Year 11 Standard · Unit 1 Describing data · Lesson 1 of 6

How do you find an average when only a table is left?

A grouped table keeps the counts and loses the exact values. This lesson asks what can still be said about the centre of the data, and how far any answer read from a table can be trusted.

Statistics and ProbabilityScientific calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 8: the mean of 4, 7, 9, 12 and 8 is ?
    Answer8. The total is 40 and there are 5 values.
  2. From Year 8: the median of 3, 9, 4, 7, 12 and 5 is ?
    Answer6. Order them: 3, 4, 5, 7, 9, 12. The middle pair is 5 and 7.
  3. From Year 7: a frequency table shows a score of 3 occurring 6 times. How much do those values add to the total?
    Answer18. 3 × 6 = 18.
Words used in this lesson
class
a group of values in a grouped table, such as 20 to under 30. For example: the interval 60 to under 64.
class boundary
the lower or upper end of a class; a value on the upper boundary belongs to the next class. For example: 64 is the start of the next class.
midpoint
halfway between the boundaries; the stand-in value for a class. For example: 62 for the class 60 to under 64.
estimated mean
the sum of frequency times midpoint divided by the total frequency. For example: the average found from a table, not from the raw data.
modal class
the class with the highest frequency, when the classes have equal width. For example: the tallest bar.
median class
the class that holds the middle value of the ordered data. For example: where the running total passes half the values.

The table that kept the counts

A cycling club in Kigali measured how long each member's commute took, and then lost the notebook. Only a table survives: how many commutes fell into 10 to under 20 minutes, how many into 20 to under 30, and so on.

Predict first: with no exact times left, can anyone still say what the mean commute was? If you think you can, would you trust the answer completely? Slide the values around in the first figure to test your ideas, then use the second to find the class that holds the middle value.

0
22.5mean of the 20 times
450sum of the 20 times

Every value at the start of its class: sum 450 ÷ 20 = 22.5.

Show me the steps
  1. Every value at the start of its class: 10 × 5 + 20 × 8 + 30 × 4 + 40 × 3 = 450, so the mean is 450 ÷ 20 = 22.5.
  2. Slide to 5: every value sits halfway through its class, the sum is 550 and the mean is 27.5. This is the midpoint estimate.
  3. Slide to 10: every value is at the very end of its class, the sum is 650 and the mean is 32.5. The real mean is somewhere between 22.5 and 32.5, and the midpoint estimate is in the middle of that band.

Every value in a class gets one stand-in: the midpoint.

  • The midpoint of a class is halfway between its two boundaries: (lower + upper) ÷ 2.
  • Estimated mean = (sum of frequency × midpoint) ÷ (total frequency).
  • The modal class is the class with the highest frequency (when the classes have equal width).
  • The median class is the class that holds the middle value: count along the running total until it passes half of the values.

All of these are estimates. The exact values are gone, so the answer is only as good as the table.

1
10 to under 20class holding this value
5values up to the end of that class

Value number 1 is in the class 10 to under 20; 5 values are in that class or below.

Show me the steps
  1. Value number 1 is the smallest. It is in 10 to under 20, where the running total reaches 5.
  2. Slide to 10 and then 11: both are in 20 to under 30, because the running total there reaches 13. The median of 20 values is halfway between the 10th and 11th, so 20 to under 30 is the median class.
  3. Slide to 18, 19 and 20: the largest values are in 40 to under 50, the last class.

The median class is not the middle class. A table with five classes does not have its median in the third one. Count the running total of frequencies and find the class where it passes half of all the values. And do not average the midpoints without the frequencies: a class with 1 value and a class with 40 values do not count equally. For a table of 60 values, the middle sits at the 30th and 31st values, so check which class holds them.

Worked example: mangoes in a market box

The masses of 40 mangoes in a Kerala market are grouped. Estimate the mean and name the median class.

Mass (g)Frequency
100 to under 1204
120 to under 14010
140 to under 16014
160 to under 1807
180 to under 2005
  1. Midpoints: 110, 130, 150, 170 and 190.
  2. f × midpoint: 440, 1300, 2100, 1190 and 950. Their total is 5980.
  3. Estimated mean = 5980 ÷ 40 = 149.5 g.
  4. Running totals: 4, 14, 28, 35, 40. The 20th and 21st values are in the third class, so the median class is 140 to under 160.

Your go. The first step is given. Waiting times at a clinic in Nairobi: 0 to under 5 minutes (6 people), 5 to under 10 (11), 10 to under 15 (8), 15 to under 20 (5). Estimate the mean waiting time.

  1. The midpoints are 2.5, 7.5, 12.5 and 17.5, and there are 30 people.
  2. Multiply each frequency by its midpoint, add them up, then divide by 30.
Check the answer

6 × 2.5 = 15, 11 × 7.5 = 82.5, 8 × 12.5 = 100 and 5 × 17.5 = 87.5. The total is 285 and 285 ÷ 30 = 9.5 minutes. A common slip is to divide by 4, the number of classes, instead of 30, the number of people.

With a calculator

Build the sum one class at a time so a slip is quick to find. Type frequency × midpoint for the first class and press the memory-add key (often M+), then do the next class, and so on. Recall the memory for the total, then divide by the total frequency.

Write each product on paper as you go. If the final mean is outside the table's range, a product was mistyped.

Check before you go on

1. In a grouped table the class 30 to under 40 has frequency 8. When you estimate the mean, what does this class add to the sum before you divide by the total frequency?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A class in a table runs from 24 to under 36 minutes. What midpoint stands in for every value in it?

3. A table has 3 values in 0 to under 10, 5 in 10 to under 20 and 2 in 20 to under 30. What is the total of frequency × midpoint?

Level 2 · Building

4. Times to finish a puzzle: 0 to under 20 minutes (4 people), 20 to under 40 (9), 40 to under 60 (12), 60 to under 80 (5). Estimate the mean time in minutes.

5. Fish landed by 50 boats, in kilograms: 0 to under 5 (3 boats), 5 to under 10 (8), 10 to under 15 (13), 15 to under 20 (9), 20 to under 25 (11), 25 to under 30 (6). Which class holds the median?

Level 3 · Stretch

6. A table has 0 to under 20 (5 values), 20 to under 40 (x values), 40 to under 60 (10 values) and 60 to under 80 (5 values). The estimated mean is 36. What is x?

7. Daily rainfall over 30 days: 0 to under 10 mm (8 days), 10 to under 30 mm (12 days), 30 to under 60 mm (10 days). Estimate the mean rainfall in mm, to one decimal place.

Something to think about

A table has one class that is 0 to under 100 and holds most of the values. Another table of the same data splits that class into ten classes of width 10. Which table gives the better estimate of the mean, and why? Say what you would need to know to be sure.

One way to think about it

The table with narrower classes. Each stand-in value is then closer to the values it replaces, so the estimate relies less on errors happening to cancel, and the band the true mean must lie in is narrower. With one huge class the midpoint can be far from where most values really sit. You can never be sure from the table alone: only the raw values would settle it.

Exit ticket

8. A table of 10 scores has 2 in 5 to under 15, 5 in 15 to under 25 and 3 in 25 to under 35. What is the estimated mean?