0580

Averages and Range

Statistics

Why it is only an estimate

  • Grouped data records how many values fall in each class interval but not the values themselves
  • Without the original values the exact total cannot be found, so the mean can only be estimated
  • The estimate assumes every value in a class sits at the midpoint of that class
  • Find a midpoint by adding the two endpoints and halving
  • The phrase "estimate of the mean" in a question is the signal that the data is grouped

The method

  • Add a midpoint column, then a frequency × midpoint column
  • Total the frequency column and the frequency × midpoint column
  • Divide the second total by the first
  • Whichever class holds the largest frequency is the modal class
Exam tip

Use the midpoint of each class, not its width or its upper end, and remember the answer is an estimate: the original values are gone. Show both totals, the total frequency and the total of frequency × midpoint, because the method marks sit in those two lines rather than in the final division.

Worked example

An estimate of the mean from grouped data

The table shows the time t minutes taken by 50 people: 8 people with 0 < t ⩽ 10, 14 with 10 < t ⩽ 20, 18 with 20 < t ⩽ 30 and 10 with 30 < t ⩽ 40. Calculate an estimate of the mean, and write down the modal class.

Solution:

  • Midpoints are (0 + 10) ÷ 2 = 5, then 15, 25 and 35
  • Multiply each frequency by its midpoint
  • 8 × 5 = 40, 14 × 15 = 210, 18 × 25 = 450, 10 × 35 = 350
  • Total of frequency × midpoint = 40 + 210 + 450 + 350 = 1050
  • Total frequency = 8 + 14 + 18 + 10 = 50
  • Estimate of the mean = 1050 ÷ 50 = 21 minutes
  • A frequency of 18 is the largest in the table, so the modal class is 20 < t ⩽ 30