6.2: The Median
Learning Objectives
 Understand the median of a set of numerical data.
 Compute the median of a given set of data.
 Understand the mean of a set of data as it applies to real world situations.
Introduction
Young players from the minor hockey league have decided to order team wind suits. They must have their measurements taken to ensure a proper fit. The waist measurement for each of the boys was taken and following are the results:
Andy –27in. Barry – 27in. Juan – 23in. Miguel – 27.5 in. Nick – 28in.
Robert – 22in. Sheldon – 24in. Trevor – 25in. Walter – 26.5in.
What is the median of these waist measurements?
You will be able to answer this question once you understand what is meant by the median of the waist measurements.
The test scores for five students were 31, 62, 66, 71 and 73. The mean mark is 60.6 which is lower than all but one of the student’s marks. The mean has been lowered by the one very low mark. A better measure of the average performance of the five students would be the middle mark of 66. The median is the middle number, that number for which there are as many above it as below it in a set of organized data. Organized data is simply the numbers arranged from smallest to largest or from largest to smallest. The median, for an odd number of data, is the value that divides the data into two halves. If \begin{align*}n\end{align*} represents the number of data and \begin{align*}n\end{align*} is an odd number, then the median will be found in position \begin{align*}\frac{n + 1}{2}\end{align*}.
If \begin{align*}n\end{align*} represents the number of data and \begin{align*}n\end{align*} is even, then the median will be the mean of the two values found before and after the \begin{align*}\frac{n + 1}{2}\end{align*} position.
Example 1: Find the median of:
a) 10, 2, 14, 6, 8, 12, 4
b) 3, 9, 2, 5, 7, 1, 6, 4, 2, 5
Solution:
a) The first step is to organize the data – arrange the numbers from smallest to largest.
\begin{align*}10, 2, 14, 6, 8, 12, 4 \qquad \rightarrow \qquad 2, 4, 6, 8, 10, 12, 14\end{align*}
The number of data is an odd number so the median will be found in the \begin{align*}\frac{n + 1}{2}\end{align*} position.
\begin{align*}\frac{n + 1}{2} = \frac{7 + 1}{2} = \frac{8}{2} = 4\end{align*}
The median is the value that is found in the \begin{align*}4^{th}\end{align*} position.
\begin{align*}2, 4, 6, \fbox{8}, 10, 12, 14\end{align*}
The median is 8.
b) The first step is to organize the data – arrange the numbers from smallest to largest.
\begin{align*}3, 9, 2, 5, 7, 1, 6, 4, 2, 5 \qquad \rightarrow \qquad 1, 2, 2, 3, 4, 5, 5, 6, 7, 9\end{align*}
The number of data is an even number so the median will be the mean of the number found before and the number found after the \begin{align*}\frac{n + 1}{2}\end{align*} position.
\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 is 5.
\begin{align*}1, 2, 2, 3, \fbox{4, 5}, 5, 6, 7, 9\end{align*}
Therefore the median is \begin{align*}\frac{4 + 5}{2} = \frac{9}{2} = 4.5\end{align*}
Example 2: The weekly earnings for workers at a local factory are as follows:
\begin{align*}& \$450 \quad \$550 \quad \$425 \quad \$600 \quad \$375 \quad \$475 \quad \$550 \quad \$500 \quad \$425\\ \\ & \$400 \quad \$500 \quad \$475 \quad \$525 \quad \$450 \quad \$575\end{align*}
What is the median of the earnings?
Solution:
\begin{align*}& \$375 \quad \$400 \quad \$425 \quad \$425 \quad \$450 \quad \$450 \quad \$475 \quad \fbox{\$475} \quad \$500\\ \\ & \$500 \quad \$525 \quad \$550 \quad \$550 \quad \$575 \quad \$600\end{align*}
There is an odd number of data so the median will be the value in the \begin{align*}8^{th}\end{align*} position.
The median of the earnings is \begin{align*}\$475\end{align*}.
Often a survey will result in a large number of data and organizing the data to determine the median can take a great deal of time. To help with this task, you can use the TI83 calculator.
Example 3: A local Internet company conducted a survey of 50 users of home computers with Internet access to estimate the number of hours they spent each week on the Internet. The following table contains the estimates provided by the users:
\begin{align*}& 12\; \quad 15\; \quad 25\; \quad 11\; \quad 8\; \quad \ 20\; \ \ 15\; \quad 14\; \quad 7\; \ \quad 10\; \\ & 18\; \quad 13\; \quad 8\; \quad \ 23\; \quad 28\; \quad 3\; \quad \ 16\; \quad 24 \quad 10\; \quad 5\; \\ & 18\; \quad 25\; \quad 12\; \quad 8\; \quad \ 13\; \quad 15 \quad 10\; \quad 12\; \quad 5\; \ \quad 10\; \\ & 14\; \quad 22\; \quad 16\; \quad 6\; \quad \ 19\; \quad 18\; \quad 4\; \ \quad 12 \quad 20\; \quad 13\; \\ & 5\;\; \ \quad 18\; \quad 24\; \quad 6\; \quad \ 3\; \quad \ 16\; \quad 21\; \quad 26\; \ \ 7\; \ \quad \ 9\;\end{align*}
What is the median number of hours the users spent on the Internet?
Solution: Using the TI83 calculator:
(Step One)
(Step Two)
Now go back to your list by repeating Step One. Your numbers are now organized  in order from smallest to largest.
\begin{align*}& 3\; \ \quad 3\; \ \quad 4 \ \quad 5\; \quad \ 5\; \quad 5\; \ \quad 6 \quad \ 6\; \quad \ 7 \ \quad 7 \\ & 8\; \quad \ 8\; \quad \ 8 \quad \ 9 \quad \ 10 \quad 10 \quad 10 \quad 10 \quad 11 \quad 12 \\ & 12 \quad 12 \quad 12 \quad 13 \quad {\color{blue}13} \quad {\color{blue}13} \quad 14 \quad 14 \quad 15 \quad 15 \\ & 15 \quad 16 \quad 16 \quad 16 \quad 18 \quad 18 \quad 18 \quad 18 \quad 19 \quad 20 \\ & 20 \quad 21 \quad 22 \quad 23 \quad 24 \quad 24 \quad 25 \quad 25 \quad 26 \quad 28\end{align*}
There is an even number of data so the median will be the mean of the number above and the number below the \begin{align*}\frac{n + 1}{2} = \frac{50 + 1}{2} = 25.5\end{align*} position. The number below is 13 and the number above is 13.
This result can be confirmed by using the TI83 calculator. You already have the data entered and sorted.
Therefore the median number of hours the users spent on the Internet was 13. Now you should be able to answer the question that was posed at the beginning of the lesson. The boys had their waist measurements taken so they could order team wind suits.
The results were:
Andy –27in. Barry – 27in. Juan – 23in. Miguel – 27.5 in. Nick – 28in.
Robert – 22in. Sheldon – 24in. Trevor – 25in. Walter – 26.5in.
Solution:
22, 23, 24, 25, 26.5, 27, 27, 27.5, 28
\begin{align*}\frac{n + 1}{2} = \frac{9 + 1}{2} = 5\end{align*} The median is the number in the \begin{align*}5^{th}\end{align*} position.
\begin{align*}22, 23, 24, 25, \fbox{26.5}, 27, 27, 27.5, 28\end{align*}
The median of the waist measurements is 26.5 inches.
Lesson Summary
The median is one of the other measures of Central Tendency and is often used in statistics. You know how to compute the median of a given set of data when there is an even number of data and when there is an odd number of data. On addition, you have also learned how to use the TI83 calculator to organize large number of data.
Points to Consider
 Is the median of a set of data useful in any other aspect of statistics?
 Is only the median of the entire set of data a useful value?
Review Questions
 Find the median of each of the following sets of numbers:
 25, 33, 38, 64, 56, 38, 35, 55, 48
 10, 20, 17, 12, 23, 22, 18, 25, 12, 21
 34, 45, 52, 37, 58, 49, 30, 29, 56, 41, 55, 38
 114, 101, 123, 112, 108, 128, 106, 118, 121
 The attendance of students in a Mathematics 10 class during one week was 31, 29, 28, 32, 33. What is the median attendance?
 The number of carrots needed to fill a ten pound bag were 169, 184, 176, 173, 171 and 181. What is the median number of carrots?
 The temperature at noon time was recorded for one week in May. The daily noon time temperatures recorded were \begin{align*}82^\circ F, 80^\circ F, 70^\circ F, 68^\circ F, 76^\circ F, 74^\circ F, 64^\circ F\end{align*}. What was the median temperature?
 A waitress received the following tips over a twoweek period: \begin{align*}& \$35.00 \quad \$28.00 \quad \$33.00 \quad \$41.00 \quad \$27.00 \quad \$46.00 \quad \$39.00 \\ & \$25.00 \quad \$31.00 \quad \$36.00 \quad \$28.00 \quad \$43.00 \quad \$48.00 \quad \$36.00\end{align*} What is the median of the tips she received?
 Two dice were thrown together fifteen times and the results are shown below:
Total Roll  Frequency 

2  2 
5  1 
4  3 
11  2 
6  1 
10  1 
12  2 
9  2 
8  1 
What is the median score?
 The price per pound of Granny Smith apples at various supermarkets was \begin{align*}\$1.79, \$1.49, \$1.55, \$1.68, \$1.75, \$1.45, \$1.59, \$1.85, \$1.70, \$1.65\end{align*} What is the median price of the apples?
 A local running club hosted a 200m race. The times of 9 of the runners were recorded as: \begin{align*}24.2s, 22.9s, 23.1s, 25.6s, 22.5s, 24.0s, 23.3s, 22.3s, 24.6s\end{align*} What is the median time of the runners?
 The weights in kilograms of eight young boys were 41, 37, 34, 37, 46, 38, 41, and 44. What is the median weight?
 A student recorded the following marks on 10 Science quizzes: \begin{align*}66, 51, 74, 69, 71, 58, 79, 82, 64, 77\end{align*} What was the median mark?
 The times in minutes taken by a girl walking to improve her lifestyle were 35, 36, 40, 39, 37, 42, and 30. What is the median time?
 A member of the Over 60 bowling team recorded the following scores during a weekend tournament: \begin{align*}88, 109, 85, 97, 89, 111, 94, 121, 99, 88, 102, 81\end{align*} What was the median score?
 A nurse who works relief at the local hospital has been recording her wages for the past eleven weeks. Her wages during this period were: \begin{align*}& \$600 \quad \$420 \quad \$725 \quad \$560 \quad \$400 \quad \$850 \quad \$675 \\ & \$590 \quad \$390 \quad \$700 \quad \$740\end{align*} What was her median wage?
 A Boys and Girls Police Club has members from 11 years of age to 16 years of age. The ages of the fifty members are shown in the following table:
Age of Members(yrs)  Number of Members 

11  5 
12  9 
13  3 
14  11 
15  10 
16  12 
Use the TI83 calculator to determine the median age of the members.
 Bonus: A set of four numbers that begins with the number 5 is arranged from smallest to largest. If the median is 7, what is a possible set of numbers?
Review Answers

 (38)
 (19)
 (43)
 (114)
 31 students
 (174.5)
 \begin{align*}74^\circ F\end{align*}
 \begin{align*}(\$35.50)\end{align*}
 8
 \begin{align*}(\$1.665 \approx \$1.67)\end{align*}
 23.3 seconds
 (39.5 kilograms)
 70 points
 (37 minutes)
 95.5 points
 \begin{align*}(\$600)\end{align*}
 14 years
 (5, 6, 8, 9)
Answer Key for Review Questions (even numbers)
2. 31 students
4. \begin{align*}74^\circ F\end{align*}
6. 8
8. 23.3 seconds
10. 70 points
12. 95.5 points
14. 14 years
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