Connecting the Dots... · Lesson 4 of 13
Outliers and Medians
“See why unusual values affect averages and learn to find the centre of sorted data.”
• Find medians for odd and even observation counts. • Recognise outliers without assuming they are mistakes. • Explain how mean and median respond differently to unusual values. • Compare centres using heights, reading counts, and page counts. • Handle repeated observations correctly.
Outliers and Medians
The mean gives equal shares of the total, but a very unusual observation can change that total greatly. Imagine comparing two families when one includes a much younger child. We need to see the heights themselves before deciding whether their means give a useful account of the comparison.
Height of a Family
Yaangba’s family has heights 169, 173, 155, 165, 160, and 164 cm. Poovizhi’s family has heights 170, 173, 165, 118, and 175 cm. Four members of Poovizhi’s family are relatively tall, but its 118 cm observation pulls the total downward.
| Family | Count | Total height (cm) | Mean height (cm) |
|---|---|---|---|
| Yaangba | 6 | 986 | 986 ÷ 6 ≈ 164.33 |
| Poovizhi | 5 | 801 | 801 ÷ 5 = 160.2 |
An observation that differs markedly from most of the other values in the dataset. Whether a value is unusual depends on the collection and context.
The 118 cm height is much smaller than the other heights in that family. Calling it an outlier does not mean it was recorded incorrectly or should be removed. It may be a completely valid height. Here it helps explain why the mean is below the heights of four of the five members.
Finding the median
Instead of distributing the total, we can organise the observations from smallest to largest and identify the centre of their order. This produces the median. Sorting is essential: the middle item in a randomly arranged list has no special meaning.
The middle value of sorted numerical data when the count is odd; the mean of the two middle values when the count is even.
Problem
Find Poovizhi’s family’s median height.
- 1.Sort all five heights: 118, 165, 170, 173, 175.
- 2.The third value is in the middle: two positions precede it and two follow it.
- 3.The median is 170 cm. It gives a different perspective from the 160.2 cm mean.
Problem
Find Yaangba’s family’s median height.
- 1.Sort the six heights: 155, 160, 164, 165, 169, 173.
- 2.The two central positions are the third and fourth, containing 164 and 165.
- 3.Take their mean: (164 + 165) ÷ 2 = 329 ÷ 2 = 164.5 cm.
With repeated values, the numbers of observations strictly below and strictly above the median need not be equal. For 1, 2, 2, 2, 9, the median is 2: only one value is strictly below it and one strictly above it. What matters is the central position in the sorted list. At least half the observations are at or below the median, and at least half are at or above it.
Problem
What happens to Poovizhi’s mean and median if we investigate the four taller family members separately?
- 1.The selected data is 165, 170, 173, 175. Its sum is 683, so mean = 683 ÷ 4 = 170.75 cm.
- 2.Its median is (170 + 173) ÷ 2 = 171.5 cm.
- 3.The original mean increased by 10.55 cm, while the median increased by 1.5 cm. State clearly that this describes a selected subgroup, not the full family.
Are you a bookworm?
Fifteen students reported how many short stories they read: 6, 3, 0, 8, 2, 5, 7, 15, 12, 10, 40, 5, 0, 1, 8. Most counts lie fairly near the lower end, while 40 is far beyond the others. Before calculating, think about which way that high value might move the equal-share value.
Problem
Find the mean and median of the story counts and compare them without 40.
- 1.The total is 122 and the count is 15. Mean = 122 ÷ 15 ≈ 8.13 stories.
- 2.Sorted data: 0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15, 40. The eighth value is 6, so median = 6.
- 3.Without 40, the total becomes 82 across 14 observations. Mean = 82 ÷ 14 ≈ 5.86.
- 4.The two middle values of the selected set are 5 and 6, giving median 5.5. The mean changes more than the median.
The mean responds to the size of each value through the sum. The median responds mainly to ordering and central positions. This explains why a high value may lift the mean much more than the median. It does not mean median never changes or that mean is always the wrong choice.
Are We on the Same Page?
A newspaper has 16, 18, 20, 22, 26, 16, and 10 pages from Monday through Sunday. Unlike the story-count example, the observations are less dominated by one distant value. Comparing the two centres helps us describe this particular collection.
Problem
Find the newspaper’s mean and median page counts.
- 1.Sum = 16 + 18 + 20 + 22 + 26 + 16 + 10 = 128. Mean = 128 ÷ 7 ≈ 18.29 pages.
- 2.Sort the seven values: 10, 16, 16, 18, 20, 22, 26. The fourth value gives median 18.
- 3.The mean and median are close in this dataset. This is an observation about these values, not a proof that every distribution with close centres has the same shape.
The idea of describing a central or representative value around which observations tend to lie. Mean and median are two measures of central tendency.
Do not automatically delete an outlier, or assume the median is unaffected by every change. Also, mean below median does not prove a low outlier exists. Inspect the observations and their context.
For 0, 10, 10, 10, 20, mean and median are both 10 even though the two ends are distant from the middle cluster. Opposite effects on the total can balance. Equal centres do not imply that all observations are close together.
Quiz
What is the median of 9, 1, 7, 3, 5?
What is the median of 2, 4, 10, 12?
What should happen before finding a median?
Which is a sound description of an outlier?
For 1, 2, 2, 2, 9, the median is:
Which conclusion is justified when mean and median are equal?
Practice Problems
- Find the median of Yahapur’s twelve onion prices.
- Find Wahapur’s median price and compare the ordering of the means and medians.
- Find mean and median for 3, 4, 5, 6, 32. Compare them after replacing 32 by 7.
- Can a median be a number not present in the observations? Give an example.
- Why is 118 cm not automatically a measurement error in the family-height example?
- Compare these two groups’ results using mean and median: A = 85, 76, 90, 85, 39, 48, 56, 95, 81, 75; B = 68, 59, 73, 86, 47, 79, 90, 93, 86.
Sorted prices: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59. The middle pair is 35 and 39, so median = (35 + 39) ÷ 2 = ₹37/kg.
Key Takeaways
• Sort every observation before finding the median. • For an even count, average the two middle values. • Repeated values must retain their positions. • Unusually high or low values can strongly affect the mean. • Median often changes less under extreme observations, but it can change. • Choose a representative measure according to the question and distribution.