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 mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center.
The 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 sample mean is an x with a bar over it (pronounced “x bar”):
The Greek letter μ (pronounced "mew") represents the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random.
To see that both ways of calculating the mean are the same, consider the sample: 1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4
\[\begin{array}{c}
\bar{x}=\dfrac{1+1+1+2+2+3+4+4+4+4+4}{11}=2.7 \\
\bar{x}=\dfrac{3(1)+2(2)+1(3)+5(4)}{11}=2.7
\end{array}\]
In the second calculation, the frequencies are 3, 2, 1, and 5.
You can quickly find the location of the median by using the expression \(\dfrac{n+1}{2}\).
The letter \(n\) is the total number of data values in the sample. If \(n\) is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If \(n\) 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 97 , then \(\dfrac{n+1}{2}=\dfrac{97+1}{2}=49\). The median is the \(49^{\text {th }}\) value in the ordered data. If the total number of data values is 100 , then \(\dfrac{n+1}{2}=\dfrac{100+1}{2}=50.5\). The median occurs midway between the \(50^{\text {th }}\) and \(51^{\text {st }}\) values. The location of the median and the value of the median are not the same. The upper case letter \(M\) is often used to represent the median. The next example illustrates the location of the median and the value of the median.
A hospital administrator keeps track of the ages (in years) of patients visiting the emergency room over a one-week period (data are sorted from smallest to largest): 3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47;
Calculate the mean and the median.
- Answer
-
The calculation for the mean is:
\[\bar{x}=\dfrac{[3+4+(8)(2)+10+11+12+13+14+(15)(2)+(16)(2)+\ldots+35+37+40+(44)(2)+47]}{40}=23.6\]
To find the median, \(M\), first use the formula for the location. The location is:
\[\dfrac{n+1}{2}=\dfrac{40+1}{2}=20.5\]
Starting at the smallest value, the median is located between the \(20^{\text {th }}\) and \(21^{\text {st }}\) values (the two 24 s ):
\[\begin{array}{l}
3 ; 4 ; 8 ; 8 ; 10 ; 11 ; 12 ; 13 ; 14 ; 15 ; 15 ; 16 ; 16 ; 17 ; 17 ; 18 ; 21 ; 22 ; 22 ; 24 ; 24 ; 25 ; 26 ; 26 ; 27 ; 27 ; 29 ; 29 ; \\
31 ; 32 ; 33 ; 33 ; 34 ; 34 ; 35 ; 37 ; 40 ; 44 ; 44 ; 47 ; \\
M=\dfrac{24+24}{2}=24
\end{array}\]
Suppose that in a study tracking the annual healthcare costs for 50 individuals with a chronic disease, one person incurs an annual cost of $5,000,000 (due to highly specialized, long-term critical care) and the other 49 each incur a much lower, though substantial, annual cost of $30,000.
Which is the better measure of the "center" of the typical annual healthcare cost: the mean or the median?
- Answer
-
\[\begin{array}{l}
\bar{x}=\dfrac{5,000,000+49(30,000)}{50}=129,400 \\
M=30,000
\end{array}\](There are 49 people who incured \(\$ 30,000\) and one person who incured \(\$ 5,000,000\).)
The median is a better measure of the "center" than the mean because 49 of the values are 30,000 and one is \(5,000,000\). The \(5,000,000\) is an outlier. The 30,000 gives us a better sense of the middle of the data.
Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal.
Body Mass Index (BMI) percentile scores for 20 patients are as follows:
50; 53; 59; 59; 63; 63; 72; 72; 72; 72; 72; 76; 78; 81; 83; 84; 84; 84; 90; 93
Find the mode.
- Answer
-
The most frequent score is 72, which occurs five times. Mode = 72.
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.
NOTE
The mode can be calculated for qualitative data as well as for quantitative data. For example, in a survey asking individuals to identify their primary health risk factor from a list of options, if the data set of responses is: High Stress, High Stress, High Stress, Sedentary Lifestyle, Sedentary Lifestyle, Poor Diet, Family History, Smoking, High Blood Pressure, the mode is High Stress. This means 'High Stress' is the most frequently reported primary risk factor in this sample, which can guide public health intervention prioritization.
The Law of Large Numbers and the Mean
The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean \(\bar{x}\) of the sample is very likely to get closer and closer to \(\mu\). This is discussed in more detail later in the text.
Sampling Distributions and Statistic of a Sampling Distribution
You can think of a sampling distribution as a relative frequency distribution with a great many samples. (See Sampling and Data for a review of relative frequency). Suppose thirty randomly selected students were asked the number of times they exercised the previous week. The results are in the relative frequency table shown below.
| # of exercise sessions | Relative Frequency |
|---|---|
| 0 | \(\dfrac{5}{30}\) |
| 1 | \(\dfrac{15}{30}\) |
| 2 | \(\dfrac{6}{30}\) |
| 3 | \(\dfrac{3}{30}\) |
| 4 | \(\dfrac{1}{30}\) |
If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.
A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean \(\bar{x}\) is an example of a statistic which estimates the population mean \(\mu\).
Calculating the Arithmetic Mean of Grouped Frequency Tables
When 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: mean \(=\dfrac{\text { data sum }}{\text { number of data values }}\) We simply need to modify the definition to fit within the restrictions of a frequency table.
Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is \(\dfrac{\text { lower boundary+upper boundary }}{2}\). We can now modify the mean definition to be
Mean of Frequency Table \(=\dfrac{\sum f m}{\sum f}\) where \(f=\) the frequency of the interval and \(m=\) the midpoint of the interval.
A frequency table displaying percentile intervals of Body Mass Index (BMI) scores in a class of Exercise Physiology students is shown. Find the best estimate of the class mean.
| BMI Percentile | Number of Students |
|---|---|
| 50–56.5 | 1 |
| 56.5–62.5 | 0 |
| 62.5–68.5 | 4 |
| 68.5–74.5 | 4 |
| 74.5–80.5 | 2 |
| 80.5–86.5 | 3 |
| 86.5–92.5 | 4 |
| 92.5–98.5 | 1 |
- Answer
-
Find the midpoints for all intervals
Table \(PageIndex{2}\) BMI Percentile Midpoint 50–56.5 53.25 56.5–62.5 59.5 62.5–68.5 65.5 68.5–74.5 71.5 74.5–80.5 77.5 80.5–86.5 83.5 86.5–92.5 89.5 92.5–98.5 95.5 Calculate the sum of the product of each interval frequency and midpoint. \(\sum f m\)
\(53.25(1)+59.5(0)+65.5(4)+71.5(4)+77.5(2)+83.5(3)+89.5(4)+95.5(1)=1460.25\)
\(\mu=\dfrac{\sum f m}{\sum f}=\dfrac{1460.25}{19}=76.86\)