The Arithmetic Mean: Raw, Tabular & Grouped Data
The mean ($\bar{x}$) represents the mathematical balance point of a dataset, incorporating the exact numerical magnitude of every single observation.
- Raw Ungrouped Data: $$\bar{x} = \frac{\sum x}{n} = \frac{x_1 + x_2 + \dots + x_n}{n}$$
- Discrete Frequency Table: $$\bar{x} = \frac{\sum fx}{\sum f} = \frac{\sum fx}{n}$$
- Grouped Continuous Data (Estimated Mean): $$\bar{x} \approx \frac{\sum f x_{\text{mid}}}{\sum f}$$ where $x_{\text{mid}} = \frac{\text{Lower Boundary} + \text{Upper Boundary}}{2}$ is the class midpoint.
- Combined (Weighted) Mean of Two Groups: $$\bar{x}_{\text{combined}} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$
The Median: Positional Central Value
The median is the physical middle value when data is arranged in ascending numerical order. It divides the distribution into two equal halves ($50\%$ below, $50\%$ above).
- Order First: Always sort the $n$ data values in ascending order.
- Position Rule: The median lies at position: $$\text{Median Position} = \frac{n + 1}{2}$$
- If $n$ is odd: The median is the single value exactly at position $\frac{n + 1}{2}$.
Example: For $n = 9$, position is $\frac{10}{2} = 5$th value. - If $n$ is even: The median is the arithmetic mean of the two central values at $\frac{n}{2}$ and $\frac{n}{2} + 1$.
Example: For $n = 10$, average the $5$th and $6$th values: $\frac{x_5 + x_6}{2}$. - Robustness: The median is resistant to outliers. An extreme score of $1{,}000{,}000$ does not pull the median away from the center.
The Mode & Modal Class
The mode is the most frequently occurring value in a dataset.
- Unimodal: Exactly one mode exists.
- Bimodal: Two values share the highest frequency.
- No Mode: Every value appears with equal frequency.
- Modal Class (Grouped Data): The class interval with the highest frequency (or highest frequency density if class widths differ).
- Categorical Data: The mode is the only measure of central tendency that can be used for nominal qualitative data (e.g., most common car color).
Comparative Selection & Linear Transformations ($y = ax + b$)
Selecting the most appropriate average depends on the distribution's symmetry and the investigation's context.
- Mean: Best for symmetric distributions without extreme outliers (e.g., standardized test scores, manufactured part dimensions).
- Median: Best for skewed distributions or datasets containing extreme outliers (e.g., household incomes, real estate property prices).
- Mode: Best for manufacturing and inventory decisions where the single most popular size or model dictates stocking orders.
If every value $x$ in a dataset undergoes the linear transformation $y = ax + b$ (where $a$ is a scaling multiplier and $b$ is a constant shift): $$\bar{y} = a\bar{x} + b$$ $$\text{Median}(y) = a\text{Median}(x) + b$$ $$\text{Mode}(y) = a\text{Mode}(x) + b$$ All three measures of central tendency shift by $+b$ and scale by $\times a$!
Symmetric vs. skewed distributions: extreme positive values pull the mean rightwards, while the median remains centrally located.
Worked Example 1: Estimated Mean of Grouped Data
The table below records the travel distances (in km) of 50 employees commuting to work:
| Distance ($d$ km) | Frequency ($f$) |
|---|---|
| $0 \le d < 10$ | $12$ |
| $10 \le d < 20$ | $18$ |
| $20 \le d < 30$ | $14$ |
| $30 \le d < 50$ | $6$ |
a) Calculate the mid-interval value for each class.
b) Calculate an estimate of the mean commuting distance.
c) State the modal class.
Step-by-step Solution:
Part a: Mid-interval values ($x_{\text{mid}}$):
$x_1 = \frac{0 + 10}{2} = 5$; $x_2 = \frac{10 + 20}{2} = 15$;
$x_3 = \frac{20 + 30}{2} = 25$; $x_4 = \frac{30 + 50}{2} = 40$. [A1]
Part b: Estimated Mean:
$\sum f x_{\text{mid}} = (12 \times 5) + (18 \times 15) + (14 \times 25) + (6 \times 40)$ [M1]
$\sum f x_{\text{mid}} = 60 + 270 + 350 + 240 = 920\text{ km}$.
Total employees $\sum f = 12 + 18 + 14 + 6 = 50$.
$\bar{x} = \frac{920}{50} = 18.4\text{ km}$. [A1]
Part c: Modal class:
Notice intervals have different widths: $w_1 = 10, w_2 = 10, w_3 = 10, w_4 = 20$.
Frequency densities: $\text{FD}_1 = 1.2$, $\text{FD}_2 = 1.8$, $\text{FD}_3 = 1.4$, $\text{FD}_4 = 0.3$.
The highest frequency density is $1.8$, so the modal class is $10 \le d < 20$. [Criterion A: Knowledge & Understanding]
Worked Example 2: Combined Mean & Linear Transformation
In an IB school, Section A ($24$ students) scored a mean of $72\%$ on a math test, while Section B ($16$ students) scored a mean of $82\%$.
a) Calculate the combined overall mean score for all $40$ students.
b) The teacher discovers that every test was mistakenly graded out of $90$ instead of $100$. To correct this, each student's mark $x$ is converted via $y = \frac{10}{9}x + 2$. Calculate the new combined mean.
Step-by-step Solution:
Part a: Combined Mean:
Total sum of marks in Section A $= 24 \times 72 = 1728$.
Total sum of marks in Section B $= 16 \times 82 = 1312$. [M1]
Combined Total $= 1728 + 1312 = 3040$.
Combined Mean $\bar{x} = \frac{3040}{40} = 76\%$. [A1]
Part b: Linear Transformation:
Applying $y = ax + b$ with $a = \frac{10}{9}$ and $b = 2$: [M1]
$\bar{y} = \frac{10}{9}\bar{x} + 2 = \frac{10}{9}(76) + 2 = \frac{760}{9} + 2 \approx 84.44 + 2 = 86.44\%$.
The new combined mean is $86.4\%$ (3 s.f.). [Criterion A: Conceptual Precision]
Tiered Practice Exercises
10 problems advancing from basic mean/median/mode calculations to combined weighted averages, algebraic unknown frequencies, and real-world outlier investigations.
Find the mean, median, and mode of the following set of 9 test scores: $$14, 18, 12, 18, 15, 21, 18, 16, 12$$
View Solution & Markscheme
Method & Marks:
Mean: $\bar{x} = \frac{14+18+12+18+15+21+18+16+12}{9} = \frac{144}{9} = 16$. [A1]
Order scores: $12, 12, 14, 15, 16, 18, 18, 18, 21$.
Median position is $\frac{9+1}{2} = 5$th value $\implies \text{Median} = 16$. [A1]
Mode is the most frequent score: $18$ (appears 3 times). [A1]
Find the median of the following set of 10 numbers: $$45, 32, 56, 41, 38, 50, 48, 36, 54, 42$$
View Solution & Markscheme
Method & Marks:
Sort in ascending order: [M1]
$32, 36, 38, 41, 42, 45, 48, 50, 54, 56$.
$n = 10$ is even. The median is the average of the 5th and 6th values:
$\text{Median} = \frac{42 + 45}{2} = \frac{87}{2} = 43.5$. [A1]
A student has scored $78, 85, 92,$ and $81$ on four biology assessments. What mark must the student achieve on the fifth assessment to attain an overall mean of $85$?
View Solution & Markscheme
Method & Marks:
Required total for 5 tests: $5 \times 85 = 425$. [M1]
Sum of first 4 tests: $78 + 85 + 92 + 81 = 336$.
Fifth test mark $x = 425 - 336 = 89$. [A1]
Calculate the mean and median from the frequency table below:
| Score ($x$) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency ($f$) | 4 | 7 | 11 | 6 | 2 |
View Solution & Markscheme
Method & Marks:
Total frequency $n = \sum f = 4 + 7 + 11 + 6 + 2 = 30$. [M1]
$\sum fx = (1 \times 4) + (2 \times 7) + (3 \times 11) + (4 \times 6) + (5 \times 2) = 4 + 14 + 33 + 24 + 10 = 85$.
Mean $\bar{x} = \frac{85}{30} \approx 2.83$. [A1]
Median position is $\frac{30+1}{2} = 15.5$, so average of 15th and 16th values.
Cumulative frequencies: $4, 11, 22, 28, 30$.
Both the 15th and 16th values fall in the $x = 3$ group $\implies \text{Median} = 3$. [A1]
A dataset of 8 numbers has a mean of $15$. When a 9th number is added, the new mean becomes $17$. What is the value of the 9th number?
View Solution & Markscheme
Method & Marks:
Sum of original 8 numbers: $8 \times 15 = 120$. [M1]
Sum of 9 numbers: $9 \times 17 = 153$. [M1]
The 9th number is $153 - 120 = 33$. [A1]
The mean of 5 positive integers $a, b, c, d, e$ is $12$. The median is $10$, the unique mode is $8$, and the range is $14$. Find the five numbers.
View Solution & Markscheme
Method & Marks:
Let $a \le b \le c \le d \le e$.
Median is $c = 10$. [M1]
Unique mode is $8$, so $a = b = 8$ (since they must be less than median 10). [M1]
Range is $14 \implies e - a = 14 \implies e = 8 + 14 = 22$. [A1]
Mean is $12 \implies \sum = 5 \times 12 = 60$.
$8 + 8 + 10 + d + 22 = 60 \implies 48 + d = 60 \implies d = 12$.
The five numbers are: $8, 8, 10, 12, 22$. [A1]
A company pays its 20 junior staff a mean salary of $\$35{,}000$, its 8 senior staff a mean salary of $\$65{,}000$, and its 2 executives a mean salary of $\$180{,}000$.
a) Calculate the combined mean salary of all 30 employees.
b) If the median salary is $\$38{,}000$, explain why the median is a more representative figure for typical employee pay than the mean.
View Solution & Markscheme
Method & Marks:
a) Total payroll: [M1]
$(20 \times 35{,}000) + (8 \times 65{,}000) + (2 \times 180{,}000) = 700{,}000 + 520{,}000 + 360{,}000 = \$1{,}580{,}000$.
Combined mean $= \frac{1{,}580{,}000}{30} \approx \$52{,}667$. [A1]
b) The mean ($\$52{,}667$) is heavily inflated by the 2 executive salaries ($\$180{,}000$). In fact, over $66\%$ of the employees earn well below the mean. The median ($\$38{,}000$) is resistant to these top outliers and reflects the true typical experience of workers. [R1: Criterion D]
The frequency distribution for the number of hours students spent studying for an exam is given below. The mean study time is known to be $4.6\text{ hours}$.
| Hours ($x$) | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| Frequency ($f$) | $8$ | $k$ | $12$ | $4$ |
View Solution & Markscheme
Method & Marks:
Total frequency: $\sum f = 8 + k + 12 + 4 = 24 + k$. [M1]
$\sum fx = (2 \times 8) + (4 \times k) + (6 \times 12) + (8 \times 4) = 16 + 4k + 72 + 32 = 120 + 4k$. [M1]
Using $\bar{x} = 4.6$:
$\frac{120 + 4k}{24 + k} = 4.6 \implies 120 + 4k = 4.6(24 + k)$
$120 + 4k = 110.4 + 4.6k \implies 120 - 110.4 = 4.6k - 4k \implies 9.6 = 0.6k$
$k = \frac{9.6}{0.6} = 16$. [A1]
Total students $= 24 + 16 = 40$. [A1]
A set of temperatures measured in Celsius has mean $\bar{x} = 25^\circ\text{C}$ and median $M = 23^\circ\text{C}$.
a) Use the linear conversion formula $F = 1.8C + 32$ to find the mean and median in degrees Fahrenheit.
b) Prove algebraically that for any general linear transformation $y = ax + b$, the mean of $y$ is always equal to $a\bar{x} + b$.
View Solution & Markscheme
Method & Marks:
a) Mean in Fahrenheit: $\bar{F} = 1.8(25) + 32 = 45 + 32 = 77^\circ\text{F}$. [A1]
Median in Fahrenheit: $M_F = 1.8(23) + 32 = 41.4 + 32 = 73.4^\circ\text{F}$. [A1]
b) Proof:
Let $y_i = ax_i + b$ for $i = 1, \dots, n$.
$\bar{y} = \frac{1}{n}\sum_{i=1}^n y_i = \frac{1}{n}\sum_{i=1}^n (ax_i + b) = \frac{1}{n}\left(a\sum x_i + \sum b\right)$. [M1]
Since $\sum_{i=1}^n b = nb$:
$\bar{y} = a\left(\frac{\sum x_i}{n}\right) + \frac{nb}{n} = a\bar{x} + b$. $\blacksquare$ [A1][R1]
Criterion B & D Investigation: An experiment records the half-life breakdown counts of an isotope every minute. The first 10 recorded values are:
$$40, 42, 45, 46, 47, 49, 50, 52, 53, 56$$
An 11th observation is recorded with count $x$.
a) Express the new mean in terms of $x$.
b) Find the value of $x$ such that the median of the 11 scores remains exactly equal to the original median.
c) Evaluate the effect on the mean versus the median if an instrument malfunction records $x = 500$.
View Solution & Markscheme
Method & Marks:
a) Sum of original 10 scores $= 40 + 42 + 45 + 46 + 47 + 49 + 50 + 52 + 53 + 56 = 480$.
New mean $\bar{x}_{11} = \frac{480 + x}{11}$. [A1]
b) Original median of 10 scores: average of 5th (47) and 6th (49) $= \frac{47 + 49}{2} = 48$.
For 11 scores, the median is the 6th score in order. To keep the median equal to $48$, setting $x = 48$ inserts it between $47$ and $49$, making the 6th score exactly $48$. [M1][A1]
c) If $x = 500$:
New mean $= \frac{480 + 500}{11} = \frac{980}{11} \approx 89.1$ (distorted by more than $40$ units above the typical range).
The new median is the 6th score, which becomes $49$ (shifts by only $1$ unit).
This dramatically demonstrates the vulnerability of the mean to extreme measurement errors and the robustness of the median. [R1: Criterion D Reflection]
Measures of Central Tendency Quick Reference
Raw Mean
$$\bar{x} = \frac{\sum x}{n}$$
Sum divided by count.Frequency Table Mean
$$\bar{x} = \frac{\sum fx}{\sum f}$$
Weighted by frequencies.Grouped Estimated Mean
$$\bar{x} \approx \frac{\sum fx_{\text{mid}}}{\sum f}$$
Uses class midpoints.Median Position
$$\text{Pos} = \frac{n + 1}{2}$$
Sort data ascending first.Combined Mean
$$\bar{x}_{\text{comb}} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$
Weighted group average.Linear Shift
$$y = ax + b \implies \bar{y} = a\bar{x} + b$$
Applies to mean, median, mode.Top 5 Common MYP Central Tendency Pitfalls
- Averaging Averages Directly: If Class A (30 students) has mean 70 and Class B (10 students) has mean 90, the overall mean is NOT $\frac{70 + 90}{2} = 80$! You must compute the weighted combined mean: $\frac{(30 \times 70) + (10 \times 90)}{40} = 75$.
- Forgetting to Order Data Before Finding the Median: Never take the middle number from an unsorted list. The definition of median requires ascending (or descending) order.
- Confusing Median Position with Median Value: For $n = 9$, $\frac{9+1}{2} = 5$ gives the position (5th number), NOT the median value itself.
- Dividing by the Number of Rows in a Frequency Table: When computing $\frac{\sum fx}{\sum f}$, divide by total frequency $\sum f$, NEVER by the number of table rows!
- Selecting the Mean for Heavily Skewed Data: When outliers are present (like real estate or salaries), quoting the mean misrepresents typical conditions. In MYP evaluations (Criterion D), always recommend the median for skewed scenarios.
Self-Grading Mini Test
5 quiz questions on mean, median, mode, and weighted averages.
What is the median of the dataset: $12, 5, 9, 18, 7, 14$?
Explanation: Sort in ascending order: $5, 7, 9, 12, 14, 18$. The middle two numbers are $9$ and $12$. Median $= \frac{9 + 12}{2} = 10.5$.
Which measure of central tendency is most resistant to extreme outliers?
Explanation: The median is purely positional (middle 50%), so extreme high or low values do not distort its position.
Group 1 has $10$ items with mean $20$. Group 2 has $30$ items with mean $40$. What is the combined mean?
Explanation: $\bar{x} = \frac{(10 \times 20) + (30 \times 40)}{10 + 30} = \frac{200 + 1200}{40} = \frac{1400}{40} = 35$.
A dataset has mean $\bar{x} = 50$. If every value is multiplied by $2$ and then increased by $5$, what is the new mean?
Explanation: Under $y = 2x + 5$, $\bar{y} = 2\bar{x} + 5 = 2(50) + 5 = 105$.
Which measure of central tendency can be used for nominal qualitative data such as blood type?
Explanation: Nominal data has no numerical values or order, so arithmetic sums or rankings are impossible. Only the most frequent category (mode) can be identified.