Lesson 8 ~25 min Unit 4 · Data & Chance +85 XP

The Mean

Calculate the arithmetic average from raw data and frequency tables — and know when a single outlier can make the mean misleading.

Today's hook: Your maths class average was 72%. That single number summarises 28 different scores. The mean is the most powerful summary statistic — and the most misunderstood. Learn to calculate it and know when NOT to use it.

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Think First

warm-up

Before you read on — quickly: Five friends earned these amounts in a week of work: $50, $60, $40, $55, $45. What would be a "fair" amount if they shared the total equally? How would you calculate it? Try it, then check your reasoning as you go.

Record your answer in your workbook.

The Big Idea

+5 XP

The mean (arithmetic average) is calculated by adding all values and dividing by the number of values. It is the balance point of the data — the value each person would have if everything were shared equally. Formula: $\bar{x} = \dfrac{\sum x}{n}$

For the earnings: 50 + 60 + 40 + 55 + 45 = 250. There are 5 friends. Mean = 250 ÷ 5 = $50. If each friend got $50, the total would still be $250. Think of it as "levelling out" the data — take from the high values and give to the low values until everyone is equal.

$\bar{x} = \dfrac{\sum x}{n} = \dfrac{250}{5} = 50$

Add ALL values first

Sum every value in the data set before dividing. Don't skip zeros.

Divide by n (count of values)

n is the total number of individual data values, not the number of categories.

Mean can be a decimal

The mean often isn't a whole number. 7 ÷ 2 = 3.5 is a valid mean.

What You'll Master

objectives

Know

The formula $\bar{x} = \frac{\sum x}{n}$ and what each symbol means
That n is the count of data values, not categories
What an outlier is and how it affects the mean

Understand

Why the mean is the balance point of the data
How to calculate mean from a frequency table
Why outliers pull the mean away from the typical value

Can Do

Calculate the mean from a list of raw data values
Calculate the mean using a frequency table
Identify when the median might be a better measure than the mean

Words You Need

vocabulary

Mean ($\bar{x}$)The arithmetic average: sum of all values divided by the number of values.

$\sum x$ (sigma x)The sum of all data values. $\sum$ is the Greek letter sigma, meaning "add them all up".

$n$The number of data values in the set (not the number of categories).

Balance pointThe mean is the point where the data "balances" — high and low values cancel out.

Frequency table meanMultiply each value by its frequency, sum the results, then divide by total frequency.

Outlier effectAn extreme value that pulls the mean higher or lower, away from the typical value.

Spot the Trap

heads-up

Wrong: In a frequency table with categories {1, 2, 3, 4, 5}, dividing the sum by 5 (the number of categories) instead of by the total frequency (e.g. 20 students).

Right: n = total number of data points = sum of all frequencies. In a class of 20 students, n = 20, not 5 (the number of score categories).

Wrong: Trusting the mean when a data set contains an outlier. Data: 5, 6, 7, 8, 100. Mean = 25.2 — but 25.2 is not at all typical of the first four values.

Right: When outliers exist, the median (7 in this case) is a much better description of the "typical" value than the mean (25.2).

Mean from Raw Data

+5 XP

To find the mean from a list: (1) add all the values to get $\sum x$, (2) count how many values there are to get $n$, (3) divide: $\bar{x} = \dfrac{\sum x}{n}$.

Test scores: 72, 85, 68, 90, 74, 81, 77, 83. Step 1: $\sum x$ = 72+85+68+90+74+81+77+83 = 630. Step 2: $n$ = 8 (eight scores). Step 3: $\bar{x}$ = 630 ÷ 8 = 78.75. The mean test score is 78.75 out of 100. Notice: this is not a whole number, and that's perfectly fine.

$\bar{x} = \frac{630}{8} = 78.75$

Don't skip zeros

If a data value is 0, it still counts as a value — include it in n.

Add carefully

Use a column method or calculator. One addition error ruins the whole answer.

Sanity check

The mean must be between the minimum and maximum values. If it's outside that range, you've made an error.

Mean from a Frequency Table

+5 XP

When data is in a frequency table, you can't just add the values column — you must multiply each value by its frequency to find the contribution of each group, then sum everything and divide by total frequency.

Score (x): 3, 4, 5. Frequency (f): 4, 6, 2.
$x \times f$: 3×4=12, 4×6=24, 5×2=10.
$\sum (x \times f) = 12+24+10 = 46$.
Total frequency $n = 4+6+2 = 12$.
Mean $= 46 \div 12 = \mathbf{3.83}$ (2 d.p.)

$\bar{x} = \frac{\sum(x \times f)}{\sum f} = \frac{46}{12} \approx 3.83$

Add an x×f column

Always add a third column to your table. Multiply each row's value by its frequency.

Divide by total frequency

n = sum of ALL frequencies (how many data values there are in total).

Check: is mean in range?

Mean must be between min and max values of x. If not, recheck your x×f column.

Effect of Outliers on the Mean

+5 XP

The mean is sensitive to outliers. A single extremely high or low value pulls the mean in its direction. When outliers are present, the mean can be misleading about what's "typical" — the median may give a better description of the centre.

House prices in a street: $400k, $420k, $380k, $410k, $395k, $2,000k. Without the mansion: mean = (400+420+380+410+395)÷5 = $401k. With the mansion: mean = (2005)÷6 = $334k... wait, that's wrong. 2005÷6 ≈ $667k. The $2M mansion makes the mean $267k higher — far above 5 out of 6 actual prices. Median = $405k, which is much more representative.

Outlier → Mean shifts far → Median stays typical

Always check for outliers

Before reporting the mean, scan the data for extreme values that look very different.

Median is resistant

The median is not affected by outliers — it's based on position, not values.

Real-world context

House prices, income data and test scores often use median for this exact reason.

Watch Me Solve It · 3 examples

Watch Me Solve It · Mean from raw data

+15 XP per step

PROBLEM

Find the mean of these 8 values: 7, 12, 9, 15, 11, 8, 14, 4.

1

Find the sum of all values ($\sum x$)

7 + 12 + 9 + 15 + 11 + 8 + 14 + 4 = 80

Add carefully. You can group pairs: (7+14)=21, (12+8)=20, (9+11)=20, (15+4)=19. Total = 21+20+20+19 = 80.
2

Count the number of values ($n$)

n = 8 (there are 8 values)
3

Apply the formula

$\bar{x} = \frac{\sum x}{n} = \frac{80}{8} = \mathbf{10}$

Check: 10 is between the min (4) and max (15). Makes sense.

Answer$\bar{x} = 80 \div 8 = 10$

Watch Me Solve It · Mean from a frequency table

+15 XP per step

PROBLEM

Number of siblings (x): 0, 1, 2, 3. Frequency (f): 5, 8, 4, 3. Find the mean number of siblings.

1

Add an x×f column

0×5=0, 1×8=8, 2×4=8, 3×3=9

Multiply each number-of-siblings value by how many students have that many siblings.
2

Sum the x×f column and count total frequency

$\sum(x\times f) = 0+8+8+9 = 25$. $n = 5+8+4+3 = 20$
3

Calculate the mean

$\bar{x} = \frac{25}{20} = \mathbf{1.25}$ siblings

Check: 1.25 is between 0 and 3. Divide by n=20 (total students), NOT by 4 (number of categories).

AnswerMean = 25 ÷ 20 = 1.25 siblings

Watch Me Solve It · Outlier effect on mean

+15 XP per step

PROBLEM

Weekly wages of 5 workers: $500, $520, $480, $510, $490. The boss earns $5,000/week. Show how adding the boss's wage affects the mean, and argue whether median would be better.

1

Calculate mean without the boss

Sum = 500+520+480+510+490 = 2500. Mean = 2500 ÷ 5 = $500

The 5 workers earn very similarly — the mean of $500 represents them well.
2

Add the boss and recalculate mean

New sum = 2500 + 5000 = 7500. New mean = 7500 ÷ 6 = $1,250
3

Compare and recommend

Mean jumped from $500 to $1,250 — an increase of $750 caused by one outlier. Median of 6 values = (500+510)÷2 = $505 — much more typical.

Five of the six people earn between $480 and $520. Median ($505) represents them. Mean ($1,250) does not.

AnswerMean with boss = $1,250 (misleading). Median = $505 (much more representative of typical wage).

Common Pitfalls

heads-up

Dividing by the number of categories, not n

In a frequency table: if there are 4 categories and 20 students, students divide the sum by 4 instead of 20. This gives an answer 5 times too large.

Fix: n = total of the frequency column (all students), not the number of rows in the table.

Not including zeros in the count

Data: 0, 3, 5, 8, 4. Students sometimes ignore the 0 and say n=4. But 0 is a valid data value that must be included. Sum = 20, n = 5, mean = 4.

Fix: Count every data value including zeros. Zero contributes 0 to the sum but still counts as one data point.

Rounding too early

Rounding 46÷12 to 3.8 in the middle of a calculation, then using 3.8 in a further step, compounds the error. Final answers can end up significantly wrong.

Fix: Keep all decimal places until the final answer. Only round at the very last step, as directed by the question.

Copy Into Your Books

Mean Formula

$\bar{x} = \frac{\sum x}{n}$
$\sum x$ = sum of all values
$n$ = number of data values
Mean is between min and max

Mean from Frequency Table

Add a column: $x \times f$
Sum all $x \times f$ values
Divide by total frequency ($\sum f$)
$\bar{x} = \frac{\sum(x \times f)}{\sum f}$

Outlier Effect

One extreme value pulls mean toward it
Median is "resistant" — unaffected by outliers
Use median when outliers are present

Common Errors

Divide by n (data count), not categories
Include zeros in n
Round only at the final step
Sanity check: mean must be in data range

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · The Mean

4 problems

Four drill problems to sharpen your mean calculation skills. Work each, then reveal the answer.

1 Find the mean of: 7, 12, 9, 15, 11, 8, 14.

Sum = 7+12+9+15+11+8+14 = 76. n = 7. Mean = 76 ÷ 7 ≈ 10.86
2 If the mean of 5 numbers is 9, what is their sum?

Mean = Sum ÷ n, so Sum = Mean × n = 9 × 5. Sum = 45
3 Frequency table: value 3 (freq 4), value 5 (freq 6), value 7 (freq 2). Find the mean.

x×f: 3×4=12, 5×6=30, 7×2=14. Sum of x×f = 56. Total freq n = 4+6+2 = 12. Mean = 56 ÷ 12 ≈ 4.67
4 Data set: 5, 6, 7, 8. Then an outlier of 100 is added. By how much does the mean change?

Original mean: (5+6+7+8)÷4 = 26÷4 = 6.5. New mean: (5+6+7+8+100)÷5 = 126÷5 = 25.2. Change = 25.2 − 6.5 = 18.7 increase

Complete in your workbook.

Quick Check · 5 questions

Find the mean of: 3, 7, 5, 9, 6.

+10 XP

The mean of 4 numbers is 8. What is their sum?

+10 XP

When finding mean from a frequency table, what do you divide by?

+10 XP

How does an outlier affect the mean?

+10 XP

Why is the median preferred over the mean when outliers exist?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Medium 3 MARKS

Q6. A student scored the following in 6 tests: 65, 72, 80, 58, 90, 75. Calculate the mean score. Show your working step by step.

Answer in your workbook.

Apply Medium 3 MARKS

Q7. Number of pets (x): 0, 1, 2, 3. Frequency (f): 6, 9, 3, 2. Calculate the mean number of pets per household. Show your x×f column.

Answer in your workbook.

Evaluate Hard 3 MARKS

Q8. A class of 10 students scored: 55, 62, 68, 70, 72, 74, 76, 78, 80, 5. Show that the mean is heavily influenced by the score of 5. Calculate both the mean with and without that score, then state which measure of centre would better represent this class and why.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — Sum = 30, n = 5, mean = 6.

2. B — Sum = mean × n = 8 × 4 = 32.

3. A — Divide by total of the frequency column (total data values).

4. D — Outlier pulls the mean in its direction, making it unrepresentative.

5. B — Median is not affected by extreme values (position-based).

Show Your Working Model Answers

Q6 (3 marks): Sum = 65+72+80+58+90+75 = 440 [1]. n = 6 [1]. Mean = 440÷6 ≈ 73.3 (1 d.p.) [1].

Q7 (3 marks): x×f: 0×6=0, 1×9=9, 2×3=6, 3×2=6 [1]. Sum(x×f)=21, Total f = 20 [1]. Mean = 21÷20 = 1.05 pets [1].

Q8 (3 marks): Sum with all 10: 5+55+62+68+70+72+74+76+78+80=640. Mean=640÷10=64 [1]. Without score of 5: sum=635, n=9, mean=635÷9≈70.6 [1]. Better measure: median (or mean without outlier). Nine of ten students scored between 55 and 80, with median around 71. The mean of 64 is dragged down by the outlier score of 5 and is not typical of the class [1].

Stretch Challenge · +25 XP, +10 coins

The Missing Value

A class of 6 students has a mean score of 74. Five of their scores are known: 68, 80, 71, 78, 65. (a) What is the sixth student's score? Show all working. (b) A seventh student joins and the new mean becomes 73. What did the seventh student score? (c) Is the new mean a good representation of all 7 scores? Justify your answer by comparing the mean with the individual scores.

Reveal solution

(a) Total for 6 = 74×6 = 444. Known sum = 68+80+71+78+65 = 362. Sixth = 444−362 = 82. (b) New total for 7 = 73×7 = 511. Previous total = 444. Seventh = 511−444 = 67. (c) Scores: 65, 67, 68, 71, 78, 80, 82. Mean = 73. All scores are relatively close to 73 (range 65–82), with no significant outliers. The mean of 73 is a reasonably good representation here.

Quick Review

Mean formula

$\bar{x} = \frac{\sum x}{n}$ — sum divided by count of values.

Frequency table mean

$\bar{x} = \frac{\sum(x \times f)}{\sum f}$ — multiply each value by its frequency first.

Zeros count

A score of 0 is still a data value. Include it in n and in the sum.

n = data values, not categories

Divide by the total number of individual data points, not the number of table rows.

Outlier effect

One extreme value pulls the mean away from the typical cluster. Use median instead.

Sanity check

Mean must lie between the min and max values. If it doesn't, recheck your sum.

Interactive: Mean Calculator

Enter your own data set and see how the mean changes as you add or remove values — especially when you add an outlier.

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