9.7 Measures of central tendency
A measure of central tendency is a single value that stands for the whole set — what is loosely called an average. The three in common use are the mean, the median and the mode.
9.7.1 The arithmetic mean
Definition 9.22. The mean of \(n\) observations \(x_1,x_2,\ldots ,x_n\) is \[\bar {x}=\frac {\sum _{i=1}^{n}x_i}{n}.\] If the values occur with frequencies \(f_1,f_2,\ldots ,f_n\), then \[\bar {x}=\frac {1}{N}\sum _{i=1}^{n}f_ix_i,\qquad \text {where } N=\sum _{i=1}^{n}f_i.\]
Example 9.23. (a) Find the mean of the scores \(12\), \(7\), \(15\), \(9\), \(11\), \(6\).
(b) The table shows the number of children in each of \(50\) households. Find the mean number of children per household.
| Number of children, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of households, \(f\) | 4 | 9 | 15 | 12 | 7 | 3 |
Solution. (a) \[\bar {x}=\frac {12+7+15+9+11+6}{6}=\frac {60}{6}=10.\]
(b) With a frequency table, each value must be counted as many times as it occurs, so multiply before adding. It is worth setting the work out in a column:
| \(x\) | \(f\) | \(fx\) |
| 0 | 4 | 0 |
| 1 | 9 | 9 |
| 2 | 15 | 30 |
| 3 | 12 | 36 |
| 4 | 7 | 28 |
| 5 | 3 | 15 |
| Total | \(N=50\) | \(\sum fx=118\) |
\[\bar {x}=\frac {\sum fx}{N}=\frac {118}{50}=2.36.\]
Note 9.24. The answer \(2.36\) children is not a possible household, and that is not a mistake. A mean need not be a value the data could actually take; it is a balance point, not an observation.
Note 9.25. Dividing by \(6\) — the number of rows — instead of by \(50\) is the standard error here. The divisor is always the total frequency.
9.7.2 Assumed mean
Large numbers that are all close together are tedious to add. Guessing a value \(A\) near the mean and working with the small differences from it is quicker and less error-prone.
Definition 9.26. If \(A\) is an assumed mean and \(d_i=x_i-A\), then \[\bar {x}=A+\frac {\sum d_i}{n},\qquad \text {or}\qquad \bar {x}=A+\frac {\sum f_id_i}{N}\] for data in a frequency table.
Example 9.27. Use a suitable assumed mean to find \(\bar {x}\) for \[748,\quad 752,\quad 747,\quad 755,\quad 749,\quad 753,\quad 750,\quad 754.\]
Solution. All eight values sit near \(750\), so take \(A=750\). The differences \(d_i=x_i-750\) are \[-2,\quad 2,\quad -3,\quad 5,\quad -1,\quad 3,\quad 0,\quad 4,\] which are small enough to add mentally: \[\sum d_i=-2+2-3+5-1+3+0+4=8.\] Therefore \[\bar {x}=A+\frac {\sum d_i}{n}=750+\frac {8}{8}=750+1=751.\]
Note 9.28. The negative differences must keep their signs. Adding the sizes and ignoring the signs gives \(20\) instead of \(8\), and a mean of \(752.5\) — and the method’s whole advantage is lost to an error it was supposed to prevent.
Note 9.29. Any value of \(A\) gives the right answer; a poor guess just makes the arithmetic heavier. Taking \(A=0\) recovers the ordinary formula exactly.
9.7.3 Mean of a grouped distribution
Once data has been grouped, the individual values are gone. The class mark is used to stand for every observation in its class, and the mean is worked out as though every member of the class sat exactly at the midpoint.
Example 9.30. The daily earnings, in kwacha, of \(60\) market traders are recorded below. Estimate the mean daily earning.
| Earnings (K) | 1 – 20 | 21 – 40 | 41 – 60 | 61 – 80 | 81 – 100 |
| Number of traders | 6 | 11 | 20 | 15 | 8 |
Solution. The class marks are the averages of the class limits: \(\frac {1+20}{2}=10.5\), \(\frac {21+40}{2}=30.5\), and so on.
| Earnings (K) | Class mark \(x\) | \(f\) | \(fx\) |
| 1 – 20 | 10.5 | 6 | 63.0 |
| 21 – 40 | 30.5 | 11 | 335.5 |
| 41 – 60 | 50.5 | 20 | 1010.0 |
| 61 – 80 | 70.5 | 15 | 1057.5 |
| 81 – 100 | 90.5 | 8 | 724.0 |
| Total | \(N=60\) | \(\sum fx=3190\) | |
\[\bar {x}=\frac {\sum fx}{N}=\frac {3190}{60}=53.17\ \text {(2 d.p.)}\]
Note 9.31. This is an estimate, and the question says so. Assuming every trader in a class earned exactly the class mark is not true, though the errors above and below the midpoint largely cancel. With the original figures the mean could be calculated exactly; from the table it cannot.
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