15.2 Measures of Central Tendency
SL 4.3
- how to calculate the mean of a list, of a frequency table, and (as an estimate) of grouped data using mid-interval values
- the difference between a parameter (\(\mu\)) and a statistic (\(\bar{x}\))
- how to find the median of a list and of a frequency table
- the mode and the modal class
- when the mean can be misleading, and which average to use
Averages
Often we want one number that describes the “centre” of a data set. Such a number is called a measure of central tendency (or an average). The three most common are the mean, the median and the mode.
Mean
The mean is the sum of all the values divided by the number of values.
Mean, \(\bar{x}\), of a set of data In the formula booklet
\[ \bar{x} = \frac{\sum_{i=1}^{k} f_i x_i}{n}, \qquad \text{where } n = \sum_{i=1}^{k} f_i \]
Here the value \(x_i\) occurs \(f_i\) times (a frequency table). For \(n\) separate values \(x_1, x_2, \ldots, x_n\) (every \(f_i = 1\)) this is
\[ \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{\sum x}{n} \]
- \(\bar{x}\): the mean
- \(f_i\): the frequency of the value \(x_i\)
- \(x_i\): the \(i\)th data value
- \(k\): the number of different values
- \(n\): the total number of data values (the total frequency)
For grouped data we do not know the exact values. We use the mid-interval value of each class as \(x_i\). The answer is then only an estimate of the mean.
Parameters and statistics Not in the formula booklet
- A number that describes a population is a parameter. It is written with a Greek letter, for example the population mean \(\mu\).
- A number that describes a sample is a statistic, for example the sample mean \(\bar{x}\).
At SL, a data set is treated as the population unless the question says otherwise, so \(\mu\) and \(\bar{x}\) are calculated in the same way.
Example 1 Find the mean of each data set.
(a) \(58, 61, 64, 69, 73\)
(b)
| Score, \(x\) | 4 | 6 | 8 | 10 |
|---|---|---|---|---|
| Frequency, \(f\) | 3 | 5 | 8 | 4 |
(c) (lengths measured to the nearest cm)
| Length (cm) | 5–9 | 10–14 | 15–19 | 20–24 |
|---|---|---|---|---|
| Frequency, \(f\) | 3 | 6 | 8 | 3 |
(a) \(\bar{x} = \dfrac{58 + 61 + 64 + 69 + 73}{5} = \dfrac{325}{5} = 65\).
(b) \(n = 3 + 5 + 8 + 4 = 20\) and \(\sum fx = 12 + 30 + 64 + 40 = 146\), so
\[ \bar{x} = \frac{146}{20} = 7.3 . \]
(c) The mid-interval values are \(7, 12, 17, 22\) (for example, 5–9 means \(4.5 \le L < 9.5\), with middle \(7\)).
\[ \sum fx = 3(7) + 6(12) + 8(17) + 3(22) = 21 + 72 + 136 + 66 = 295, \qquad n = 20 \]
\[ \bar{x} \approx \frac{295}{20} = 14.75 \text{ cm}. \]
This is an estimate, because we used mid-interval values instead of the real lengths.
Lists & Spreadsheet: type the values (or mid-interval values) in column A and the frequencies in column B. Then Menu → Statistics → Stat Calculations → One-Variable Statistics, choose 1 list, set X1 List to a[] and Frequency List to b[]. The mean is shown as \(\bar{x}\); \(\Sigma x\) and \(n\) are also shown, so you can check your table. For (c) above the GDC gives \(\bar{x} = 14.75\).
When the mean is misleading
The mean uses every value, so one very large or very small value can pull it far away from the main body of the data.
Example 2 Find the mean of \(8, 5, 7, 6, 74\). Does the mean describe the centre of the data well?
\(\bar{x} = \dfrac{8 + 5 + 7 + 6 + 74}{5} = \dfrac{100}{5} = 20\).
Four of the five values are between 5 and 8, so \(20\) does not describe the centre well. The value \(74\) is far from the others and pulls the mean up. The median, \(7\), is a better average here.
Median
Median Not in the formula booklet
The median is the middle value when the data are written in order.
For \(n\) values, the median is the \(\left(\dfrac{n + 1}{2}\right)\)th value. If \(n\) is even, this is halfway between two values: take the mean of those two values.
The median is not affected much by extreme values.
Example 3 Find the median of each data set.
(a) \(9, 4, 6, 4, 11, 8, 3\) \(\qquad\) (b) \(12, 15, 9, 20, 14, 18\)
(a) In order: \(3, 4, 4, \mathbf{6}, 8, 9, 11\). There are 7 values, so the median is the 4th value: 6.
(b) In order: \(9, 12, \mathbf{14}, \mathbf{15}, 18, 20\). There are 6 values, so the median is halfway between the 3rd and 4th:
\[ \frac{14 + 15}{2} = 14.5 . \]
For a frequency table, use a running total of the frequencies to find where the middle value is.
Example 4 A spinner with the numbers 1 to 6 is spun 50 times. Find the median score.
| Score, \(x\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency, \(f\) | 7 | 10 | 9 | 6 | 11 | 7 |
\(n = 50\), so the median is halfway between the 25th and 26th values.
Running totals: \(7\), \(7 + 10 = 17\), \(17 + 9 = 26\), …
The 18th to 26th values are all \(3\). So the 25th and 26th values are both \(3\), and the median is 3.
Which average is better? The mean uses all the data and is usually more stable from one sample to another, so it is preferred for further work. The median is better when there are extreme values.
Mode
Mode and modal class Not in the formula booklet
- The mode is the value that occurs most often (with the highest frequency).
- For grouped data with equal class widths, the modal class is the class with the highest frequency.
A data set can have two modes (bimodal), or no mode if every value occurs equally often.
Example 5 The marks (out of 10) of 15 students are
\[ 6,\ 3,\ 7,\ 7,\ 9,\ 2,\ 7,\ 5,\ 6,\ 10,\ 7,\ 6,\ 4,\ 8,\ 8 . \]
Find the mode.
The mark \(7\) occurs 4 times; \(6\) occurs 3 times; \(8\) occurs twice; every other mark occurs once. The mode is 7.
Example 6 The times, \(t\) minutes, that 50 customers waited at a bank are shown.
| Time, \(t\) (min) | \(0 \le t < 10\) | \(10 \le t < 20\) | \(20 \le t < 30\) | \(30 \le t < 40\) | \(40 \le t < 50\) |
|---|---|---|---|---|---|
| Frequency | 5 | 12 | 18 | 10 | 5 |
(a) Write down the modal class.
(b) Find an estimate of the mean waiting time.
(a) \(20 \le t < 30\) (frequency 18).
(b) Mid-interval values: \(5, 15, 25, 35, 45\).
\[ \sum fx = 5(5) + 12(15) + 18(25) + 10(35) + 5(45) = 25 + 180 + 450 + 350 + 225 = 1230 \]
\[ \bar{x} \approx \frac{1230}{50} = 24.6 \text{ minutes}. \]
Example 7 The table shows the number of books read by some students last month. The mean is \(3\). Find \(k\).
| Books, \(x\) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency, \(f\) | 4 | \(k\) | 9 | 6 | 3 |
\(\sum f = 22 + k\) and \(\sum fx = 4 + 2k + 27 + 24 + 15 = 70 + 2k\).
\[ \frac{70 + 2k}{22 + k} = 3 \;\Longrightarrow\; 70 + 2k = 66 + 3k \;\Longrightarrow\; k = 4 . \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Median of \(9, 4, 6, 4, 11, 8, 3\) is \(4\) | Order the data first: median \(= 6\) | The median is the middle of the ordered data. |
| \(\bar{x} = \dfrac{\sum fx}{\text{number of classes}}\) | Divide by \(n = \sum f\) | \(n\) is the number of data values. |
| Mean of a frequency table \(= \dfrac{\sum x}{n}\) | \(\dfrac{\sum fx}{n}\) | Each value must be counted \(f\) times. |
| Mean of grouped data is exact | It is an estimate | Mid-interval values replace the real values. |
| Mode \(= 18\) in Example 6 | Modal class \(20 \le t < 30\) | The mode is a value, not a frequency. |
Exercises
1 Find the mean, median and mode of \(12, 7, 9, 15, 7, 10, 8, 12, 7\).
Sum \(= 87\), \(n = 9\): mean \(= \dfrac{87}{9} = 9.67\) (3 s.f.).
In order: \(7, 7, 7, 8, \mathbf{9}, 10, 12, 12, 15\): median \(= 9\). Mode \(= 7\).
2 Find the mean and median of \(31, 24, 28, 35, 22, 30\).
Mean \(= \dfrac{170}{6} = 28.3\) (3 s.f.). In order: \(22, 24, 28, 30, 31, 35\); median \(= \dfrac{28 + 30}{2} = 29\).
3 Find the mean, median and mode of the number of cars owned by 30 families.
| Cars, \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency, \(f\) | 6 | 11 | 8 | 3 | 2 |
\(\sum fx = 0 + 11 + 16 + 9 + 8 = 44\), mean \(= \dfrac{44}{30} = 1.47\) (3 s.f.).
Median: between the 15th and 16th values. Running totals \(6, 17, \ldots\), so both are \(1\): median \(= 1\).
Mode \(= 1\).
4 The mean of \(4, 7, x, 10, 2x\) is \(9\). Find \(x\), then find the median.
\(\dfrac{21 + 3x}{5} = 9 \Rightarrow 21 + 3x = 45 \Rightarrow x = 8\).
The data are \(4, 7, 8, 10, 16\), so the median is \(8\).
5 The masses, \(m\) kg, of 40 parcels are shown.
| Mass, \(m\) (kg) | \(40 \le m < 50\) | \(50 \le m < 60\) | \(60 \le m < 70\) | \(70 \le m < 80\) | \(80 \le m < 90\) |
|---|---|---|---|---|---|
| Frequency | 4 | 9 | 15 | 8 | 4 |
(a) Write down the modal class.
(b) Estimate the mean mass.
(c) Explain why your answer to (b) is only an estimate.
(a) \(60 \le m < 70\).
(b) \(\sum fx = 4(45) + 9(55) + 15(65) + 8(75) + 4(85) = 180 + 495 + 975 + 600 + 340 = 2590\); mean \(\approx \dfrac{2590}{40} = 64.75\) kg.
(c) The real masses are not known; each class is represented by its mid-interval value.
6 The mean of 8 numbers is \(12\). One number is removed and the mean of the other 7 numbers is \(11\). Find the number that was removed.
Total of 8 numbers \(= 96\); total of 7 numbers \(= 77\). The removed number is \(96 - 77 = 19\).
7 In a test, the 20 students of class A have a mean mark of \(64\) and the 25 students of class B have a mean mark of \(70\). Find the mean mark of all 45 students.
Totals: \(20 \times 64 = 1280\) and \(25 \times 70 = 1750\).
Mean \(= \dfrac{1280 + 1750}{45} = \dfrac{3030}{45} = 67.3\) (3 s.f.). (It is not \(\frac{64 + 70}{2} = 67\), because the classes have different sizes.)
8 The prices (in thousands of dollars) of five flats sold in one street were \(320, 350, 340, 365, 2100\). Find the mean and the median. Which is the better average here? Explain.
Mean \(= \dfrac{3475}{5} = 695\); median \(= 350\).
The median is better: one very expensive flat pulls the mean far above the prices of the other four.
9 The scores of 30 players in a game are grouped as shown. Estimate the mean score.
| Score | 0–9 | 10–19 | 20–29 | 30–39 | 40–49 |
|---|---|---|---|---|---|
| Frequency | 3 | 8 | 12 | 5 | 2 |
Scores are whole numbers, so the mid-interval values are \(4.5, 14.5, 24.5, 34.5, 44.5\).
\(\sum fx = 13.5 + 116 + 294 + 172.5 + 89 = 685\); mean \(\approx \dfrac{685}{30} = 22.8\) (3 s.f.).
10 The table shows the number of goals, \(x\), scored by a team in each match of a season. The mean number of goals is \(1.8\). Find \(p\).
| Goals, \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency, \(f\) | 3 | 7 | \(p\) | 4 | 2 |
\(\sum f = 16 + p\) and \(\sum fx = 0 + 7 + 2p + 12 + 8 = 27 + 2p\).
\[ \frac{27 + 2p}{16 + p} = 1.8 \;\Longrightarrow\; 27 + 2p = 28.8 + 1.8p \;\Longrightarrow\; 0.2p = 1.8 \;\Longrightarrow\; p = 9 . \]