• 4 correct answers: 15 students
• 5 correct answers: 24 students
• 6 correct answers: 22 students
• 7 correct answers: 24 students
• 8 correct answers: 15 students
• 9 correct answers: 7 students
• 10 correct answers: 2 students

#### Practice 1

What is the variance of the distribution of scores for the hypothetical quiz whose results are described above? Round off the answer to three decimal places.

#### Solution 1

Let the independent variable be represented as x. There are n = 130 individual quiz scores, one for each student. Let's call these individual scores xi, where i is an index number that can range from 1 to 130 inclusive, so there are 130 xi's ranging from x1 to x130. Let's call the absolute frequency for each particular numerical score fj, where j is an index number that can range from 0 to 10 inclusive, so there are 11 fj's ranging from f0 to f10. The population mean, μp, is approximately 5.623, as we have already found. We are looking for the variance of x, symbolized Var(x). We learned that the variance of a set of n values x1 through xn with population mean μ is given by the following formula:

Var(x) = (1/n)[(x1 – μ)2 + (x2 – μ)2 + … + (xn – μ)2]

It's messy to figure this out in our hypothetical scenario where there are 130 individual test scores, grouped according to the number of students attaining each score. It helps to compile a table as we grind our way through the above formula, and fill in the values as calculations are made. Table 8-7 is our crutch here. The possible test scores are listed in the first (far-left-hand) column.

Table 8-7 Table for Practice 1. The population mean μp, is 5.623. The individual test scores are denoted xi. There are 130 students; i ranges from 1 to 130 inclusive. The absolute frewuency for each score is denoted fi. There are 11 possible scores; j ranges from 0 through 10 inclusive.

In the second column, the absolute frequency for each score is shown. In the third column, the population mean, 5.623, is subtracted from the score. This gives us a tabulation of the differences between μ and each particular score xi. In the fourth column, each of these differences is squared. In the fifth column, the numbers in the fourth column are multiplied by the absolute frequencies. In the sixth column, the numbers in the fifth column are cumulatively added up. This is the equivalent of summing all the formula terms inside the square brackets, and gives us a grand total of 546.55 at the lower-right corner of the table. This number must be multiplied by 1/n (or divided by n) to get the variance. In our case, n = 130. Therefore:

Var(x) = 546:55=130

= 4:204

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