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Prices, location and spread
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2.1 The mean of a combined batch

This first subsection looks at how a mean can be calculated when two unequally weighted batches are combined.

Example 7 Alan’s and Beena’s biscuits

Suppose we are conducting a survey to investigate the general level of prices in some locality. Two colleagues, Alan and Beena, have each visited several shops and collected information on the price of a standard packet of a particular brand of biscuits. They report as follows (Figure 9).

  • Alan visited five shops, and calculated that the mean price of the standard packet at these shops was 81.6p.

  • Beena visited eight shops, and calculated that the mean price of the standard packet at these shops was 74.0p.

Described image
Figure 9 Means of biscuit prices

If we had all the individual prices, five from Alan and eight from Beena, then they could be amalgamated into a single batch of 13 prices, and from this combined batch we could calculate the mean price of the standard packet at all 13 shops. However, our two investigators have unfortunately not written down, nor can they fully remember, the prices from individual shops. Is there anything we can do to calculate the mean of the combined batch?

Fortunately there is, as long as we are interested in arithmetic means. (If they had recorded the medians instead, then there would have been very little we could do.)

The mean of the combined batch of all 13 prices will be calculated as

fraction sum open bracket of the combined batch prices close bracket over size open bracket of the combined batch close bracket end .

We already know that the size of the combined batch is the sum of the sizes of the two original batches; that is, 5+8=13 . The problem here is how to find the sum of the combined batch of Alan’s and Beena’s prices. The solution is to rearrange the familiar formula

mean = fraction sum over size end

so that it reads

sum = mean times size .

This will allow us to find the sums of Alan’s five prices and Beena’s eight prices separately. Adding the results will produce the sum of the combined batch prices. Finally, dividing by 13 completes the calculation of finding the combined batch mean.

Let us call the sum of Alan’s prices ‘sum(A)’ and the sum of Beena’s prices ‘sum(B)’.

For Alan: mean = 81.6 and size = 5 , so sum open bracket A close bracket = 81.6 times 5=408 .

For Beena: mean =74.0 and size =8 , so sum open bracket B close bracket =74.0 times 8 =592 .

For the combined batch:

mean = fraction combined sum over combined size end = fraction 408 +592 over 13 end = fraction 1000 over 13 end simeq 76.9

Here, the result has been rounded to give the same number of digits as in the two original means.

The process that we have used above is an important one. It will be used several times in the rest of this course. The box below summarises the method, using symbols.

Mean of a combined batch

The formula for the mean overline x subscript uppercase C end of a combined batch uppercase C is

overline x subscript uppercase C end = fraction overline x subscript uppercase A end n subscript uppercase A end + overline x subscript uppercase B end n subscript uppercase B end over n subscript uppercase A end + n subscript uppercase B end end comma

where batch uppercase C consists of batch uppercase A combined with batch uppercase B , and

overline x subscript uppercase A end = mean of batch uppercase A comma n subscript uppercase A end = size of batch uppercase A comma overline x subscript uppercase B end = mean of batch uppercase B comma n subscript uppercase B end = size of batch uppercase B.

For our survey in Example 7,

overline x subscript uppercase A end = 81.6 comma n subscript uppercase A end = 5 comma overline x subscript uppercase B end = 74.0 comma n subscript uppercase B end = 8.

The formula summarises the calculations we did as

overline x subscript uppercase C end = fraction open bracket 81 .6 times 5 close bracket + open bracket 74 .0 times 8 close bracket over 5 +8 end .

This expression is an example of a weighted mean. The numbers 5 and 8 are the weights. We call this expression the weighted mean of 81.6 and 74.0 with weights 5 and 8, respectively.

To see why the term weighted mean is used for such an expression, imagine that Figure 10 shows a horizontal bar with two weights, of sizes 5 and 8, hanging on it at the points 81.6 and 74.0, and that you need to find the point at which the bar will balance. This point is at the weighted mean: approximately 76.9.

Described image
Figure 10 Point of balance at the weighted mean

This physical analogy illustrates several important facts about weighted means.

  • It does not matter whether the weights are 5 kg and 8 kg or 5 tonnes and 8 tonnes; the point of balance will be in the same place. It will also remain in the same place if we use weights of 10 kg and 16 kg or 40 kg and 64 kg – it is only the relative sizes (i.e. the ratio) of the weights that matter.

  • The point of balance must be between the points where we hang the weights, and it is nearer to the point with the larger weight.

  • If the weights are equal, then the point of balance is halfway between the points.

This gives the following rules.

Rules for weighted means

Rule 1 The weighted mean depends on the relative sizes (i.e. the ratio) of the weights.

Rule 2 The weighted mean of two numbers always lies between the numbers and it is nearer the number that has the larger weight.

Rule 3 If the weights are equal, then the weighted mean of two numbers is the number halfway between them.

Example 8 Two batches of small televisions

Suppose that we have two batches of prices (in pounds) for small televisions:

Batch uppercase A has mean 119 and size 7. Batch uppercase B has mean 185 and size 13 .

To find the mean of the combined batch we use the formula above, with

overline x subscript uppercase A end = 119 comma n subscript uppercase A end = 7 comma overline x subscript uppercase B end = 185 comma n subscript uppercase B end = 13.

This gives

overline x subscript uppercase C end = fraction open bracket 119 times 7 close bracket + open bracket 185 times 13 close bracket over 7 +13 end = fraction 833 +2405 over 20 end = fraction 3238 over 20 end = 161.9 simeq 162.

Note that this is the weighted mean of 119 and 185 with weights 7 and 13 respectively. It lies between 119 and 185 but it is nearer to 185 because this has the greater weight: 13 compared with 7.

Example 8 is the subject of the following screencast. [Note that references to ‘the unit’ and ‘the units’ should be interpreted as ‘this course’. The original wording refers to the Open University course from which this material is adapted.]

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Screencast 2 Calculating a weighted mean
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