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Center and spread

Mean, median, range and mean absolute deviation, and choosing the center that fits the data.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

In this lesson you calculate the mean and the median, measure spread with the range and the mean absolute deviation, and choose the center that fits the data. The choice is the skill: one extreme value drags the mean and leaves the median where it is, so skewed data calls for the median. Reporting a mean where a median belongs is one of the most common ways numbers mislead, usually without anyone meaning to.

2. What you already know

You can describe a distribution by its center, its spread and its shape, and you can spot an outlier on a dot plot. You found the middle value of a data set by counting to the middle position, and the range by subtracting the lowest value from the highest. You can add a list of numbers and divide with decimals. This lesson turns 'center' and 'spread' into numbers you can calculate, compare and choose between.

3. Words this lesson uses

4. Two ways to find the center, and one way to measure the spread

A measure of center squeezes a whole data set into one typical number. There are two you will use.

The mean is the fair share. Add all the values and divide by how many there are. If five friends have 2, 3, 4, 6 and 10 stickers, together they have 25. Shared out equally, each would get $25 \div 5 = 5$. The mean is 5 stickers, even though nobody has exactly 5.

The median is the middle. Put the values in order and take the one in the middle: in 2, 3, 4, 6, 10 the median is 4. Half the values are at or below it and half are at or above it.

A measure of variability says how spread out the values are. The range (greatest minus least) is the quickest, $10 - 2 = 8$ here, but it only looks at two values. The mean absolute deviation, or MAD, uses every value. Find how far each value is from the mean: 3, 2, 1, 1 and 5. The mean of those distances is $12 \div 5 = 2.4$. A typical friend's sticker count is about 2.4 away from the mean of 5.

A center always travels with a spread. 'The mean is 5' says where the data is; 'the MAD is 2.4' says how tightly it gathers there.

5. Finding the mean

To find the mean:

1. Add all the values to get the total.
2. Count the values.
3. Divide the total by the count.

The mean of 12, 15, 9, 18 and 16 is $70 \div 5 = 14$. The mean does not have to be one of the values, and it does not have to be a whole number: the mean of 3, 4 and 4 is $11 \div 3 \approx 3.67$.

Why does dividing give a fair share? Division splits a total into equal groups. If every value were replaced by the mean, the total would stay the same: $5 \times 14 = 70$. That gives a quick check, and it also lets you work backwards. If a player needs a mean of 14 points over 5 games, she needs $5 \times 14 = 70$ points in all.

The mean is also the balance point. Add the distances of the values below the mean, then the distances of the values above it. The two totals are always equal. For 12, 15, 9, 18, 16 with mean 14: below, $2 + 5 = 7$; above, $1 + 4 + 2 = 7$. If your totals do not match, the mean is wrong.

6. Finding the median

To find the median:

1. Put the values in order from least to greatest.
2. If the count is odd, the median is the middle value, in position $(\text{count} + 1) \div 2$.
3. If the count is even, there are two middle values. The median is halfway between them: add them and divide by 2.

For 7 values the median is the 4th; for 8 values it is halfway between the 4th and the 5th. The ordering step is the one people skip. The middle of the list as it was written down means nothing, because the list could have been written in any order.

The median only cares about order. Change the largest value from 18 to 1,800 and the median of 9, 12, 15, 16, 18 is still 15. That is why the median is called resistant: an outlier cannot drag it far.

7. Choosing between the mean and the median

Both are honest measures of center, but they answer slightly different questions. The mean asks, 'If everything were shared equally, how much would each get?' The median asks, 'What does the one in the middle have?'

• When the data is roughly symmetric with no outliers, the mean and median are close together, and either one works. The mean is often preferred because it uses every value.
• When the data is skewed or has an outlier, the mean gets pulled toward the long tail. The median stays with the bulk of the data, so it describes a typical value better.

Five workers at a small shop earn 30, 32, 35 and 38 thousand dollars a year, and the owner earns 165 thousand. The mean is 60 thousand, more than four of the five people earn. The median, 35 thousand, is what a typical worker actually takes home. An advertisement saying 'average pay: 60 thousand dollars' would be true and still misleading. Whenever you read an average, ask which one it is and whether the data has a long tail.

8. Measuring spread: the range and the MAD

The range is easy to find, but one outlier can make it huge. The mean absolute deviation is steadier, because it averages the distances of all the values.

To find the MAD:

1. Find the mean.
2. Find each value's distance from the mean. Distances are always positive: a value 3 below the mean and a value 3 above are both 3 away.
3. Add the distances.
4. Divide by the number of values.

A small MAD means the values huddle close to the mean, so the mean is a good summary. A large MAD means the values are scattered, and any single number describes them less well. Two classes with the same mean test score of 80 can have a MAD of 2 points (nearly everyone scored 78 to 82) or a MAD of 15 points (scores all over the place). Same center, very different classes. The MAD is in the same units as the data: if the data is in inches, so is the MAD.

9. The method, step by step, and how to check it

To summarize a data set with numbers:

1. Order the values and count them.
2. Find the median from the middle position.
3. Find the mean: total divided by count.
4. Find the range: greatest minus least.
5. Find the MAD if the mean is your center: the mean of the distances from the mean.
6. Choose the center that fits: the mean for symmetric data, the median for skewed data or data with outliers.
7. Report a center with a spread, in the units of the data.

To check:

• The mean and the median must both lie between the least and greatest values.
• Mean times count must give back the total.
• The distances below the mean must add up to the distances above it.
• The MAD can never be bigger than the range, and it is zero only if every value is the same.
• If the mean and median are far apart, look for an outlier or a long tail: the mean has moved toward it.

10. In the world: household income

The U.S. Census Bureau reports that the median household income in the United States was 80,610 dollars in 2023. Why the median and not the mean? Incomes are strongly skewed right. Most households earn somewhere between a few thousand and a couple of hundred thousand dollars, but a small number earn millions, and they form a very long tail. Those few huge incomes pull the mean household income well above 100,000 dollars, a figure that more than half of all households never reach. You can see the same thing on a small scale: if nine households each earn 60 thousand dollars and one earns 2 million, the mean is $(9 \times 60 + 2{,}000) \div 10 = 254$ thousand dollars, while the median is still 60 thousand. When you read a news story about a 'typical' income, home price or salary, check which center it uses.

11. In the world: normal temperatures

Weather forecasters compare each day with its 'normal' high temperature. In the United States, the National Oceanic and Atmospheric Administration computes these normals as means over 30 years, currently 1991 to 2020. Suppose a town's normal high for a spring week is 64 degrees Fahrenheit, and this week's highs were 58, 61, 66, 70, 63, 64 and 66. Their total is 448, and $448 \div 7 = 64$, right at normal. But the distances from 64 are 6, 3, 2, 6, 1, 0 and 2, which add to 20, so the MAD is $20 \div 7 \approx 2.9$ degrees. A week that swung from 58 to 70 feels very different from a week of steady 64s, even with the same mean. That is why forecasters report both how warm it was on average and how far each day was above or below normal.

12. Mistakes to watch for

Taking the middle of an unordered list. Always sort first.

Dividing by the wrong count. The mean divides by how many values there are, including any zeros. A score of 0 is a value.

Forgetting the two middle values. With an even count, the median is halfway between them.

Using the mean for skewed data. One huge value pulls the mean far from the typical value. Use the median.

Letting distances go negative. In the MAD every distance is positive. If you add signed differences you always get 0.

13. The mean of five basketball games

1. A player scores 12, 15, 9, 18 and 16 points in five games. Add the points.

   $12 + 15 + 9 + 18 + 16 = 70$

   The mean begins with the total.

2. Count the games.

   $5 \text{ games}$

   The total will be shared among this many values.

3. Divide the total by the count.

   $70 \div 5 = 14 \text{ points}$

   Shared out equally, each game would have 14 points.

4. Check the balance.

   $\text{below: } 2 + 5 = 7, \qquad \text{above: } 1 + 4 + 2 = 7$

   12 and 9 are 2 and 5 below 14; 15, 18 and 16 are 1, 4 and 2 above. Equal totals mean 14 is the balance point.

5. Say what the mean tells you.

   $\text{mean} = 14 \text{ points per game}$

   The player's scoring averages 14 points, though she never scored exactly 14.

14. The median of eight temperatures

1. The daily high temperatures for eight days were 71, 64, 68, 75, 80, 66, 73 and 69 degrees Fahrenheit. Put them in order.

   $64, \; 66, \; 68, \; 69, \; 71, \; 73, \; 75, \; 80$

   The median is found only in an ordered list.

2. Count the values.

   $8 \text{ values, an even count}$

   An even count has two middle values, not one.

3. Pick out the two middle values.

   $\text{4th} = 69, \qquad \text{5th} = 71$

   Three values sit below the 4th and three above the 5th.

4. Find the number halfway between them.

   $\dfrac{69 + 71}{2} = \dfrac{140}{2} = 70$

   The median of an even count is the mean of the two middle values.

5. Find the range.

   $80 - 64 = 16 \,^{\circ}\text{F}$

   Greatest minus least.

6. Report the center with the spread.

   $\text{median } 70 \,^{\circ}\text{F}, \quad \text{range } 16 \,^{\circ}\text{F}$

   Half the days were 70 degrees or cooler, and the highs varied over 16 degrees.

15. Pay at a small shop: mean or median?

1. Five people at a shop earn 30, 32, 35, 38 and 165 thousand dollars a year. The values are in order; find the median.

   $\text{median} = 35$

   It is the 3rd of 5 values, with two below and two above.

2. Add the salaries.

   $30 + 32 + 35 + 38 + 165 = 300$

   The mean starts from the total.

3. Divide by the number of people.

   $300 \div 5 = 60$

   The mean salary is 60 thousand dollars.

4. Compare the mean with the data.

   $30, 32, 35, 38 < 60$

   Four of the five people earn less than the mean.

5. Find the mean without the owner's salary.

   $(30 + 32 + 35 + 38) \div 4 = 135 \div 4 = 33.75$

   Removing the one outlier drops the mean by more than 26 thousand dollars.

6. Find the median without it.

   $\dfrac{32 + 35}{2} = 33.5$

   The median moves by only 1.5 thousand: it resists the outlier.

7. Choose the center.

   $\text{typical pay} \approx 35 \text{ thousand dollars}$

   The data has one far-out value, so the median describes a typical worker; the mean describes nobody.

16. Pumpkin weights: the mean absolute deviation

1. Five pumpkins weigh 8, 10, 12, 13 and 17 pounds. Add the weights.

   $8 + 10 + 12 + 13 + 17 = 60$

   The MAD is built on the mean, so start there.

2. Divide by 5 for the mean.

   $60 \div 5 = 12 \text{ pounds}$

   The fair-share weight.

3. Find the distances of the lighter pumpkins.

   $12 - 8 = 4, \qquad 12 - 10 = 2$

   Subtract the smaller number from the larger so the distance is positive.

4. Find the distances of the others.

   $12 - 12 = 0, \quad 13 - 12 = 1, \quad 17 - 12 = 5$

   A pumpkin that weighs exactly the mean is 0 away.

5. Add all five distances.

   $4 + 2 + 0 + 1 + 5 = 12$

   This is the total amount the pumpkins stray from the mean.

6. Divide by the number of pumpkins.

   $12 \div 5 = 2.4 \text{ pounds}$

   The MAD is the mean of the distances.

7. Check the balance.

   $\text{below: } 4 + 2 = 6, \qquad \text{above: } 1 + 5 = 6$

   The distances below and above the mean match, so the mean of 12 was right.

8. Say what the MAD tells you.

   $\text{mean } 12 \text{ lb}, \quad \text{MAD } 2.4 \text{ lb}$

   A typical pumpkin weighs about 2.4 pounds more or less than 12 pounds.

17. Your turn: find the mean and the median of 4, 7, 7, 9 and 13

1. Add the values.

   $4 + 7 + 7 + 9 + 13 = 40$

   The mean starts from the total.

2. Divide by the count.

   $40 \div 5 = 8$

   There are five values.

3. Your turn: work this step out. Its working is at the end of the packet.

   Take the middle of the ordered list.

18. Guided practice

Five friends earned 6, 9, 57, 5 and 8 dollars last week. The one who earned 57 dollars runs a dog-walking business. Which number best describes what a typical friend earned?

• the median, 8 dollars
• the mean, 17 dollars
• the highest amount, 57 dollars

19. Guided practice

Five sunflowers are 42, 45, 49, 51 and 53 inches tall. Complete the worked solution to find their mean and their mean absolute deviation.

1. Add the five heights.

   $42 + 45 + 49 + 51 + 53 =$ _s

   The mean starts from the total of all the values.

2. Divide the total by the number of sunflowers.

   $\text{total} \div 5 =$ _mean inches

   The mean shares the total equally among the five flowers.

3. Find each flower's distance from the mean.

   $6, \; 3, \; 1, \; 3, \; 5$

   Two flowers are shorter than the mean and three are taller; each distance is counted as positive.

4. Add the distances.

   $6 + 3 + 1 + 3 + 5 =$ _t

   This is how far the flowers stray from the mean altogether.

5. Divide by the number of flowers.

   $\text{total distance} \div 5 =$ _mad inches

   The mean absolute deviation is the average distance from the mean.

20. Guided practice

Five homerooms collected cans for a food drive: 19, 21, 44, 26 and 50 cans. What is the total, and what is the mean number of cans per homeroom?

The total is _total cans, and the mean is _mean cans per homeroom.

21. Practice

Seven students timed how many minutes they took to finish the same puzzle: 26, 12, 34, 22, 36, 18 and 29. Find the median time and the range.

The median is _med minutes, and the range is _r minutes.

22. Practice

Six students counted their jumping jacks in one minute: 61, 81, 51, 70, 58 and 68. Find the median and the range.

The median is _med jumping jacks, and the range is _r.

23. Practice

Four players scored 9, 1, 10 and 4 points in a video game. Move the marker to the mean score.

0 |——————————| 20

Mark the position with a cross, then write the value:

24. Practice

Five homes sold on one street this year for 295, 229, 1068, 308 and 255 thousand dollars. The 1068 thousand dollar home is a mansion on a large lot. Find the median price, the mean price, and how many of the five homes sold for less than the mean.

The median is _med thousand dollars, the mean is _mean thousand dollars, and _below of the homes sold for less than the mean.

25. Somewhere new

A bowler scored 125, 144 and 159 in her first three games. She wants a mean of 138 over four games. What total does she need, what does she have so far, and what must she score in the fourth game?

She needs a total of _need, has _have so far, and must score _x in the fourth game.

26. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

27. Test question

A bakery checks that its loaves of bread are the same size. Four loaves weigh 491, 502, 496 and 507 grams. Find the mean weight, the total distance of the loaves from the mean, and the mean absolute deviation.

The mean is _mean grams, the distances add up to _tot grams, and the mean absolute deviation is _mad grams.

28. What you can do now

You can compute the mean, median, range and mean absolute deviation, and choose between the centers. Without looking: which center suits data with one very large value, and why? What does a small mean absolute deviation tell you?

Working for the steps left to you

17. Your turn: find the mean and the median of 4, 7, 7, 9 and 13, step 3