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The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.

Notice that instead of dividing by n=20, the calculation divided by n-1=20-1=19 because the data is a sample. For the sample variance, we divide by the sample size minus one ( n-1 ). Why not divide by n ? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (n-1) gives a better estimate of the population variance.

Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.

The standard deviation, s or σ , is either zero or larger than zero. When the standard deviation is 0, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large.

The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data .

Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:

  • 33
  • 42
  • 49
  • 49
  • 53
  • 55
  • 55
  • 61
  • 63
  • 67
  • 68
  • 68
  • 69
  • 69
  • 72
  • 73
  • 74
  • 78
  • 80
  • 83
  • 88
  • 88
  • 88
  • 90
  • 92
  • 94
  • 94
  • 94
  • 94
  • 96
  • 100

  • Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
  • Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:
    • The sample mean
    • The sample standard deviation
    • The median
    • The first quartile
    • The third quartile
    • IQR
  • Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.
  • Data Frequency Relative Frequency Cumulative Relative Frequency
    33 1 0.032 0.032
    42 1 0.032 0.064
    49 2 0.065 0.129
    53 1 0.032 0.161
    55 2 0.065 0.226
    61 1 0.032 0.258
    63 1 0.032 0.29
    67 1 0.032 0.322
    68 2 0.065 0.387
    69 2 0.065 0.452
    72 1 0.032 0.484
    73 1 0.032 0.516
    74 1 0.032 0.548
    78 1 0.032 0.580
    80 1 0.032 0.612
    83 1 0.032 0.644
    88 3 0.097 0.741
    90 1 0.032 0.773
    92 1 0.032 0.805
    94 4 0.129 0.934
    96 1 0.032 0.966
    100 1 0.032 0.998 (Why isn't this value 1?)
    • The sample mean = 73.5
    • The sample standard deviation = 17.9
    • The median = 73
    • The first quartile = 61
    • The third quartile = 90
    • IQR = 90 - 61 = 29
  • The x-axis goes from 32.5 to 100.5; y-axis goes from -2.4 to 15 for the histogram; number of intervals is 5 for the histogram so the width of an interval is (100.5 - 32.5) divided by 5 which is equal to 13.6. Endpoints of the intervals: starting point is 32.5, 32.5+13.6 = 46.1, 46.1+13.6 = 59.7, 59.7+13.6 = 73.3, 73.3+13.6 = 86.9, 86.9+13.6 = 100.5 = the ending value; No data values fall on an interval boundary.
    A hybrid image displaying both a histogram and box plot described in detail in the answer solution above.

The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73-33=40) than the spread in the upper 50% (100-73=27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR=29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.

Comparing values from different data sets

The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, it can be misleading to compare the data values directly.

  • For each data value, calculate how many standard deviations the value is away from its mean.
  • Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.
  • # ofSTDEVs = value - mean standard deviation
  • Compare the results of this calculation.

#ofSTDEVs is often called a "z-score"; we can use the symbol z. In symbols, the formulas become:

Sample x = x + z s z = x - x s
Population x = μ + z σ z = x - μ σ

Two students, John and Ali, from different high schools, wanted to find out who had the highest grade point average (GPA) when compared to his school. Which student had the highest GPA when compared to his school?

Student GPA School Mean GPA School Standard Deviation
John 2.85 3.0 0.7
Ali 77 80 10

For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.

# ofSTDEVs = value - mean standard deviation ; z = x - μ σ

For John, z = # ofSTDEVs = 2.85 - 3.0 0.7 = - 0.21

For Ali, z = # ofSTDEVs = 77 - 80 10 = - 0.3

John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his mean while Ali's GPA is 0.3 standard deviations below his mean.

John's z-score of −0.21 is higher than Ali's z-score of −0.3 . For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.

The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.

    For any data set, no matter what shape the distribution of the data is:

  • At least 75% of the data is within 2 standard deviations of the mean.
  • At least 89% of the data is within 3 standard deviations of the mean.
  • At least 95% of the data is within 4 1/2 standard deviations of the mean.
  • This is known as Chebyshev's Rule.

    For data having a distribution that is mound-shaped and symmetric:

  • Approximately 68% of the data is within 1 standard deviation of the mean.
  • Approximately 95% of the data is within 2 standard deviations of the mean.
  • More than 99% of the data is within 3 standard deviation of the mean.
  • This is known as the Empirical Rule.
  • It is important to note that this rule only applies when the shape of the distribution of the data is mound-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.
Practice Key Terms 2

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Source:  OpenStax, Collaborative statistics: custom version modified by r. bloom. OpenStax CNX. Nov 15, 2010 Download for free at http://legacy.cnx.org/content/col10617/1.4
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