Mean, median and range in continuous distributions | Year 10

01 / LEARN

Understand the idea.

Centre and spread answer different questions

A continuous variable measures a quantity such as time or mass. Rounding a time to tenths does not make the underlying variable discrete. The mean shares the total equally; the median locates the middle of the ordered observations.

Order the values before taking the median, averaging the middle pair when there are an even number. Calculate the range from the largest and smallest values. All three summaries retain the measurement unit, but an extreme value can pull the mean away from most observations.

\bar{x} = \frac{\sum x}{n}

Grouped measurements give estimates

A grouped table hides the individual measurements. Replace each class by its midpoint, then multiply by its frequency: a midpoint representing five observations must contribute five times. Divide the sum of these products by the total frequency.

This produces an estimated mean, because the actual observations need not equal their class midpoints. Cumulative counts can locate the class containing the middle observation, but grouped data alone generally cannot determine its exact value.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find all three summaries

Race times are 12.4, 12.8, 13.1, 13.1, 14.6 and 15.0 seconds. Find the mean, median and range.

Worked solution

1
Find the total and count.
$12.4 + 12.8 + 13.1 + 13.1 + 14.6 + 15.0 = 81.0 s, n = 6$
The mean uses every race time once.
2
Share the total equally.
$\bar{x} = \frac{81.0}{6} = 13.5 s$
Divide the total by the number of observations.
3
Average the middle pair.
$median = \frac{13.1 + 13.1}{2} = 13.1 s$
The ordered list has six values, so positions three and four are central.
4
Subtract the extreme times.
$range = 15.0 - 12.4 = 2.6 s$
Range measures the full observed span.

Answer

Mean 13.5 s; median 13.1 s; range 2.6 s.

EXAMPLE 02

Estimate a mean from intervals

Estimate the mean journey time using the table.

Journey times
Time interval(min)	Frequency
0≤t<10	2
10≤t<20	5
20≤t<30	3

Worked solution

1
Find the class midpoints.
$\frac{0 + 10}{2} = 5, \frac{10 + 20}{2} = 15, \frac{20 + 30}{2} = 25 min$
Each midpoint represents the observations in its interval.
2
Weight each midpoint by its count.
$\sum f m = 2 (5) + 5 (15) + 3 (25) = 160 min$
Classes with more journeys must contribute more to the estimated total.
3
Divide by the total frequency.
$\bar{t} \approx \frac{160}{2 + 5 + 3} = 16 min$
There are ten journeys; the unknown individual times make this an estimate.

Answer

Estimated mean 16 minutes.

03 / YOUR TURN

Try it yourself.

Check your understanding

Times are 4.2, 4.7, 5.1 and 6.0 minutes. What is their median?

Show answer & reasoning

4.9 minutes

The four values are already ordered. Average positions two and three: $\frac{4.7 + 5.1}{2} = 4.9 min$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Find the mean of the lengths 2.4, 2.8, 3.2 and 3.6 cm.
2 marks
Find the median and range of 8.2, 7.5, 9.0, 8.4 and 7.9 seconds.
2 marks
Find the mean and median of 1.2, 1.4, 1.5, 1.7 and 9.2 kg. Which is more typical of the four smaller masses?
3 marks
Estimate the mean waiting time for the table. Explain why it is not exact.
3 marks
Waiting times
Interval(min) Frequency
$0 \leq t < 10$ 7
$10 \leq t < 20$ 2
$20 \leq t < 30$ 1
One class has mean height 1.55 m. Another has median height 1.55 m. Does this prove their height distributions are the same?
2 marks
A runner's five times have mean 14.2 s. Four times total 55.9 s. Find the fifth time.
3 marks

Waiting times
Interval(min)	Frequency
$0 \leq t < 10$	7
$10 \leq t < 20$	2
$20 \leq t < 30$	1

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the mean of the lengths 2.4, 2.8, 3.2 and 3.6 cm.
2 marks
The total is 12.0 cm.
$12.0 \div 4 = 3.0$ cm.
3.0 cm
Find the median and range of 8.2, 7.5, 9.0, 8.4 and 7.9 seconds.
2 marks
Order 7.5, 7.9, 8.2, 8.4, 9.0.
The middle is 8.2 and $9.0 - 7.5 = 1.5$ .
Median 8.2 s; range 1.5 s
Find the mean and median of 1.2, 1.4, 1.5, 1.7 and 9.2 kg. Which is more typical of the four smaller masses?
3 marks
The sum is 15.0, so the mean is 3.0.
The middle value is 1.5; 9.2 pulls the mean upward.
Mean 3.0 kg; median 1.5 kg; median is more typical.
Estimate the mean waiting time for the table. Explain why it is not exact.
3 marks
Waiting times
Interval(min) Frequency
$0 \leq t < 10$ 7
$10 \leq t < 20$ 2
$20 \leq t < 30$ 1
Midpoints are 5, 15, 25.
$\frac{5 \times 7 + 15 \times 2 + 25 \times 1}{10} = \frac{90}{10} = 9$ .
Estimated mean 9 minutes; interval midpoints replace unknown individual times.
One class has mean height 1.55 m. Another has median height 1.55 m. Does this prove their height distributions are the same?
2 marks
The summaries measure different features.
Different data sets can share a summary without sharing spread or shape.
No.
A runner's five times have mean 14.2 s. Four times total 55.9 s. Find the fifth time.
3 marks
The five times total $5 \times 14.2 = 71.0$ s.
$71.0 - 55.9 = 15.1$ s.
15.1 s

Waiting times
Interval(min)	Frequency
$0 \leq t < 10$	7
$10 \leq t < 20$	2
$20 \leq t < 30$	1

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Curriculum reference & lesson scope

Australian Curriculum v 9 and SA Curriculum Mathematics

Year 10 continuous distributions, bivariate data, two-way categorical tables, bias and investigations, with the approved percentile and standard-deviation content included in the normal chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Centre and spread answer different questions

Grouped measurements give estimates

See the method in action.

Find all three summaries

Find the total and count.

Share the total equally.

Average the middle pair.

Subtract the extreme times.

Estimate a mean from intervals

Find the class midpoints.

Weight each midpoint by its count.

Divide by the total frequency.

Try it yourself.

Check your understanding

Keep practising