# Mr pryors first test finding percentiles problem use

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Mr. Pryor’s First TestFinding percentilesPROBLEM:Use the scores on Mr. Pryor’s first statistics test to find the percentiles for thefollowing students:(a)Norman, who earned a 72.(b)Katie, who scored 93.(c)The two students who earned scores of 80.SOLUTION:(a)Only 1 of the 25 scores in the class is below Norman’s 72. His percentile is computed asfollows: 1/25=0.04, or 4%. So Norman scored at the 4th percentile on this test.(b)Katie’s 93 puts her at the 96th percentile, because 24 out of 25 test scores fall below herresult.(c)Two students scored an 80 on Mr. Pryor’s first test. Because 12 of the 25 scores in the classwere less than 80, these two students are at the 48th percentile.EXAMPLEFor PracticeTry Exercise1AgeFrequency40–44245–49750–541355–591260–64765–693Let’s expand this table to include columns for relative frequency, cumulativefrequency, and cumulative relative frequency.To get the values in therelative frequencycolumn, divide the count in eachclass by 44, the total number of presidents. Multiply by 100 to convert to apercent.To fill in thecumulative frequencycolumn, add the counts in the frequencycolumn for the current class and all classes with smaller values of the variable.For thecumulative relative frequencycolumn, divide the entries in the cumu-lative frequency column by 44, the total number of individuals. Multiply by100 to convert to a percent.
87Here is the original frequency table with the relative frequency, cumulative fre-quency, and cumulative relative frequency columns added.AgeFrequencyRelativefrequencyCumulativefrequencyCumulative relativefrequency40–4422/44=0.045, or 4.5%22/44=0.045, or 4.5%45–4977/44=0.159, or 15.9%99/44=0.205, or 20.5%50–541313/44=0.295, or 29.5%2222/44=0.500, or 50.0%55–591212/44=0.273, or 27.3%3434/44=0.773, or 77.3%60–6477/44=0.159, or 15.9%4141/44=0.932, or 93.2%65–6933/44=0.068, or 6.8%4444/44=1.000, or 100%To make acumulative relative frequency graph,we plot a point correspond-ing to the cumulative relative frequency in each class at the smallest value of thenextclass. For example, for the 40 to 44 class, we plot a point at a height of 4.5%above the age value of 45. This means that 4.5% of presidents were inauguratedbeforethey were 45 years old. (In other words, age 45 is the 4.5thpercentile of the inauguration age distribution.)It is customary to start a cumulative relative frequency graph witha point at a height of 0% at the smallest value of the first class (in thiscase, 40). The last point we plot should be at a height of 100%. Weconnect consecutive points with a line segment to form the graph.Figure 2.1 shows the completed cumulative relative frequency graph.Here’s an example that shows how to interpret a cumulativerelative frequency graph.FIGURE 2.1Cumulative relative frequency graph forthe ages of U.S. presidents at inauguration.Cumulative relative frequency (%)02040608010040706560555045Age at inaugurationSome people refer to cumulativerelative frequency graphs as “ogives”(pronounced “o-jives”).

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