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The formula for the standard deviation is at the end of the chapter.

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 G.P.A. when compared to his school. Which student had the highest G.P.A. 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 G.P.A. when compared to his school because his G.P.A. is 0.21 standard deviations below his school's mean while Ali's G.P.A. is 0.3 standard deviations below his school's 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 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 deviations 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.

**With contributions from Roberta Bloom

Questions & Answers

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Introduction about quantum dots in nanotechnology
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s. Reply
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Source:  OpenStax, Collaborative statistics for mt230. OpenStax CNX. Aug 18, 2011 Download for free at http://legacy.cnx.org/content/col11345/1.2
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