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# Median

## Middle number of ascending values.

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Median - Probability and Statistics

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.

### The Median

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 \begin{align*}\frac{1}{2}\end{align*} of the data is greater than the median, and \begin{align*}\frac{1}{2}\end{align*} 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.

#### Finding the Median

1. Find the median of \begin{align*}z\end{align*}.

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

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 7th  and 8th numbers in the series: \begin{align*}\frac{7+8}{2}=7.5\end{align*}

\begin{align*}\therefore 7.5 \ \text{is the median of set} \ z\end{align*}

2. Find the median of set \begin{align*}x\end{align*}.

\begin{align*}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 \}\end{align*}

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

\begin{align*}\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}\end{align*} : Total of 9 values

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

\begin{align*}\therefore \ \frac{31}{42} \ \text{is the median of set} \ x\end{align*}

#### Real-World Application: Job Searching

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?

Credit: Nicholas Eckhart
Source: https://www.flickr.com/photos/fanofretail/13957466674

 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

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.

#### Earlier 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.

### Examples

Identify the median of each set.

#### Example 1

\begin{align*}\{12, 17, 32, 63, 85, 12, 54, 23, 39 \}\end{align*}

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.

#### Example 2

\begin{align*}\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} \end{align*}

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.

#### Example 3

\begin{align*}\{93, 91, 95, 92, 92, 93, 95, 94, 97, 93, 86, 92, 94, 89, 92, 91, 92, 93, 94, 100 \}\end{align*}

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.

#### Example 4

\begin{align*}\{275, 281, 269, 280, 268, 278, 279, 274, 275, 281, 285, 285, 278, 269, \\ ~~~~~~~283, 263, 277, 276, 269, 281, 272, 275, 276 \}\end{align*}

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.

### Review

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

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

\begin{align*}\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} \end{align*}

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### Vocabulary Language: English

TermDefinition
cumulative frequency Cumulative frequency is used to determine the number of observations that lie above (or below) a particular value in a data set.
Median The median of a data set is the middle value of an organized data set.
normal distributed If data is normally distributed, the data set creates a symmetric histogram that looks like a bell.
outliers An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.