This note covers the measures that describe where a data set is centred (mean, median, mode) and how spread out it is (range, interquartile range, variance, standard deviation). You will also see how these measures change when every value is shifted or scaled, which is useful for unit conversions.
Key ideas
Measures of central tendency.
- Mean \(\bar{x}\): the sum of the values divided by how many there are. It uses every value, so one extreme value can pull it a long way.
- Median: the middle value when the data is in order (or the mean of the two middle values if \(n\) is even). It is not affected much by outliers.
- Mode: the most frequent value. For grouped data, use the modal class, the class with the highest frequency (for equal class widths).
If the data is skewed or has outliers, the median is usually the better description of a typical value.
Frequency tables. For values \(x_i\) with frequencies \(f_i\), where \(n=\sum f_i\), the mean is \(\bar{x}=\dfrac{\sum f_ix_i}{n}\). For grouped data use the midpoint of each class as \(x_i\). The result is only an estimate, because the exact values inside each class are unknown.
Measures of dispersion.
- Range \(=\) largest \(-\) smallest. It depends only on the two extreme values.
- Interquartile range \(=Q_3-Q_1\). It describes the middle 50% of the data and ignores outliers.
- Standard deviation \(\sigma\): roughly the typical distance of the values from the mean. For a data set, \(\sigma=\sqrt{\dfrac{\sum f_i(x_i-\bar{x})^2}{n}}\). On your GDC, use the one-variable statistics feature and read \(\sigma_x\) (the population standard deviation).
- Variance \(=\sigma^2\).
The standard deviation is measured in the same units as the data, while the variance is in squared units. This is why the standard deviation is usually quoted when describing a context.
A larger \(\sigma\) means the data is more spread out. Two sets can have the same mean but very different standard deviations.
Effect of changing every value. Suppose a data set has mean \(\bar{x}\) and standard deviation \(\sigma\).
- Adding a constant \(k\) to every value: the mean becomes \(\bar{x}+k\); \(\sigma\) is unchanged (the whole set shifts but the spread does not). The median and quartiles shift by \(k\); the IQR is unchanged.
- Multiplying every value by a constant \(k\gt0\): the mean becomes \(k\bar{x}\) and the standard deviation becomes \(k\sigma\). The variance becomes \(k^2\sigma^2\).
- Together, for new values \(y=ax+b\): mean \(a\bar{x}+b\), standard deviation \(|a|\sigma\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\bar{x}=\dfrac{\sum f_ix_i}{n}\) | Mean from a frequency table; \(n=\sum f_i\). Use class midpoints for grouped data. |
| Median position \(\tfrac{n+1}{2}\) | Place of the median in an ordered list of \(n\) values. |
| Range | Largest minus smallest. |
| \(\text{IQR}=Q_3-Q_1\) | Spread of the middle half of the data. |
| \(\sigma=\sqrt{\dfrac{\sum f_i(x_i-\bar{x})^2}{n}}\) | Standard deviation. Normally found with the GDC. |
| Variance \(=\sigma^2\) | Square of the standard deviation. |
| \(y=ax+b\) | New mean \(a\bar{x}+b\); new standard deviation \(|a|\sigma\). |
Worked examples
Example 1 A tennis coach records the number of double faults, \(x\), in each of 30 matches: \(x=2,3,4,5,6\) with frequencies \(3,7,10,6,4\). Find the mean, median, mode and standard deviation.
\(n=3+7+10+6+4=30\).
\(\sum fx=2(3)+3(7)+4(10)+5(6)+6(4)=6+21+40+30+24=121\).
Mean \(=\dfrac{121}{30}=4.03\) (3 s.f.).
Median: \(n=30\), so the median is the mean of the 15th and 16th values. Cumulative frequencies are 3, 10, 20, so both are 4. Median \(=4\).
Mode: the highest frequency is 10, so the mode is 4.
Standard deviation: enter \(x\) in list 1 and \(f\) in list 2, then use one-variable statistics with the frequency list. The GDC gives \(\sigma_x=1.17\).
Answer: mean 4.03, median 4, mode 4, \(\sigma=1.17\)
Example 2 The times taken by 50 couriers: \(10\le t\lt20\): 4; \(20\le t\lt30\): 11; \(30\le t\lt40\): 20; \(40\le t\lt50\): 9; \(50\le t\lt60\): 6. Estimate the mean and standard deviation, and state the modal class.
The class midpoints are \(15,25,35,45,55\).
\(\sum fx=4(15)+11(25)+20(35)+9(45)+6(55)=60+275+700+405+330=1770\).
Mean \(\approx\dfrac{1770}{50}=35.4\).
Enter midpoints and frequencies into the GDC lists and run one-variable statistics: \(\sigma_x\approx10.9\).
The highest frequency is 20, so the modal class is \(30\le t\lt40\).
Answer: mean \(\approx35.4\) min, \(\sigma\approx10.9\) min, modal class \(30\le t\lt40\)
Example 3 Daily temperatures in a city have mean \(20\ {}^\circ\text{C}\) and standard deviation \(3\ {}^\circ\text{C}\). Temperature in degrees Fahrenheit is \(F=1.8C+32\). Find the mean and standard deviation in \(^\circ\text{F}\), and the variance in \(^\circ\text{F}^2\).
Mean: \(1.8\times20+32=36+32=68\).
Standard deviation: the \(+32\) does not change the spread, and multiplying by 1.8 scales it: \(1.8\times3=5.4\).
Variance: \(5.4^2=29.16\). (Check: \(1.8^2\times3^2=3.24\times9=29.16\).)
Answer: mean \(68\ {}^\circ\text{F}\), \(\sigma=5.4\ {}^\circ\text{F}\), variance \(29.2\) to 3 s.f. (exactly 29.16)
Common errors
- Common error: using class boundaries (such as 20 or 30) instead of the midpoint when estimating a mean from grouped data.
- Common error: adding a constant to every value and also changing the standard deviation. Shifting does not change spread.
- Common error: multiplying the variance by \(k\) instead of \(k^2\) when the data is multiplied by \(k\).
- Common error: forgetting to put the frequency list into the GDC, so it treats each value as occurring once.
- Common error: giving a grouped mean as an exact value. It is an estimate; say so.
Exam tip: Use \(\sigma_x\) from your GDC unless the question says otherwise, and say that you used the GDC. Quote answers to 3 significant figures.
Check yourself
- For the data 3, 5, 8, 8, 11, find the mean, median, mode and range.
Answer
Mean \(=\tfrac{35}{5}=7\). Median \(=8\). Mode \(=8\). Range \(=11-3=8\). - The mean of five numbers is 12. Four of them are 8, 10, 15 and 11. Find the fifth.
Answer
Total \(=5\times12=60\). The four known add to 44. Fifth \(=60-44=16\). - A data set has mean 40 and standard deviation 5. Every value is doubled and then 10 is subtracted. Find the new mean and standard deviation.
Answer
New mean \(=2\times40-10=70\). New \(\sigma=2\times5=10\). - For the data 2, 3, 3, 4, 40, find the mean and median. Which better describes a typical value?
Answer
Mean \(=\tfrac{52}{5}=10.4\). Median \(=3\). The median is better, because the value 40 pulls the mean far above most of the data.