## Learning Outcomes- Recognize, describe, and calculate the measures of the center of data: mean, median, and mode.
The center of a data set is also a way of describing location. The two most widely used measures of the center of the data are the ## NoteThe words mean and average are often used interchangeably. The substitution of one word for the other is common practice. The technical term is arithmetic mean and average is technically a center location. However, in practice among non-statisticians, average is commonly accepted for arithmetic mean. When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the The Greek letter [latex]μ[/latex] (pronounced mew) represents the To see that both ways of calculating the mean are the same, consider the sample: [latex]1[/latex]; [latex]1[/latex]; [latex]1[/latex]; [latex]2[/latex]; [latex]2[/latex]; [latex]3[/latex]; [latex]4[/latex]; [latex]4[/latex]; [latex]4[/latex]; [latex]4[/latex]; [latex]4[/latex] [latex]\displaystyle\overline{{x}}=\frac{{{1}+{1}+{1}+{2}+{2}+{3}+{4}+{4}+{4}+{4}+{4}}}{{11}}={2.7}[/latex][latex]\displaystyle\overline{{x}}=\frac{{{3}{({1})}+{2}{({2})}+{1}{({3})}+{5}{({4})}}}{{11}}={2.7}[/latex]In the second example, the frequencies are [latex]3[/latex], [latex]2[/latex], [latex]1[/latex], and [latex]5[/latex]. You can quickly find the location of the median by using the expression [latex]\displaystyle\frac{{{n}+{1}}}{{2}}[/latex]. The letter [latex]n[/latex] is the total number of data values in the sample. If [latex]n[/latex] is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If [latex]n[/latex] is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered. For example, if the total number of data values is [latex]97[/latex], then [latex]\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{97}+{1}}}{{2}}={49}[/latex]. The median is the [latex]49[/latex]th value in the ordered data. If the total number of data values is [latex]100[/latex], then [latex]\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{100}+{1}}}{{2}}[/latex] = [latex]50.5[/latex]. The median occurs midway between the [latex]50[/latex]th and [latex]51[/latex]st values. The location of the median and the value of the median are ## ExampleAIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest): [latex]3[/latex]; [latex]4[/latex]; [latex]8[/latex]; [latex]8[/latex]; [latex]10[/latex]; [latex]11[/latex]; [latex]12[/latex]; [latex]13[/latex]; [latex]14[/latex]; [latex]15[/latex]; [latex]15[/latex]; [latex]16[/latex]; [latex]16[/latex]; [latex]17[/latex]; [latex]17[/latex]; [latex]18[/latex]; [latex]21[/latex]; [latex]22[/latex]; [latex]22[/latex]; [latex]24[/latex]; [latex]24[/latex]; [latex]25[/latex]; [latex]26[/latex]; [latex]26[/latex]; [latex]27[/latex]; [latex]27[/latex]; [latex]29[/latex]; [latex]29[/latex]; [latex]31[/latex]; [latex]32[/latex]; [latex]33[/latex]; [latex]33[/latex]; [latex]34[/latex]; [latex]34[/latex]; [latex]35[/latex]; [latex]37[/latex]; [latex]40[/latex]; [latex]44[/latex]; [latex]44[/latex]; [latex]47[/latex] Calculate the mean and the median. Show Solution The calculation for the mean is [latex]\displaystyle\overline{{x}}=\frac{{{[{3}+{4}+{({8})}{({2})}+{10}+{11}+{12}+{13}+{14}+{({15})}{({2})}+{({16})}{({2})}+\ldots+{35}+{37}+{40}+{({44})}{({2})}+{47}]}}}{{40}}={23.6}[/latex] To find the median, [latex]M[/latex], first use the formula for the location. The location is: [latex]\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{40}+{1}}}{{2}}={20.5}[/latex] Starting at the smallest value, the median is located between the [latex]20[/latex]th and [latex]21[/latex]st values (the two [latex]24[/latex]s): [latex]3[/latex]; [latex]4[/latex]; [latex]8[/latex]; [latex]8[/latex]; [latex]10[/latex]; [latex]11[/latex]; [latex]12[/latex]; [latex]13[/latex]; [latex]14[/latex]; [latex]15[/latex]; [latex]15[/latex]; [latex]16[/latex]; [latex]16[/latex]; [latex]17[/latex]; [latex]17[/latex]; [latex]18[/latex]; [latex]21[/latex]; [latex]22[/latex]; [latex]22[/latex]; [latex]24[/latex]; [latex]24[/latex]; [latex]25[/latex]; [latex]26[/latex]; [latex]26[/latex]; [latex]27[/latex]; [latex]27[/latex]; [latex]29[/latex]; [latex]29[/latex]; [latex]31[/latex]; [latex]32[/latex]; [latex]33[/latex]; [latex]33[/latex]; [latex]34[/latex]; [latex]34[/latex]; [latex]35[/latex]; [latex]37[/latex]; [latex]40[/latex]; [latex]44[/latex]; [latex]44[/latex]; [latex]47[/latex] [latex]\displaystyle{M}=\frac{{{24}+{24}}}{{2}}={24}[/latex] ## Finding the Mean and the Median Using the TI-83, 83+, 84, 84+ CalculatorClear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER. Enter data into the list editor. Press STAT 1:EDIT. Put the data values into list L1. Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER. Press the down and up arrow keys to scroll. [latex]\displaystyle\overline{{x}}[/latex]= [latex]23.6[/latex], [latex]M[/latex] = [latex]24[/latex] ## Try ItThe following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median. [latex]3[/latex]; [latex]4[/latex]; [latex]5[/latex]; [latex]7[/latex]; [latex]7[/latex]; [latex]7[/latex]; [latex]7[/latex]; [latex]8[/latex]; [latex]8[/latex]; [latex]9[/latex]; [latex]9[/latex]; [latex]10[/latex]; [latex]10[/latex]; [latex]10[/latex]; [latex]10[/latex]; [latex]10[/latex]; [latex]11[/latex]; [latex]12[/latex]; [latex]12[/latex]; [latex]13[/latex]; [latex]14[/latex]; [latex]14[/latex]; [latex]15[/latex]; [latex]15[/latex]; [latex]17[/latex]; [latex]17[/latex]; [latex]18[/latex]; [latex]19[/latex]; [latex]19[/latex]; [latex]19[/latex]; [latex]21[/latex]; [latex]21[/latex]; [latex]22[/latex]; [latex]22[/latex]; [latex]23[/latex]; [latex]24[/latex]; [latex]24[/latex]; [latex]24[/latex]; [latex]24[/latex] Show Solution Mean: [latex]3 + 4 + 5 + 7 + 7 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 11 + 12 + 12 + 13 + 14 + 14 + 15 + 15 + 17 + 17 + 18 + 19 + 19 + 19 + 21 + 21 + 22 + 22 + 23 + 24 + 24 + 24 = 544 [/latex] [latex]\displaystyle\frac{{544}}{{39}}={13.95}[/latex] Median: Starting at the smallest value, the median is the [latex]20 [/latex]th term, which is [latex]13[/latex]. ## exampleSuppose that in a small town of [latex]50[/latex] people, one person earns $[latex]5,000,000[/latex] per year and the other [latex]49[/latex] each earn $[latex]30,000 [/latex]. Which is the better measure of the center: the mean or the median? Show Solution [latex]\displaystyle\overline{{x}}=\frac{{5,000,000}+{49}({30,000})}{{50}}={129400}[/latex], [latex]\\{M}={30000}[/latex] (There are [latex]49[/latex] people who earn $[latex]30,000[/latex] and one person who earns $[latex]5,000,000[/latex].) The median is a better measure of the center than the mean because [latex]49[/latex] of the values are [latex]30,000[/latex] and one is [latex]5,000,000[/latex]. The [latex]5,000,000[/latex] is an outlier. The [latex]30,000[/latex] gives us a better sense of the middle of the data. ## Try ItIn a sample of [latex]60[/latex] households, one house is worth $[latex]2,500,000[/latex]. Half of the rest are worth $[latex]280,000[/latex], and all the others are worth $[latex]315,000[/latex]. Which is the better measure of the center: the mean or the median? Show Solution The median is the better measure of the center than the mean because [latex]59[/latex] of the values are $[latex]280,000[/latex] and one is $[latex]2,500,000[/latex]. The $[latex]2,500,000[/latex] is an outlier. Either $[latex]280,000[/latex] or $[latex]315,000[/latex] gives us a better sense of the middle of the data. Another measure of the center is the mode. The ## ExampleStatistics exam scores for [latex]20[/latex] students are as follows: [latex]50[/latex], [latex]53[/latex], [latex]59[/latex], [latex]59[/latex], [latex]63[/latex], [latex]63[/latex], [latex]72[/latex], [latex]72[/latex], [latex]72[/latex], [latex]72[/latex], [latex]72[/latex], [latex]76[/latex], [latex]78[/latex], [latex]81[/latex], [latex]83[/latex], [latex]84[/latex], [latex]84[/latex], [latex]84[/latex], [latex]90[/latex], [latex]93[/latex] Find the mode. Show Solution The most frequent score is [latex]72[/latex], which occurs five times. Mode = [latex]72[/latex]. ## Try ItThe number of books checked out from the library from [latex]25[/latex] students are as follows: [latex]0[/latex], [latex]0[/latex], [latex]0[/latex], [latex]1[/latex], [latex]2[/latex], [latex]3[/latex], [latex]3[/latex], [latex]4[/latex], [latex]4[/latex], [latex]5[/latex], [latex]5[/latex], [latex]7[/latex], [latex]7[/latex], [latex]7[/latex], [latex]7[/latex], [latex]8[/latex], [latex]8[/latex], [latex]8[/latex], [latex]9[/latex], [latex]10[/latex], [latex]10[/latex], [latex]11[/latex], [latex]11[/latex], [latex]12[/latex], [latex]12[/latex] Find the mode. Show Solution The most frequent number of books is [latex]7[/latex], which occurs four times. Mode = [latex]7[/latex]. ## ExampleFive real estate exam scores are [latex]430[/latex], [latex]430[/latex], [latex]480[/latex], [latex]480[/latex], [latex]495[/latex]. The data set is bimodal because the scores [latex]430[/latex] and [latex]480[/latex] each occur twice. When is the mode the best measure of the center? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing. ## NoteThe mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red. Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software. ## Try ItFive credit scores are [latex]680[/latex], [latex]680[/latex], [latex]700[/latex], [latex]720[/latex], [latex]720[/latex]. The data set is bimodal because the scores [latex]680[/latex] and [latex]720[/latex] each occur twice. Consider the annual earnings of workers at a factory. The mode is $[latex]25,000[/latex] and occurs [latex]150[/latex] times out of [latex]301[/latex]. The median is $[latex]50,000[/latex] and the mean is $[latex]47,500[/latex]. What would be the best measure of the center? Show Solution Because $[latex]25,000[/latex] occurs nearly half the time, the mode would be the best measure of the center because the median and mean dont represent what most people make at the factory. Watch the following video from Khan Academy on finding the mean, median and mode of a set of data. ## The Law of Large Numbers and the MeanThe Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean [latex]\displaystyle\overline{{x}}[/latex] of the sample is very likely to get closer and closer to [latex]µ[/latex]. This is discussed in more detail later in the text. ## Sampling Distributions and Statistic of a Sampling DistributionYou can think of a
A ## Calculating the Mean of Grouped Frequency TablesWhen only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean: Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is [latex]\displaystyle\frac{{\text{lower boundary } + \text{ upper boundary}}}{{2}}[/latex] We can now modify the mean definition to be [latex]\displaystyle\text{Mean of Frequency Table} = \frac{\sum\nolimits{fm}}{\sum\nolimits{f}}[/latex] where [latex]f[/latex] = the frequency of the interval and [latex]m[/latex] = the midpoint of the interval. ## exampleA frequency table displaying professor Blounts last statistic test is shown. Find the best estimate of the class mean. Grade IntervalNumber of Students[latex]5056.5[/latex][latex]1[/latex][latex]56.562.5[/latex][latex]0[/latex][latex]62.568.5[/latex][latex]4[/latex][latex]68.574.5[/latex][latex]4[/latex][latex]74.580.5[/latex][latex]2[/latex][latex]80.586.5[/latex][latex]3[/latex][latex]86.592.5[/latex][latex]4[/latex][latex]92.598.5[/latex][latex]1[/latex]Show Solution Find the midpoints for all intervals Grade IntervalMidpoint[latex]5056.5[/latex][latex]53.25[/latex][latex]56.562.5[/latex][latex]59.5[/latex][latex]62.568.5[/latex][latex]65.5[/latex][latex]68.574.5[/latex][latex]71.5[/latex][latex]74.580.5[/latex][latex]77.5[/latex][latex]80.586.5[/latex][latex]83.5[/latex][latex]86.592.5[/latex][latex]89.5[/latex][latex]92.598.5[/latex][latex]95.5[/latex]- Calculate the sum of the product of each interval frequency and midpoint. [latex]\sum\nolimits{fm}[/latex] [latex]53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.2[/latex]
- [latex]\displaystyle\mu=\frac{{\sum{f}{m}}}{{\sum{f}}} =\frac{{1460.25}}{{19}}={76.86}[/latex]
## Try ItMaris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data: Hours Teenagers Spend on Video GamesNumber of Teenagers[latex]03.5[/latex][latex]3[/latex][latex]3.57.5[/latex][latex]7[/latex][latex]7.511.5[/latex][latex]12[/latex][latex]11.515.5[/latex][latex]7[/latex][latex]15.519.5[/latex][latex]9[/latex]What is the best estimate for the mean number of hours spent playing video games? Show Solution Find the midpoint of each interval, multiply by the corresponding number of teenagers, add the results and then divide by the total number of teenagers The midpoints are [latex]1.75[/latex], [latex]5.5[/latex], [latex]9.5[/latex], [latex]13.5[/latex], [latex]17.5[/latex]. [latex]Mean = (1.75)(3) + (5.5)(7) + (9.5)(12) + (13.5)(7) + (17.5)(9) = 409.75[/latex] ## ReviewThe mean and the median can be calculated to help you find the center of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set. ## Formula Review[latex]\displaystyle\mu=\frac{{\sum{f}{m}}}{{\sum{f}}}[/latex] Where [latex]f[/latex] = interval frequencies and [latex]m[/latex] = interval midpoints. ## ReferencesData from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013). Demographics: Obesity adult prevalence rate. Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2228&l=en (accessed April 3, 2013). ## Video |