When data come in a frequency table, each value stands for several people. Multiply before you add, count down the table to find the median, and, on Extended, use midpoints to estimate the mean of grouped data.
Mean \(= \dfrac{\sum fx}{\sum f}\): multiply each value by its frequency, add, then divide by the TOTAL frequency. For grouped data (Extended) use the midpoint of each class as \(x\), and call the answer an estimate.
Section 01
Mean, median and mode from a table.
Open the lab's frequency table of goals scored. Move one frequency slider and watch the \(f \times x\) column, its total \(\sum fx\), \(\sum f\) and the mean change together. Then use the cumulative frequency column to find the median: the first running total that reaches the middle position. Switch to Grouped for the Extended version.
Mean: add an \(fx\) column. Mean \(= \dfrac{\sum fx}{\sum f}\), where \(\sum f\) is the number of values.
Median: find the position \(\dfrac{n+1}{2}\), then count down the frequencies (a running total) until you reach it.
Mode: the VALUE with the highest frequency, not the frequency itself.
Range: largest value \(-\) smallest value that actually has a frequency.
Worked example All four from a table
The numbers of goals scored in 25 matches are shown.
Divide \(\sum fx\) by the total FREQUENCY (25 matches), not by the number of rows (5). And the mode is the value 1, not its frequency 8.
Section 02
Grouped data: an estimate.
Extended only. When the data are grouped into classes, the exact values are lost, so you cannot find the exact mean. Assume every value sits at the midpoint of its class: the mean you get is an estimate.
Midpoint of a class \(= \dfrac{\text{lower bound} + \text{upper bound}}{2}\); the class \(20 < t \le 30\) has midpoint 25.
Estimated mean \(= \dfrac{\sum f \times \text{midpoint}}{\sum f}\).
Modal class: the class with the highest frequency (for equal class widths).
Class containing the median: count down the running totals to position \(\dfrac{n+1}{2}\) (or \(\dfrac{n}{2}\) for large \(n\)).
Worked example Estimate the mean (Extended)
The times, \(t\) minutes, that 30 people waited are shown.
Time
Frequency
\(0 < t \le 10\)
4
\(10 < t \le 20\)
9
\(20 < t \le 30\)
12
\(30 < t \le 40\)
5
Calculate an estimate of the mean, and state the modal class.
Use the MIDPOINT, not the upper bound of each class. Using 10, 20, 30, 40 in the example gives 26 minutes, which is too big. Note also that the last class here is 20 wide, so its midpoint is 40, not 35.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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A table has values 1, 2, 3 with frequencies 4, 5, 1. How many values are there?
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Values 0, 1, 2, 3 have frequencies 3, 8, 6, 2. What is the mode?
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(Extended) What midpoint is used for the class \(20 < t \le 30\)?
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(Extended) Why is the mean of grouped data only an estimate?
Section 03
Working backwards with totals.
Many exam questions give the mean and ask for something else. The key move is always the same: total = mean \(\times\) number of values. Totals can be added and subtracted; means cannot.
Worked example Combining two groups
Class P has 20 students with a mean mark of 64. Class Q has 30 students with a mean mark of 59. Find the mean mark of all 50 students.
Total for P: \(20 \times 64 = 1280\). Total for Q: \(30 \times 59 = 1770\).
Combined total: \(1280 + 1770 = 3050\).
Mean: \(\dfrac{3050}{50} = 61\).
Answer: 61
Examiner trap
The mean of the two means, \(\dfrac{64 + 59}{2} = 61.5\), is wrong because the classes are different sizes. Combine TOTALS, then divide by the total number of students.
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
Exam style+20 XP
Shoe size
3
4
5
6
Frequency
2
7
8
3
Calculate the mean shoe size.
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Children in family
0
1
2
3
4
Frequency
6
11
9
3
1
Find the median number of children.
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(Extended) The heights, \(h\) cm, of 40 students are grouped: \(140 < h \le 150\): 5, \(150 < h \le 160\): 12, \(160 < h \le 170\): 15, \(170 < h \le 180\): 8. Calculate an estimate of the mean height.
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\(x\)
0
1
2
3
\(f\)
4
7
\(k\)
3
The mean is 1.5. Find \(k\).
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Frequency tables, averaged.
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