# How do you find the best measure of center and variability

### 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 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.

### Note

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 (read $x$ bar): $\displaystyle\overline{{x}}$.

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:

$1$; $1$; $1$; $2$; $2$; $3$; $4$; $4$; $4$; $4$; $4$

$\displaystyle\overline{{x}}=\frac{{{1}+{1}+{1}+{2}+{2}+{3}+{4}+{4}+{4}+{4}+{4}}}{{11}}={2.7}$$\displaystyle\overline{{x}}=\frac{{{3}{({1})}+{2}{({2})}+{1}{({3})}+{5}{({4})}}}{{11}}={2.7}$

In the second example, the frequencies are $3$, $2$, $1$, and $5$.

You can quickly find the location of the median by using the expression $\displaystyle\frac{{{n}+{1}}}{{2}}$.

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 $\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{97}+{1}}}{{2}}={49}$. The median is the $49$th value in the ordered data. If the total number of data values is $100$, then $\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{100}+{1}}}{{2}}$ = $50.5$. The median occurs midway between the $50$th and $51$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.

### Example

AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (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.

Show Solution

The calculation for the mean is $\displaystyle\overline{{x}}=\frac{{{[{3}+{4}+{({8})}{({2})}+{10}+{11}+{12}+{13}+{14}+{({15})}{({2})}+{({16})}{({2})}+\ldots+{35}+{37}+{40}+{({44})}{({2})}+{47}]}}}{{40}}={23.6}$

To find the median, $M$, first use the formula for the location. The location is: $\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{40}+{1}}}{{2}}={20.5}$

Starting at the smallest value, the median is located between the $20$th and $21$st values (the two $24$s):

$3$; $4$; $8$; $8$; $10$; $11$; $12$; $13$; $14$; $15$; $15$; $16$; $16$; $17$; $17$; $18$; $21$; $22$; $22$; $24$; $24$; $25$; $26$; $26$; $27$; $27$; $29$; $29$; $31$; $32$; $33$; $33$; $34$; $34$; $35$; $37$; $40$; $44$; $44$; $47$

$\displaystyle{M}=\frac{{{24}+{24}}}{{2}}={24}$

## Finding the Mean and the Median Using the TI-83, 83+, 84, 84+ Calculator

Clear 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.

$\displaystyle\overline{{x}}$= $23.6$, $M$ = $24$

### Try It

The 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.

$3$; $4$; $5$; $7$; $7$; $7$; $7$; $8$; $8$; $9$; $9$; $10$; $10$; $10$; $10$; $10$; $11$; $12$; $12$; $13$; $14$; $14$; $15$; $15$; $17$; $17$; $18$; $19$; $19$; $19$; $21$; $21$; $22$; $22$; $23$; $24$; $24$; $24$; $24$

Show Solution
Mean: $3 + 4 + 5 + 7 + 7 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 11 + 12 + 12 + 13 + 14 + 14 + 15 + 15 + 17 + 17 + 18 + 19 + 19 + 19 + 21 + 21 + 22 + 22 + 23 + 24 + 24 + 24 = 544$

$\displaystyle\frac{{544}}{{39}}={13.95}$

Median: Starting at the smallest value, the median is the $20$th term, which is $13$.

### example

Suppose that in a small town of $50$ people, one person earns $$5,000,000$ per year and the other $49$ each earn$$30,000$. Which is the better measure of the center: the mean or the median?

Show Solution

$\displaystyle\overline{{x}}=\frac{{5,000,000}+{49}({30,000})}{{50}}={129400}$, $\\{M}={30000}$

(There are $49$ people who earn $$30,000$ and one person who earns$$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.

### Try It

In a sample of $60$ households, one house is worth $$2,500,000$. Half of the rest are worth$$280,000$, and all the others are worth $$315,000$. Which is the better measure of the center: the mean or the median? Show Solution The median is the better measure of the center than the mean because $59$ of the values are$$280,000$ and one is $$2,500,000$. The$$2,500,000$ is an outlier. Either $$280,000$ or$$315,000$ 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.

### Example

Statistics exam scores for $20$ students 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.

Show Solution

The most frequent score is $72$, which occurs five times. Mode = $72$.

### Try It

The number of books checked out from the library from $25$ students are as follows:

$0$, $0$, $0$, $1$, $2$, $3$, $3$, $4$, $4$, $5$, $5$, $7$, $7$, $7$, $7$, $8$, $8$, $8$, $9$, $10$, $10$, $11$, $11$, $12$, $12$

Find the mode.

Show Solution
The most frequent number of books is $7$, which occurs four times. Mode = $7$.

### Example

Five real estate exam scores are $430$, $430$, $480$, $480$, $495$. The data set is bimodal because the scores $430$ and $480$ 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.

### Note

The 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 It

Five credit scores are $680$, $680$, $700$, $720$, $720$. The data set is bimodal because the scores $680$ and $720$ each occur twice. Consider the annual earnings of workers at a factory. The mode is $$25,000$ and occurs $150$ times out of $301$. The median is$$50,000$ and the mean is $$47,500$. What would be the best measure of the center? Show Solution Because$$25,000$ 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 Mean

The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean $\displaystyle\overline{{x}}$ of the sample is very likely to get closer and closer to $µ$. 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. Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below.

# of moviesRelative Frequency$0$$\displaystyle\frac{{5}}{{30}}$$1$$\displaystyle\frac{{15}}{{30}}$$2$$\displaystyle\frac{{6}}{{30}}$$3$$\displaystyle\frac{{3}}{{30}}$$4$$\displaystyle\frac{{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 $\displaystyle\overline{{x}}$ is an example of a statistic which estimates the population mean $μ$.

### Calculating the 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:
$\displaystyle\text{mean}=\frac{{\text{data sum}}}{{\text{number of data values}}}$. 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 $\displaystyle\frac{{\text{lower boundary } + \text{ upper boundary}}}{{2}}$ We can now modify the mean definition to be $\displaystyle\text{Mean of Frequency Table} = \frac{\sum\nolimits{fm}}{\sum\nolimits{f}}$ where $f$ = the frequency of the interval and $m$ = the midpoint of the interval.

### example

A frequency table displaying professor Blounts last statistic test is shown. Find the best estimate of the class mean.

Grade IntervalNumber of Students$5056.5$$1$$56.562.5$$0$$62.568.5$$4$$68.574.5$$4$$74.580.5$$2$$80.586.5$$3$$86.592.5$$4$$92.598.5$$1$
Show Solution

Find the midpoints for all intervals

Grade IntervalMidpoint$5056.5$$53.25$$56.562.5$$59.5$$62.568.5$$65.5$$68.574.5$$71.5$$74.580.5$$77.5$$80.586.5$$83.5$$86.592.5$$89.5$$92.598.5$$95.5$
• Calculate the sum of the product of each interval frequency and midpoint. $\sum\nolimits{fm}$
• $53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.2$
• $\displaystyle\mu=\frac{{\sum{f}{m}}}{{\sum{f}}} =\frac{{1460.25}}{{19}}={76.86}$

### Try It

Maris 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$03.5$$3$$3.57.5$$7$$7.511.5$$12$$11.515.5$$7$$15.519.5$$9$

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 $1.75$, $5.5$, $9.5$, $13.5$, $17.5$. $Mean = (1.75)(3) + (5.5)(7) + (9.5)(12) + (13.5)(7) + (17.5)(9) = 409.75$

## Review

The 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

$\displaystyle\mu=\frac{{\sum{f}{m}}}{{\sum{f}}}$

Where $f$ = interval frequencies and $m$ = interval midpoints.

## References

Data 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).