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Statisitcs Definitions Help (page 3)

By — McGraw-Hill Professional
Updated on Apr 25, 2014

Variance

There is still another way in which the nature of a distribution can be described. This is a measure of the extent to which the values are spread out. There is something inherently different about a distribution of test scores like those portrayed in Table 2-7, compared with a distribution where every score is almost equally ''popular.'' The test results portrayed in Table 2-7 are also qualitatively different than a distribution where almost every student got the same score, say 7 answers correct.

More Definitions

In the scenario of Table 2-7, call the variable x, and let the 100 individual scores be called x1 through x100. Suppose we find the extent to which each individual score xi (where i is an integer between 1 and 100) differs from the mean score for the whole population (μp). This gives us 100 ''distances from the mean,'' d1 through d100, as follows:

The vertical lines on each side of an expression represent the absolute value. For any real number r, |r| = r if r ≥ 0, and |r| = –r if r < 0. The absolute value of a number is the extent to which it differs from 0. It avoids the occurrence of negative numbers as the result of a mathematical process.

Now, let's square each of these ''distances from the mean,'' getting this set of numbers:

The absolute-value signs are not needed in these expressions, because for any real number r, r2 is never negative.

Next, let's average all the ''squares of the distances from the mean,'' di2. This means we add them all up, and then divide by 100, the total number of scores, obtaining the ''average of the squares of the distances from the mean.'' This is called the variance of the variable x, written Var(x):

  • Var(x) = (1/100)(d12 + d22 + . . . + d1002) = (1/100)[(x1 – μp)2 + (x2 – μp)2 + . . . + (x100 – μp)2]

The variance of a set of n values whose population mean is μp is given by the following formula:

  • Var(x) = (1/n)[(x1 – μp)2 + (x2 – μp)2 + . . . + (xn – μp)2]

Standard Deviation

Standard deviation, like variance, is an expression of the extent to which values are spread out with respect to the mean. The standard deviation is the square root of the variance, and is symbolized by the italicized, lowercase Greek letter sigma (σ). (Conversely, the variance is equal to the square of the standard deviation, and is often symbolized σ2.) In the scenario of our test:

  • σ = [(1/100)(d12 + d22 + . . . + d1002)]1/2 = {(1/100)[(x1 – μp)2 + (x2 – μp)2 + . . . + (x100 – μp)2]}1/2

The standard deviation of a set of n values whose population mean is μp is given by the following formula:

  • σ = {(1/n)[(x1 – μp)2 + (x2 – μp)2 + . . . + (xn – μp)2]}1/2

These expressions are a little messy. It's easier for some people to remember these verbal definitions:

  • Variance is the average of the squares of the ''distances'' of each value from the mean.
  • Standard deviation is the square root of the variance.

Variance and standard deviation are sometimes called measures of dispersion. In this sense the term ''dispersion'' means ''spread-outedness.''

Statistics Definitions Practice Problems

Practice 1

What are the sample means for each grade in the test whose results are tabulated in Table 2-7? Use rounding to determine the answers to two decimal places.

More Definitions

Solution 1

Let's call the sample means μsa for the grade of A, μsb for the grade of B, and so on down to μsf for the grade of F.

To calculate μsa, note that 5 students received scores of 10, while 6 students got scores of 9, both scores good enough for an A. This is a total of 5 + 6, or 11, students getting the grade of A. Therefore:

    • μsa = [(5 × 10) + (6 × 9)]/11
    •       = (50 + 54)/11
    •       = 104/11
    •       = 9.45

To find μsb, observe that 19 students scored 8, and 17 students scored 7. Thus, 19 + 17, or 36, students received grades of B. Calculating:

    • μsb = [(19 × 8) + (17 × 7)]/36
    •       = (152 + 119)/36
    •       = 271/36
    •       = 7.53

To determine μsc, check the table to see that 18 students scored 6, while 11 students scored 5. Therefore, 18 + 11, or 29, students did well enough for a C. Grinding out the numbers yields this result:

    • μsc = [(18 × 6) + (11 × 5)]/29
    •       = (108 + 55)/29
    •       = 163/29
    •       = 5.62

To calculate μsd, note that 6 students scored 4, while 4 students scored 3. This means that 6 + 4, or 10, students got grades of D:

    • μsd = [(6 × 4) + (4 × 3)]/10
    •       = (24 + 12)/10
    •       = 36/10
    •       = 3.60

Finally, we determine μsf. Observe that 4 students got scores of 2, 7 students got scores of 1, and 3 students got scores of 0. Thus, 4 + 7 + 3, or 14, students failed the test:

    • μsf = [(4 × 2) + (7 × 1) + (3 × 0)]/14
    •       = (8 + 7 + 0)/14
    •       = 15/14
    •       = 1.07
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