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Lec18p2 - JUST THE MATHS UNIT NUMBER 18.2 STATISTICS...

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“JUST THE MATHS” UNIT NUMBER 18.2 STATISTICS 2 (Measures of central tendency) by A.J.Hobson 18.2.1 Introduction 18.2.2 The arithmetic mean (by coding) 18.2.3 The median 18.2.4 The mode 18.2.5 Quantiles 18.2.6 Exercises 18.2.7 Answers to exercises

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UNIT 18.2 - STATISTICS 2 MEASURES OF CENTRAL TENDENCY 18.2.1 INTRODUCTION Having shown, in Unit 18.1, how statistical data may be presented in a clear and concise form, we shall now be concerned with the methods of analysing the data in order to obtain the maximum amount of information from it. In the previous Unit, it was stated that statistical problems may be either “descriptive problems” (in which all the data is known and can be analysed) or “inference problems” (in which data collected from a “sample” population is used to infer properties of a larger population). In both types of problem, it is useful to be able to measure some value around which all items in the data may be considered to cluster. This is called “a measure of central tendency” ; and we find it by using several types of average value as follows: 18.2.2 THE ARITHMETIC MEAN (BY CODING) To obtain the “arithmetic mean” of a finite collection of n numbers, we may simply add all the numbers together and then divide by n . This elementary rule applies even if some of the numbers occur more than once and even if some of the numbers are negative. However, the purpose of this section is to introduce some short-cuts (called “coding” ) in the calculation of the arithmetic mean of large collections of data. The methods will be illustrated by the following example, in which the number of items of data is not over-large: EXAMPLE The solid contents, x , of water (in parts per million) was measured in eleven samples and the following data was obtained: 4520 4490 4500 4500 4570 4540 4520 4590 4520 4570 4520 Determine the arithmetic mean, x , of the data. 1
Solution (i) Direct Calculation By adding together the eleven numbers, then dividing by 11, we obtain x = 49840 ÷ 11 4530 . 91 (ii) Using Frequencies We could first make a frequency table having a column of distinct values x i , ( i = 1 , 2 , 3 , ...... , 11), a column of frequencies f i , ( i = 1 , 2 , 3 ...... , 11) and a column of corresponding values f i x i . The arithmetic mean is then calculated from the formula x = 1 11 11 i =1 f i x i . In the present example, the table would be x i f i f i x i 4490 1 4490 4500 2 9000 4520 4 18080 4540 1 4540 4570 2 9140 4590 1 4590 Total 49840 The arithmetic mean is then x = 49840 ÷ 11 4530 . 91 as before.

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Lec18p2 - JUST THE MATHS UNIT NUMBER 18.2 STATISTICS...

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