Univariate Data

NZ Level 5

Median

Lesson

The median is a measure of **central tendency**. In other words, it's one way of describing a value that represents the middle or the centre of a data set. The median (which kind of sounds like medium) is the **middle score** in a data set.

Remember!

The data must be ordered (usually in ascending order) to calculate the median.

Which term is in the middle?

Say we have five numbers in our data set: $4$4, $11$11, $15$15, $20$20 and $24$24.

The median would be $15$15 because it is right in the middle. There are two numbers on either side of it.

4, 11, 15, 20, 24

However, if we have a larger data set, we may not be able to see straight away which term is in the middle. There are two methods we can use to help us work this out.

Basically, once your data is ordered, you can cross out a high number and a low number until you only have one number left. Let's check out this process using an example. Here is a data set with nine numbers: $1$1, $1$1, $3$3, $5$5, $7$7, $9$9, $9$9, $10$10, $15$15.

1. Check all your numbers are in ascending order (ie. in order from smallest to largest).

2. Cross the smallest and the largest number out like so:

3. Repeat the process from step 2, making sure you work from the outside in, taking the smallest number and the largest number each time until you have one term left. We can see in this example that the median is 7.

NB. You will only be left with one term if there are an odd number of terms to start with. If there are an even number of terms, you will be left with two terms (if you cross them all out, you've gone too far)! All you need to do if find the average of these two terms.

You can also work out which term will be the middle number using the following formula:

Let $n$`n` be the number of terms:

$\text{middle term }=\frac{n+1}{2}$middle term =`n`+12th term

So if we use the same set of numbers from the previous example:

$1$1, $1$1, $3$3, $5$5, $7$7, $9$9, $9$9, $10$10, $15$15, there are nine numbers in the set. So to work out which value is in the middle:

$\text{middle term }$middle term | $=$= | $\frac{9+1}{2}$9+12 |

$=$= | $5$5 |

This means the fifth term will be the median: $1$1, $1$1, $3$3, $5$5, 7, $9$9, $9$9, $10$10, $15$15.

So again, we find that the median is $7$7.

Let's try that with an even number of terms. Let's look at this data set with four terms: $8$8, $12$12, $17$17, $20$20.

$\text{middle term }$middle term | $=$= | $\frac{4+1}{2}$4+12 |

$=$= | $2.5$2.5th term |

But what is the $2.5$2.5^{th} term?? Just like using the "cross out" method, the $2.5$2.5^{th} term means the average between the second and the third term. Again, remember your data must be in order before you count the terms. So in this example, the median will be the average of $12$12 and $17$17.

$\text{median }$median | $=$= | $\frac{12+17}{2}$12+172 |

$=$= | $14.5$14.5 |

Find the median from the frequency distribution table:

Score | Frequency |
---|---|

$23$23 | $2$2 |

$24$24 | $26$26 |

$25$25 | $37$37 |

$26$26 | $24$24 |

$27$27 | $25$25 |

Write down $4$4 consecutive odd numbers whose median is $40$40.

Write all solutions on the same line separated by a comma.

Solve the following using the bar graph:

Find the total number of scores.

Find the median.

Plan and conduct surveys and experiments using the statistical enquiry cycle:– determining appropriate variables and measures;– considering sources of variation;– gathering and cleaning data;– using multiple displays, and re-categorising data to find patterns, variations, relationships, and trends in multivariate data sets;– comparing sample distributions visually, using measures of centre, spread, and proportion;– presenting a report of findings