2.7 Measures of the spread of the data -- rrc math 1020  (Page 5/25)

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

1. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
2. Calculate the following to one decimal place using a calculator:
   1. The sample mean
   2. The sample standard deviation
   3. The median
   4. The first quartile
   5. The third quartile
   6. IQR
3. Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.

1. See [link]
2. 1. The sample mean = 73.5
   2. The sample standard deviation = 17.9
   3. The median = 73
   4. The first quartile = 61
   5. The third quartile = 90
   6. IQR = 90 – 61 = 29
3. The x -axis goes from 32.5 to 100.5; y -axis goes from –2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is (100.5 – 32.5) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the 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.

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.

Standard deviation of grouped frequency tables

Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula: $MeanofFrequencyTable=\frac{{\displaystyle \sum fm}}{{\displaystyle \sum f}}$
where $f=$ interval frequencies and m = interval midpoints.

Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.