Range and Averages

Overview Range and Averages

I find this sec­tion not too demand­ing. You have to learn some terms (range, mean, mode and median) and one or two tech­niques (for exam­ple, how to esti­mate the mean from a grouped fre­quency table).

There may be a Grade A ques­tion about the most appro­pri­ate aver­age to use for a cer­tain set of data. This is not as demand­ing as it seems and requires just a lit­tle com­mon sense.

As pre­vi­ously, the best way for me to learn and mem­o­rise this sub­ject is to set some ques­tions, explain the approach and then detail the answer.

Range, Dif­fer­ent types of Aver­age and Appro­pri­ate Averages


A small pri­mary school has a staff of 14 employ­ees. The indi­vid­ual salaries of the staff are as follows:

£5,000, £18,000, £23,000, £5000, £16,000, £30,000, £175,000, £4,800, £24,000, £17,000, £21,000, £28,000, £13,000 and £6,000.


  1. The range of the salaries.
  2. The mode of the salaries.
  3. The median of the salaries.
  4. The mean of the salaries.

Dis­cuss which the most appro­pri­ate aver­age to use for this set of data.


Tip: Where you have a rea­son­ably large num­ber of val­ues, say more than 5, it is use­ful to write the val­ues down in ascend­ing order (small­est to largest). This helps you to cal­cu­late the range, mode and median quickly and with less chance of mak­ing a mistake.


The range of the data is cal­cu­lated as :- Largest Value — Small­est Value

The mode (some­times called the modal value) is the most fre­quent value.

The median is the mid­dle value when all the data points are sorted in ascend­ing order. There is a for­mula for this:

Median Value= (n+1)/2  n= num­ber of val­ues in the data set.

NB if you have an odd num­ber of val­ues this is quite straight­for­ward as the median value will always be a whole num­ber. How­ever if you an even num­ber of val­ues the median value will always be x.5. In this case you add the two val­ues either side of the median and divide by two. In this way you still get a value where the num­ber of data points below the median is equal to the num­ber of data points above the median. I’m sure I’m mak­ing this sound far more dif­fi­cult than it is! best to have a look at the answer below.

The mean is the mean­est aver­age as it takes the longest to cal­cu­late. The for­mula is:-

Sum of all the values/Number of Values

The most appro­pri­ate aver­age to use is the aver­age that best rep­re­sents the full range of val­ues. If there are very small or very large val­ues that drag the mean down or up then the median could be more appro­pri­ate. Sim­i­larly if the modal value is towards the bot­tom or top of a range then it is not likely to be the most appro­pri­ate aver­age. Gen­er­ally the larger the num­ber of val­ues and the more evenly they are dis­trib­uted the more likely it is that the mean will be the most appro­pri­ate average.


Sort­ing the val­ues into ascend­ing order we have:-

£4,800, £5,000, £5,000, £6,000, £13,000, £16,000, £17,000, £18,000, £21,000, £23,000, £24,000, £28,000, £30,000, £175,000

1. The range of salaries =

£175,000 — £4,800 = £170,200

2. The mode of the salaries = £5,000. This is the only salary that is repeated.

3. The median of the salaries=

(Num­ber of Val­ues + 1)/2 = (14+1)÷2 =7.5th value

7th Value = £17,000, 8th Value = £18,000

Median = (£17,000 + £18,000)/2 = £17,500

4. The mean of the salaries is:

Sum of all the salaries/Number of val­ues = £385,800/15 = £25,720

The most appro­pri­ate aver­age to use in this case is the median. The mode is not appro­pri­ate as only one value is repeated and that value is the sec­ond small­est in the range. The mean is not appro­pri­ate as it is skewed by the largest value of £175,000. Only three salaries are in excess of the mean whereas eleven salaries are below the mean. With the mode and mean aver­ages at the low and high ends of the range, this leaves the median which tells you that half of the salaries are below £17,500 and half the salaries are above £17,500.

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