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# 5.2: The Median

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Understand the median of a set of data as being an important measure of central tendency.
• Determine the median of a set of numerical data when there is an odd number of values and an even number of values.
• Understand the application of the median to real-world problems.

Before class begins, bring out the blocks that you and your classmates chose from the pail for the lesson on mean. In addition, have the grid paper on display where each student in your class posted his or her number of blocks.

To begin the class, refer to the comments on the measures of central tendency that were recorded from the previous lesson, when the brainstorming session occurred. Highlight the comments that were made with regard to the median of a set of data and discuss this measure of central tendency with your classmates. Once the discussion has been completed, choose a handful of blocks like you did before, with your classmates doing the same.

Next, form a line with your classmates as a representation of the data from smallest to largest. In other words, those with the largest number of blocks should be at one end of the line, and those with the fewest number of blocks should be at the other end of the line. Have those at the ends of the line move out from the line 2 at a time, with 1 from each end leaving the line at the same time. This movement from the ends should enforce the concept of what is meant by center, and as you and your classmates move away, you will see that the last 1 or 2 people remaining in the line actually represent the center of the data.

A similar activity can be done by using the grid paper chart. Instead of you and your classmates moving from a line, you could simply remove your post-it notes the same way.

Another simple activity to reenforce the concept of center is to place 5 desks in a single row and to have one of your classmates sit in the middle desk.

Your classmate should be sitting in desk number 3. The other students in your class will quickly notice that there are 2 desks in front of your classmate and 2 desks behind your classmate. Therefore, your classmate is sitting in the desk in the middle position, which is the median of the desks.

From the discussion, the activity with the blocks, and the activity with the desks, you and your classmates should have an understanding of the meaning of the median with respect to a set of data. Here is another example. The test scores for 7 students were 25, 55, 58, 64, 66, 68 and 70. The mean mark is 48.6, which is lower than all but 1 of the student’s marks. The one very low mark of 25 has caused the mean to be skewed. A better measure of the average performance of the 7 students would be the middle mark of 64. The median is the number in the middle position once the data has been organized. Organized data is simply the numbers arranged from smallest to largest or from largest to smallest. 64 is the only number for which there are as many values above it as below it in the set of organized data, so it is the median. The median for an odd number of data values is the value that divides the data into 2 halves. If n\begin{align*}n\end{align*} represents the number of data values and n\begin{align*}n\end{align*} is an odd number, then the median will be found in the n+12\begin{align*}\frac{n+1}{2}\end{align*} position.

Example 10

Find the median of the following data:

a) 12, 2, 16, 8, 14, 10, 6

b) 7, 9, 3, 4, 11, 1, 8, 6, 1, 4

Solution:

a) The first step is to organize the data, or arrange the numbers from smallest to largest.

12,2,16,8,14,10,62,6,8,10,12,14,16\begin{align*}12, 2, 16, 8, 14, 10, 6 \qquad \rightarrow \qquad 2, 6, 8, 10, 12, 14, 16\end{align*}

The number of data values is 7, which is an odd number. Therefore, the median will be found in the n+12\begin{align*}\frac{n+1}{2}\end{align*} position.

n+12=7+12=82=4\begin{align*}\frac{n+1}{2}=\frac{7+1}{2}=\frac{8}{2}=4\end{align*}

In this case, the median is the value that is found in the 4th\begin{align*}4^{\text{th}}\end{align*} position of the organized data.

2,6,8,10,12,14,16\begin{align*}2, 6, 8, \boxed{10}, 12, 14, 16\end{align*}

This means that the median is 10.

b) The first step is to organize the data, or arrange the numbers from smallest to largest.

7,9,3,4,11,1,8,6,1,41,1,3,4,4,6,7,8,9,11\begin{align*}7, 9, 3, 4, 11, 1, 8, 6, 1, 4 \qquad \rightarrow \qquad 1, 1, 3, 4, 4, 6, 7, 8, 9, 11\end{align*}

The number of data values is 10, which is an even number. Therefore, the median will be the mean of the number found before the n+12\begin{align*}\frac{n+1}{2}\end{align*} position and the number found after the n+12\begin{align*}\frac{n+1}{2}\end{align*} position.

n+12=10+12=112=5.5\begin{align*}\frac{n+1}{2}=\frac{10+1}{2}=\frac{11}{2}=5.5\end{align*}

The number found before the 5.5 position is 4, and the number found after the 5.5 position is 6.

1,1,3,4,4,6,7,8,9,11\begin{align*}1, 1, 3, 4, \boxed{4, 6}, 7, 8, 9, 11\end{align*}

This means that the median is 4+62=102=5\begin{align*}\frac{4+6}{2}=\frac{10}{2}=5\end{align*}.

Example 11

The amount of money spent by each of 15 high school girls for a prom dress is shown below:

$250$150$175$375$325$300$195$400$450$225$300$360$275$350425\begin{align*}& \250 && \175 && \325 && \195 && \450 && \300 && \275 && \350 && \425\\ & \150 && \375 && \300 && \400 && \225 && \360\end{align*} What is the median price spent on a prom dress? Solution:150$400$175$425$195$450$225$250$275$300$300$325$350$360$375\begin{align*}& \150 && \175 && \195 && \225 && \250 && \275 && \300 && \boxed{\300} && \325 && \350 && \360 && \375\\ & \400 && \425 && \450\end{align*}

The prices have been organized from least to greatest, and the number of prices is an odd number. Therefore, the median will be in the n+12\begin{align*}\frac{n+1}{2}\end{align*} position: n+12=15+12=162=8\begin{align*}\frac{n+1}{2}=\frac{15+1}{2}=\frac{16}{2}=8\end{align*}.

The median price is \$300, which is the 8th\begin{align*}8^{\text{th}}\end{align*} position.

Example 12

The students from a local high school volunteered to clean up the playground as an act of community service. The numbers of pop cans collected by 20 of the students are shown in the following table:

1622108 1412361812103426446 203125159 13\begin{align*}& 16 \quad 22 \quad 10 \quad 8 \quad \ 14\\ & 12 \quad 36 \quad 18 \quad 12 \quad 10\\ & 34 \quad 26 \quad 44 \quad 6 \quad \ 20\\ & 31 \quad 25 \quad 15 \quad 9 \quad \ 13\end{align*}

What is the median number of pop cans collected by a student?

Solution:

6  8 9 1010121213141516182022252631343644\begin{align*}& 6 \quad \ \ 8 \quad \ 9 \quad \ 10 \quad 10\\ & 12 \quad 12 \quad 13 \quad 14 \quad 15\\ & 16 \quad 18 \quad 20 \quad 22 \quad 25\\ & 26 \quad 31 \quad 34 \quad 36 \quad 44\end{align*}

There is an even number of data values in the table, so the median will be the mean of the number before the n+12\begin{align*}\frac{n+1}{2}\end{align*} position and the number after the n+12\begin{align*}\frac{n+1}{2}\end{align*} position: n+12=20+12=212=10.5\begin{align*}\frac{n+1}{2}=\frac{20+1}{2}=\frac{21}{2}=10.5\end{align*}.

The number before the 10.5 position is 15, and the number after the 10.5 position is 16. Therefore, the median number of pop cans collected by a student is 15+162=312=15.5\begin{align*}\frac{15+16}{2}=\frac{31}{2}=15.5\end{align*}.

Often, the number of data values is quite large, and the task of organizing the data can take a great deal of time. To help organize data, the TI-83 calculator can be used. The following example will show you how to use the calculator to organize data and find the median for the data values.

Example 13

The local police department spent the holiday weekend ticketing drivers who were speeding. 50 locations within the state were targeted as being ideal spots for drivers to exceed the posted speed limit. The number of tickets issued during the weekend in each of the locations is shown in the following table. What is the median number of speeding tickets issued?

3212158 1642 9 18111024186 172141 3  5 352713261628313  7 37101923337 253640 1521384617379  233 41 23291940\begin{align*}& 32 \quad 12 \quad 15 \quad 8 \quad \ 16 \quad 42 \ \quad 9 \quad \ 18 \quad 11 \quad 10\\ & 24 \quad 18 \quad 6 \quad \ 17 \quad 21 \quad 41 \quad \ 3 \ \quad \ 5 \quad \ 35 \quad 27\\ & 13 \quad 26 \quad 16 \quad 28 \quad 31 \quad 3 \quad \ \ 7 \quad \ 37 \quad 10 \quad 19\\ & 23 \quad 33 \quad 7 \quad \ 25 \quad 36 \quad 40 \quad \ 15 \quad 21 \quad 38 \quad 46\\ & 17 \quad 37 \quad 9 \quad \ \ 2 \quad 33 \quad \ 41 \quad \ 23 \quad 29 \quad 19 \quad 40\end{align*}

Solution:

Using the TI-83 calculator:

Step 1:

Step 2:

The numbers that you entered into L1 are now sorted from smallest to largest.

Step 3:

You can now scroll down the list to reveal the ordered numbers.

There are 50 data values in the table. The median will be the mean of the number before the n+12\begin{align*}\frac{n+1}{2}\end{align*} position and the number after the n+12\begin{align*}\frac{n+1}{2}\end{align*} position: n+12=50+12=512=25.5\begin{align*}\frac{n+1}{2}=\frac{50+1}{2}=\frac{51}{2}=25.5\end{align*}. The number before the 25.5 position is 19, and the number after the 25.5 position is 21. This means that the median is 19+212=402=20\begin{align*}\frac{19+21}{2}=\frac{40}{2}=20\end{align*}.

2 3 3  5 67  7 8  9  91010111213 15 15161617171818191921212323242526272829  31323333353637373840  4041414246\begin{align*}& \ 2 \quad \ 3 \ \quad 3 \ \ \quad 5 \quad \ 6 \qquad 7 \quad \ \ 7 \quad \ 8 \quad \ \ 9 \quad \ \ 9\\ & 10 \quad 10 \quad 11 \quad 12 \quad 13 \quad \ 15 \ \quad 15 \quad 16 \quad 16 \quad 17\\ & 17 \quad 18 \quad 18 \quad 19 \quad \boxed{19 \quad 21} \quad 21 \quad 23 \quad 23 \quad 24\\ & 25 \quad 26 \quad 27 \quad 28 \quad 29 \quad \ \ 31 \quad 32 \quad 33 \quad 33 \quad 35\\ & 36 \quad 37 \quad 37 \quad 38 \quad 40 \quad \ \ 40 \quad 41 \quad 41 \quad 42 \quad 46\end{align*}

The median of the data can also be determined without even sorting the data. The following are 2 ways that you can use the TI-83 to determine the median of the values without sorting the data.

Method One:

All of the data values have been entered into L1. The median of the data values can now be determined by using the TI-83 as follows:

Med = 20 indicates that the median of the data values in L1 is 20.

Method Two:

Above the 0 key is the word CATALOG, and this function acts like the yellow pages of a telephone book. When you press 2ND\begin{align*}\boxed{\text{2ND}}\end{align*} 0\begin{align*}\boxed{\text{0}}\end{align*} to access the CATALOG menu, an alphabetical list of terms appears. You can either scroll down to the word median (this will take a long time) or press the blue ÷\begin{align*}\div\end{align*} to access all terms beginning with the letter ‘m’.

Again, the median of the values in L1 is 20.

Whichever method you use, the result will be same. Using technology will save you time when you are determining the median of a set of data values. When a set of data values is given in the form of a frequency table, technology is often used to determine the median.

Example 14

The following data values represent the numbers of chocolate bars sold by students during a recent fundraising campaign. What was the median number of chocolate bars sold?

Number of Bars Number of Students
11 6
12 8
13 5
14 13
15 17
16 15

Solution:

Using the TI-83 calculator:

When data is entered into a frequency table, a column that displays the cumulative frequency is often included. This column is simply the sum of the frequencies up to and including that frequency. The median can be determined by using the information that is presented in the cumulative frequency column.

Example 15

The following table shows the number of goals scored by Ashton during each of 26 hockey games. What is the median number of goals scored by Ashton during a game?

Solution:

There are 26 data values, which is an even number. The middle position is n+12=26+12=272=13.5\begin{align*}\frac{n+1}{2}=\frac{26+1}{2}=\frac{27}{2}=13.5\end{align*}, and the median is the sum of the numbers above and below position 13.5 divided by 2. According to the table, the numbers in the 13th\begin{align*}13^{\text{th}}\end{align*} and 14th\begin{align*}14^{\text{th}}\end{align*} positions are 2’s. Therefore, the median is 2+22=42=2\begin{align*}\frac{2+2}{2}=\frac{4}{2}=2\end{align*} goals.

Lesson Summary

You have learned the significance of the median as it applies to a set of numerical data. You have also learned how to calculate the median of a set of data values, whether the number of values is an odd number or an even number. In addition, you learned that if an outlier affects the mean of a set of data values, then the median is the better measure of central tendency to use. The use of technology in calculating the median was also demonstrated in this lesson.

Points to Consider

• The median of a set of data values cuts the data in half. Is only the median of an entire set of data a useful value?
• Is the median of a set of data useful in any other aspect of statistics?
• Other than as a numerical value, can the median be used to represent data in any other way?

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