The middle of an ordered list
Five students take a quiz and write down their scores in the order they finish: 8, 3, 9, 5 and 6. What is a typical score? The mean is one answer: add the scores and divide by how many there are. The median is another. It is the middle value once the values are in order.
The order matters. Read the list as it was written, and the value in the middle is 9, which is the highest score of all. It sits in the middle of the list only because of the order the students finished in. So put the scores in order from smallest to largest first: 3, 5, 6, 8, 9. Now the middle value is 6, with two scores below it and two scores above it. The median is 6.
One way to find the middle is to walk in from both ends. Cross off the smallest and the largest value together, 3 and 9. Then cross off the next pair, 5 and 8. The one value left is 6.
For a long list, work out the position instead. With n values in order, the median is the value in position . Here n = 5, and , so the median is the 3rd value in the ordered list, which is 6.
On a number line the scores fall into order by themselves. Two dots lie to the left of 6 and two lie to the right, so 6 is the median.
An even number of values
A sixth student finishes late and scores 10. In order, the six scores are 3, 5, 6, 8, 9, 10. Walking in from both ends crosses off 3 and 10, then 5 and 9, and leaves two values in the middle, 6 and 8.
The position rule says the same thing: , so the middle falls halfway between the 3rd value and the 4th. When there are two middle values, the median is the number halfway between them, which is their mean: .
No student scored 7. A median found from two middle values need not be one of the values, and it can be a half: the median of 3, 5, 6 and 9 is .
Three dots lie below 7 and three lie above it. The median, 7, falls in the gap between the two middle scores, 6 and 8.
The mode
Six students say how many books they read over the summer: 5, 2, 9, 5, 4 and 5. The mode is the value that occurs most often. Here 5 occurs three times and every other value occurs once, so the mode is 5 books.
A dot plot shows the mode at a glance. Each value gets one dot, and a value that repeats gets a stack of dots, one above another. The mode is the value with the tallest stack.
The mode is the value, not the number of times it occurs. The answer is 5 books; the 3 is how many students read 5 books.
The stack at 5 is three dots tall, taller than any other, so 5 books is the mode.
No mode, or more than one
A list does not always have exactly one mode. In 1, 3, 4, 7 every value occurs once, so no value occurs most often and the list has no mode. The answer is “no mode”, not 0: 0 is not in the list at all.
In 2, 2, 3, 6, 6 the values 2 and 6 each occur twice, more often than 3. Both are modes. A list with two modes is called bimodal.
The median never has these problems. Every list of numbers has exactly one middle, so every list has exactly one median.
When one value is far from the rest
The mean and the median do not always agree. Go back to the quiz scores 3, 5, 6, 8, 9. Their mean is (3 + 5 + 6 + 8 + 9) ÷ 5 = 31 ÷ 5 = 6.2, close to the median, 6.
Now suppose the top score was 49 instead of 9. The mean becomes (3 + 5 + 6 + 8 + 49) ÷ 5 = 71 ÷ 5 = 14.2. The mean adds every value into its total, so raising one score by 40 raises the total by 40 and the mean by 40 ÷ 5 = 8. The median is still 6: the scores in order are 3, 5, 6, 8, 49, and the middle one has not changed. Four of the five scores are below 14.2, so the mean no longer describes a typical score. The median does.
Worked example: Seven House Prices in One Street, One of Them a Mansion
Question Seven houses in one street were sold last year. The prices were $250 000, $320 000, $380 000, $380 000, $450 000, $520 000 and $1 200 000. (a) Find the mean, the median and the mode of the seven prices. (b) An estate agent advertises the street using the mean price. How many of the seven houses sold for less than that, and which average describes an ordinary house in the street better?
1.Put the prices in order and work in thousands of dollars, which keeps the arithmetic short: 250, 320, 380, 380, 450, 520, 1200. The answers are turned back into dollars at the end.
The seven prices, in thousands of dollars, on one scale. Six of them sit together and one is far to the right. 2.Add them for the mean: 250 + 320 + 380 + 380 + 450 + 520 + 1200 = 3500 thousand. There are 7 prices, so the mean is 3500 ÷ 7 = 500 thousand, that is $500 000.
The prices total 3500 thousand, so the mean is 3500 ÷ 7 = 500 thousand, that is $500 000. 3.(a) With 7 prices in order the median is the 4th of them, which is 380 thousand. The price 380 thousand occurs twice and every other price once, so the mode is 380 thousand as well. The mean is $500 000, the median is $380 000 and the mode is $380 000.
(a) The median is the 4th price, $380 000, and $380 000 occurs twice, so it is the mode as well. 4.Count the prices below the mean. The prices 250, 320, 380, 380 and 450 are all less than 500, and only 520 and 1200 are above it, so 5 of the 7 houses sold for less than the advertised mean.
Every price below the mean is marked: 250, 320, 380, 380 and 450 thousand, which is 5 of the 7 houses. 5.(b) The median, $380 000, describes an ordinary house in the street better. The one price of $1 200 000 is 700 thousand above the next highest, and on its own it lifts the mean by 700 ÷ 7 = 100 thousand. Check: leave that house out and the other six average 2300 ÷ 6 ≈ 383 thousand, which is close to the median.
(b) The median describes an ordinary house better. The one price of $1 200 000 lifts the mean by 700 ÷ 7 = 100 thousand on its own.
Answer: (a) the mean is $500 000, the median is $380 000 and the mode is $380 000; (b) 5 of the 7 houses sold for less than the mean, and the median describes an ordinary house better
Common mistakes
- Taking the median to be halfway between the lowest and the highest price, 250 + 12002 = 725 thousand. That is the midrange, and it uses only the two most extreme prices. The median is the middle value of the ordered list, so it must be counted to.
- Choosing the mean because it is the only average that uses every price. It does use every price, but one price far from the rest pulls it a long way, and here it lands above all but two of the houses. An average that most of the data sits below is not describing the data.