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Statistics guide, Unit 1, step 5 of 22Previous stepNext step

Step 5: Describing Data with Numbers

SyllabusAP Stats 1.7–1.9
In 30 seconds
  • Centre: the mean \( \bar{x} = \Sigma x \div n \) uses every value but is pulled by outliers; the median is the resistant middle; the mode is the most common value.
  • Spread: the standard deviation is roughly the typical distance from the mean. For a sample, divide by \( n – 1 \), then take the square root.
  • A z-score \( = (x – \bar{x}) \div s \) tells you how many standard deviations a value is from the mean. Outliers lie beyond \( Q_1 – 1.5 \times \text{IQR} \) or \( Q_3 + 1.5 \times \text{IQR} \).
  • Exam rule: report mean and SD for symmetric data, median and IQR for skewed data or data with outliers.
Video lesson coming soonA short animated lesson for this step will appear here.

Measures of centre: mean, median and mode

Picture a town of 1,000 households with a typical income of about 50,000 dollars. A billionaire registers as a resident. The mean income jumps to over a million dollars, yet no family is a cent richer, and the household in the middle still earns about 50,000. The mean was dragged by one extreme value; the median did not move.

Key termMean: the ordinary average, found by adding all the values and dividing by how many there are.
\[ \bar{x} = \frac{\Sigma x}{n} \]

Here \( \Sigma \) (capital sigma) means “add up”, \( \bar{x} \) (“x-bar”) is the sample mean and \( n \) is the number of values. For a whole population the mean is written \( \mu \) (“mu”) and the size \( N \); the arithmetic is the same. For example, five students who studied 4, 8, 6, 5 and 12 hours have a mean of \( 35 \div 5 = 7 \) hours.

The median is the middle value once the data is in order. If \( n \) is odd, it is the single middle value; if \( n \) is even, it is the average of the two middle values. For 4, 6, 9, 10 the median is \( (6 + 9) \div 2 = 7.5 \). The mode is the most frequent value, and it is the only measure of centre that works for categorical data, such as the most common eye colour.

Common mistakeFinding the “middle” of a list that hasn’t been put in order. Always sort the data first, then find the median.

On symmetric data the three measures nearly coincide. On skewed data they separate in a predictable way: the mean is pulled towards the long tail, the median stays near the bulk of the data, and the mode sits at the peak. A quick trick:

  • mean > median: probably right-skewed (incomes, house prices);
  • mean < median: probably left-skewed;
  • mean ≈ median: roughly symmetric.

Measures of position: percentiles and quartiles

If you are “in the 90th percentile” for an exam, about 90% of students scored at or below you.

Key termQuartiles: the values that cut ordered data into quarters; \( Q_1 \) is the 25th percentile, \( Q_2 \) is the median and \( Q_3 \) is the 75th percentile.

A simple hand method: find the median, then \( Q_1 \) is the median of the lower half and \( Q_3 \) is the median of the upper half. The five-number summary (minimum, \( Q_1 \), median, \( Q_3 \), maximum) is the skeleton behind a boxplot.

Worked example

Find the five-number summary and IQR of: 4, 7, 8, 10, 12, 14, 15, 18, 21.

Step 1\( n = 9 \), so the median is the 5th value: \( 12 \).
Step 2Lower half 4, 7, 8, 10: \( Q_1 = (7 + 8) \div 2 = 7.5 \). Upper half 14, 15, 18, 21: \( Q_3 = (15 + 18) \div 2 = 16.5 \).
Step 3Summary: \( 4,\ 7.5,\ 12,\ 16.5,\ 21 \). IQR \( = 16.5 – 7.5 = 9 \).

The middle half of the data is spread across 9 units, from 7.5 to 16.5.

Measures of spread: range, IQR and standard deviation

Two classes could both average 65%, but in one everyone scored 60 to 70 while in the other scores ran from 20 to 100. A measure of spread captures that difference.

  • The range = maximum − minimum. It uses only two values, so one outlier can blow it up.
  • The interquartile range (IQR) = \( Q_3 – Q_1 \), the spread of the middle 50%. It ignores the extremes, so it pairs naturally with the median.

The most important measure is the standard deviation. Start with each value’s deviation from the mean, \( x – \bar{x} \). If you simply averaged these they would cancel to zero. So you square each deviation first (making it positive), average the squares to get the variance, then take the square root to get back to the original units.

Key termStandard deviation: roughly the typical distance of a value from the mean, found as the square root of the variance.
\[ s^2 = \frac{\Sigma (x – \bar{x})^2}{n – 1} \qquad s = \sqrt{s^2} \qquad\qquad \sigma^2 = \frac{\Sigma (x – \mu)^2}{N} \qquad \sigma = \sqrt{\sigma^2} \]

Why \( n – 1 \)? A sample sits slightly closer to its own mean than to the population mean, so dividing by \( n \) would underestimate the spread.

Worked example

Find the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9.

Step 1Mean: \( \Sigma x = 40,\ n = 8,\ \bar{x} = 40 \div 8 = 5 \)
Step 2Deviations: \( -3, -1, -1, -1, 0, 0, 2, 4 \) (they sum to 0, a useful check).
Step 3Squares: \( 9, 1, 1, 1, 0, 0, 4, 16 \), so \( \Sigma(x – \bar{x})^2 = 32 \)
Step 4As a population: \( \sigma^2 = 32 \div 8 = 4,\ \sigma = \sqrt{4} = 2 \)
Step 5As a sample: \( s^2 = 32 \div 7 \approx 4.57,\ s = \sqrt{4.57} \approx 2.14 \)

A typical value sits about 2 units from the mean of 5.

Common mistakeTwo slips cost the most marks: dividing by \( n \) for a sample (use \( n – 1 \)), and stopping at the variance. Variance is in squared units, such as dollars-squared, which mean nothing on their own. Always take the square root.

The coefficient of variation, \( \text{CV} = (s \div \bar{x}) \times 100\% \), expresses spread as a percentage of the mean, so you can compare things measured in different units. Coffee sales averaging 200 cups with SD 30 have CV \( 30 \div 200 = 15\% \); revenue averaging 4,000 dollars with SD 480 has CV \( 480 \div 4000 = 12\% \), so coffee sales are relatively more variable.

Z-scores and the empirical rule

A student scores 82 in maths (class mean 70, SD 8) and 78 in English (class mean 60, SD 12). Which result is better? A z-score puts both on the same scale.

Key termZ-score: the number of standard deviations a value lies above (positive) or below (negative) the mean.
\[ z = \frac{x – \bar{x}}{s} \]
Worked example

Compare the student’s maths and English results using z-scores.

Step 1Maths: \( z = \dfrac{82 – 70}{8} = \dfrac{12}{8} = +1.5 \)
Step 2English: \( z = \dfrac{78 – 60}{12} = \dfrac{18}{12} = +1.5 \)

Both results sit exactly 1.5 standard deviations above the class mean, so her relative performance is identical.

For data that is roughly bell-shaped (symmetric with one peak), the empirical rule (the 68–95–99.7 rule) says about 68% of values lie within 1 SD of the mean, 95% within 2 SD and 99.7% within 3 SD. IQ scores have mean 100 and SD 15, so about 95% of people score between 70 and 130, and only about 2.5% score above 130 (half of the 5% outside).

Outliers and choosing what to report

To decide whether a value is “far from the rest”, use the IQR fences:

\[ \text{lower fence} = Q_1 – 1.5 \times \text{IQR} \qquad \text{upper fence} = Q_3 + 1.5 \times \text{IQR} \]

Anything beyond a fence is an outlier. For the summary above, \( 1.5 \times 9 = 13.5 \), so the fences are \( 7.5 – 13.5 = -6 \) and \( 16.5 + 13.5 = 30 \). The maximum, 21, is inside; a value of 35 would be flagged.

Common mistakeDeleting outliers on sight. An outlier is a flag, not a verdict. It might be a typo worth fixing, or the most important observation you have, such as one fraudulent transaction. Investigate it first.

The key idea when choosing summaries is resistance: whether a measure is thrown off by outliers.

MeasureTypeResistant to outliers?
MeanCentreNo
MedianCentreYes
RangeSpreadNo
Standard deviationSpreadNo
IQRSpreadYes

Keep the pairs together: mean with SD, median with IQR. And when you change units, adding a constant shifts the mean but not the SD, while multiplying by a constant scales both.

Exam tipShow every stage of a standard deviation calculation in a table (values, deviations, squared deviations) and state whether you divided by \( n \) or \( n – 1 \). When asked which summary to report, name the pair and justify it from the shape or the outlier, for example: “median and IQR, because the 240 is an outlier that would inflate the mean and SD.”

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find the mean of 12, 15, 20, 22, 31 and the median of 7, 3, 9, 4, 8, 2. Show worked answer
Mean \( = (12 + 15 + 20 + 22 + 31) \div 5 = 100 \div 5 = 20 \).
Median: order first: 2, 3, 4, 7, 8, 9. With \( n = 6 \), average the 3rd and 4th values: \( (4 + 7) \div 2 = 5.5 \).
EasyQ2. The mean of five numbers is 12. Four of them are 8, 11, 14 and 15. Find the fifth. Show worked answer
Total must be \( 5 \times 12 = 60 \).
Known four: \( 8 + 11 + 14 + 15 = 48 \).
Fifth number \( = 60 – 48 = 12 \).
MediumQ3. Find the sample standard deviation of 10, 12, 14, 16, 18. Show worked answer
Mean \( = 70 \div 5 = 14 \).
Deviations: \( -4, -2, 0, 2, 4 \). Squares: \( 16, 4, 0, 4, 16 \); sum \( = 40 \).
Sample variance \( s^2 = 40 \div (5 – 1) = 10 \).
\( s = \sqrt{10} \approx 3.16 \).
MediumQ4. Test A: score 40, mean 30, SD 5. Test B: score 65, mean 50, SD 12. On which test did the student do relatively better? Show worked answer
Test A: \( z = (40 – 30) \div 5 = 10 \div 5 = +2.0 \).
Test B: \( z = (65 – 50) \div 12 = 15 \div 12 = +1.25 \).
Test A is relatively better: the score is 2 SDs above the mean, compared with 1.25 SDs on Test B.
MediumQ5. A dataset has \( Q_1 = 20 \) and \( Q_3 = 32 \). Find the outlier fences. Are 5 and 55 outliers? Show worked answer
IQR \( = 32 – 20 = 12 \); \( 1.5 \times 12 = 18 \).
Lower fence \( = 20 – 18 = 2 \); upper fence \( = 32 + 18 = 50 \).
5 is above 2, so it is not an outlier. 55 is above 50, so it is an outlier.
ExamQ6. A café’s daily customer counts for one week are 60, 62, 65, 61, 63, 240, 64 (240 was a one-off festival day). (a) Find the mean and median. (b) Use the 1.5 × IQR rule to show that 240 is an outlier. (c) Which pair of summaries should the café report? [6 marks] Show worked answer
(a) Sum \( = 615 \); mean \( = 615 \div 7 \approx 87.9 \) customers [1]. Ordered: 60, 61, 62, 63, 64, 65, 240; median \( = 63 \) [1].
(b) Lower half 60, 61, 62 gives \( Q_1 = 61 \); upper half 64, 65, 240 gives \( Q_3 = 65 \); IQR \( = 4 \) [1]. Upper fence \( = 65 + 1.5 \times 4 = 71 \) [1]. Since 240 > 71, it is an outlier [1].
(c) Median and IQR, because the outlier pulls the mean far above a typical day and would inflate the SD; the median (63) represents a normal day [1].

Check yourself

1. A dataset’s mean is much larger than its median. The distribution is most likely…

A long right tail pulls the mean above the median.

2. Every value in a dataset is increased by 10. What happens to the standard deviation?

Adding a constant shifts every value equally, so distances from the mean do not change.

3. Heights are bell-shaped with mean 170 cm and SD 7 cm. About 95% of people are between…

95% lie within 2 SD: \( 170 \pm 14 \) gives 156 to 184 cm.

Your score: 0 out of 3

The short version

  • Mean, median, mode: the mean is pulled by outliers, the median resists them, the mode works for categories.
  • Quartiles give the five-number summary; IQR \( = Q_3 – Q_1 \) is the resistant spread.
  • Standard deviation: square the deviations, divide by \( n – 1 \) for a sample, take the square root.
  • Z-scores compare across scales; the 68–95–99.7 rule applies to bell-shaped data; outliers lie beyond the 1.5 × IQR fences.
  • Report mean and SD for symmetric data, median and IQR for skewed data. The full chapter in Statistics Made Simple has over 50 practice questions with worked answers.
Want more questions like these, with worked answers?
The practice book has hundreds more, graded from easy to exam level.
Next: step 6 of 22Two-Way Tables and Simulation