Estimated Mean From a Grouped Table: Use the Midpoints

The question

A clinic recorded how long, $t$ minutes, each of 50 patients waited before seeing a doctor one morning. The results are shown in the table.

Waiting time ( $t$ minutes)	Frequency
$0 < t \le 10$	7
$10 < t \le 20$	12
$20 < t \le 30$	16
$30 < t \le 45$	10
$45 < t \le 60$	5

(a) Calculate an estimate of the mean waiting time. [4 marks]

(b) Explain why your answer to part (a) is only an estimate. [1 mark]

Cambridge IGCSE note: an estimate of the mean for grouped data is Extended content in 0580.

The common wrong answer

A common approach is to take the end of each class as the value for everyone in it, because those are the numbers written in the table. So the student multiplies each frequency by the upper boundary:

\begin{aligned} 10 \times 7 &= 70 \\ 20 \times 12 &= 240 \\ 30 \times 16 &= 480 \\ 45 \times 10 &= 450 \\ 60 \times 5 &= 300 \end{aligned}

The total is $70 + 240 + 480 + 450 + 300 = 1540$ , so

\text{mean} = \frac{1540}{50} = 30.8\ \text{minutes}

The method is right (add up $f \times$ value, then divide by the total frequency), but the value used for each class is wrong. Using the upper boundary assumes that every patient in a class waited as long as possible, for example that all 7 patients in $0 < t \le 10$ waited the full 10 minutes. That pushes the answer up every time.

Cambridge examiner reports describe this exact slip: candidates use the interval boundaries, or the class widths, instead of the midpoints. Using the class widths here gives $11.5$ minutes. Another slip is dividing by 5, the number of classes, instead of 50, the number of patients, which gives an “average” wait of over four hours.

The correct working

(a) Find the midpoint of each class by adding the two boundaries and halving. For example, the midpoint of $30 < t \le 45$ is

\frac{30 + 45}{2} = 37.5

Then multiply each midpoint, $x$ , by its frequency, $f$ :

Waiting time ( $t$ minutes)	Midpoint $x$	Frequency $f$	$fx$
$0 < t \le 10$	$5$	$7$	$35$
$10 < t \le 20$	$15$	$12$	$180$
$20 < t \le 30$	$25$	$16$	$400$
$30 < t \le 45$	$37.5$	$10$	$375$
$45 < t \le 60$	$52.5$	$5$	$262.5$
Total		$50$	$1252.5$

Divide the total of $fx$ by the total frequency:

\text{mean} = \frac{\sum fx}{\sum f} = \frac{1252.5}{50} = 25.05

So the estimated mean waiting time is $25.1$ minutes (to 3 significant figures).

(b) The table doesn’t show each patient’s actual waiting time, only the class it falls in. We assumed every patient waited the midpoint time of their class, so the answer is an estimate, not the exact mean.

Why it works

Once data is grouped, the exact times are lost. All you know is that, say, 10 patients waited somewhere between 30 and 45 minutes. Some waited nearer 30, some nearer 45. The fairest single value to represent the whole class is the one in the middle, $37.5$ . Some patients will be above it and some below, and these roughly cancel out.

The upper boundary is the most extreme choice in each class, so the errors all point the same way. Here it raised the estimate by $30.8 - 25.05 = 5.75$ minutes, almost a quarter of the true estimate.

Notice that the classes are not all the same width. The midpoints of the last two classes are $37.5$ and $52.5$ , not " $35$ " and " $45$ ". Always work each midpoint out from its own boundaries rather than continuing a pattern.

How to spot it next time

Add a midpoint column to the table before you do anything else, and use only those values in $fx$ .
Divide by the total frequency (the number of people or items), never by the number of classes.
Sense-check: the estimated mean must lie inside the range of the data, and usually near the classes with the biggest frequencies. Here $25.1$ sits in $20 < t \le 30$ , the busiest class.