<meta http-equiv="refresh" content="1; url=/nojavascript/"> Median ( Read ) | Statistics | CK-12 Foundation
You are viewing an older version of this Concept. Go to the latest version.

# Median

%
Progress
Practice Median
Progress
%
Median - Probability and Statistics

#### Objective

Here you will investigate the calculation and use of the median value of a data set.

#### Concept

What is the median value of a data set? Why would you need to know how to find it if you already know how to find the mean? Is there really any difference? Can you think of a situation where the mean and median are exactly the same number?

In the lesson below we will discuss all of these concepts, and you should have no problem with answering these questions when we review them after example C.

#### Watch This

http://youtu.be/kZcuGPoqM9c ehow – Math Definitions: What is a Median Value?

#### Guidance

The median is defined as the value representing the “middle number” of a data set that has been ordered by increasing value, meaning that exactly  $\frac{1}{2}$ of the data is greater than the median, and $\frac{1}{2}$ is less. Finding the median value of a set is a very simple process, requiring very little or no actual calculation at all. Regardless, it can be an excellent representation of the central tendency of a set. The median is particularly useful when evaluating sets with multiple non-representative outliers, since the median is not sensitive to very large or very small values at the extremes of a data set.

To identify the median:

• First, organized your set in ascending numerical order and count the values:
• Second:
• If there are an odd number of values, the median is the middle number in the series.
• If there is an even number of values, the median is the arithmetic mean of the two middle numbers in the series.

Example A

Find the median of $z$ .

$z=\{1, 10, 3, 8, 5, 6, 7, 4, 9, 2, 13, 12, 15, 19 \}$

Solution: First, organize the numbers in ascending numerical order and count the values:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 19: total of 14 values

Since there is an even number of values (14 of them), the median is the arithmetic mean of the 7 th   and 8 th numbers in the series: $\frac{7+8}{2}=7.5$

$\therefore 7.5 \ \text{is the median of set} \ z$

Example B

Find the median of set $x$ .

$x=\left \{\frac{1}{8}, \frac{3}{4}, \frac{4}{5}, \frac{19}{21}, \frac{17}{20}, \frac{31}{42}, \frac{2}{3}, \frac{5}{8}, \frac{3}{8}, \right \}$

Solution: First, organize the values in ascending numerical order and count them:

$\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{2}{3}, \frac{31}{42}, \frac{3}{4}, \frac{4}{5}, \frac{17}{20}, \frac{19}{21}$  : Total of 9 values

• Since there are an odd number of values (9 of them), the median is the 5 th value: $\frac{31}{42}$ .

$\therefore \ \frac{31}{42} \ \text{is the median of set} \ x$

Example C

Brian is looking for an entry-level job at a retail store, and is debating between two options. If he wants to apply for the highest paying job, which of the two employers described below should he choose? Why?

 Employer #1 Employer #2 Company’s mean salary: $48,000 Company’s mean salary:$42,000 Company’s median salary $22,000 Company’s median salary:$28,000

Solution: Since Brian is looking for an entry-level position, he would likely earn more income with Employer #2.

Recall that the mean salary is the average of all the salaries in the company, including any managers or owners on the payroll. Since those positions are undoubtedly higher paid, they would artificially raise the mean income well above that of most employees.

By contrast, the median income is probably closer to the salary of a typical salesman, since there would be many more salesmen than managers or owners.

Concept Problem Revisited

What is the median value of a set? Why would you need to know how to find it if you already know how to find the mean? Is there really any difference? Can you think of a situation where the mean and median are exactly the same number?

The median value is the middle number when the set is organized in ascending numerical order. Depending on the composition of the set, there can actually be quite a large difference between the mean and the median, the mean being much more influenced by outliers. In a normally distributed set, the mean, median, and mode are all the same.

#### Vocabulary

The median is the middle number in a set that has been organized in ascending numerical order.

Central tendency is a measure of the central or typical value in a set.

A normally distributed set has, among other characteristics, an equal mean, median, and mode.

#### Guided Practice

Identify the median of each set in Q’s 1 - 4:

1. .

$\{12, 17, 32, 63, 85, 12, 54, 23, 39 \}$

2.  .

$\begin{array}{c|c c c c c c c c c c c c c c c c c c c} 6 & 7 & 7 & 9 \\7&3&4&4&6&8&9 \\8&0&1&2&3&3&3&3&4&4&6&7&7\\9&0\end{array}$

3. .

$\{93, 91, 95, 92, 92, 93, 95, 94, 97, 93, 86, 92, 94, 89, 92, 91, 92, 93, 94, 100 \}$

4. .

$\{275, 281, 269, 280, 268, 278, 279, 274, 275, 281, 285, 285, 278, 269, \\~~~~~~~283, 263, 277, 276, 269, 281, 272, 275, 276 \}$

Solutions:

1. Put the values in numerical order first: 12, 12, 17, 23, 32, 39, 54, 63, 85. There are nine values, an odd count, so we just take the middle number. 32 is the median.

2. First, put the values in numerical order, remembering that, as a stem plot, this data is listed with the tens place to the left of the vertical, and each separate ones place to the right. That means our values are: 67, 67, 69, 73, 74, 74, 76, 78, 79, 80, 81, 82, 83, 83, 83, 83, 84, 84, 86, 87, 87, 90. Since there are 22 values, an even number, we find the arithmetic mean of the middle two: 81 and 82, to get the median. The median is 81.5.

3. Put the values in numerical order: 86, 89, 91, 91, 92, 92, 92, 92, 92, 93, 93, 93, 93, 94, 94, 94, 95, 95, 97, 100. There are 20 numbers, an even count, so we average the middle two, which are both 93. The median is 93.

4. Put the values in numerical order: 263, 268, 269, 269, 269, 272, 274, 275, 275, 275, 276, 276, 277, 278, 278, 279, 280, 281, 281, 281, 283, 285, 285. There are 23 numbers, an odd count, so we just take the middle value. The median is 276.

#### Practice

Find the median value for each set in questions 1 – 11.

1. $\{326, 314, 325, 315, 315, 307, 318, 318, 320, 322, 325, 321, 322, 320, 312, 325, 326, 325 \}$
2. $\{35, 37, 30, 42, 32, 42, 30, 45, 34, 43, 37, 43, 27, 41, 27, 45, 31, 44, 28, 45 \}$
3. $\{123, 167, 150, 132, 152, 128, 129, 160, 140, 121 \}$
4. $\{2120, 3040, 2180, 1892, 923, 9231, 8231 \}$
5. $\{1, 23, 41, 23, 61, 130, 210, 109, 592, 203, 12 \}$
6. $\{23.43, 32.52, 23.92, 32.25, 23.43, 29.55, 28.30, 31.54 \}$
7. $\left \{ \frac{1}{2}, \frac{4}{9}, \frac{3}{7}, \frac{2}{5}, \frac{21}{23}, \frac{16}{27}, \frac{3}{4} \right \}$
8. $\left \{.57, \frac{23}{100}, .05, \frac{17}{100}, \frac{52}{100}, .42, .44, \frac{45}{100} \right \}$
9. $\{12, 1.2, .12, 102, 120, .012, .120, 1202 \}$
10. $\left \{123, 12.3, \frac{12}{30}, \frac{123}{120}, \frac{120}{123}, \frac{30}{12}, \frac{1}{123}, 1.23, .123 \right \}$
11. . .

$\begin{array}{c|c c c c c c c c c c c c c c c c c c c} 2 &2 & 2 \\3&3&5&5&6&9&9&9 \\4&1&2&2&3&3&4&4&8\\5&2&4&5&5&7&7&7\end{array}$