C9.3; E9.3 - Mean, median, mode and range

C9.3; E9.3 - Mean, median, mode and range

One number can summarise a set of data, but different summaries answer different questions. You will calculate mean, median, mode and range from lists and exact-value frequency tables, then choose a summary that suits the purpose and recognise the effect of an extreme observation.

Mean from lists and frequency tables

Can one unusually large value change the mean while leaving the middle value unchanged? You will see why this happens by keeping the jobs of the different averages separate.

Imagine redistributing 2525 counters equally among five people. Each receives five counters. The mean makes the same calculation with data: add every value, then divide by the number of observations. It is the equal share of the total, even if some observations are above it and some below.

Mean

For at least one numerical observation,

mean=sum of all valuesnumber of values\text{mean}=\frac{\text{sum of all values}}{\text{number of values}}

For exact values xx with frequencies ff,

mean=sum of fxsum of f\text{mean}=\frac{\text{sum of }fx}{\text{sum of }f}

Each fxfx counts that value repeatedly. These formulas are not given in the List of formulas: learn them.

A mean need not be an observed value or a whole number. A mean of 1.71.7 goals is possible, although a single match cannot contain 1.71.7 goals. The mean describes the set, rather than an individual match.

Worked example 1: mean from individual values

Question. Find the mean of 2,4,4,6,92,4,4,6,9.

Plan. Add all five observations, including both fours, then divide by five.

Sum=2+4+4+6+9=25every observation contributesMean=255=5equal share of the total\begin{aligned} \text{Sum}&=2+4+4+6+9=25 && \text{every observation contributes}\\ \text{Mean}&=\frac{25}{5}=5 && \text{equal share of the total} \end{aligned}

Answer. The mean is 55.

Check. Relative to five, the differences are −3,−1,−1,1,4-3,-1,-1,1,4, which add to zero. The amounts below five balance the amounts above it.

Keep the same observations but present the repeated value as a frequency. The denominator must still count five observations, rather than four distinct values.

Worked example 2: mean from a frequency table

Question. Calculate the mean of this distribution.

Value, xxFrequency, fffxfx
22112×1=22\times1=2
44224×2=84\times2=8
66116×1=66\times1=6
99119×1=99\times1=9

Plan. Multiply each value by how many times it occurs, add those products, then divide by total frequency.

Total frequency=1+2+1+1=5count observationsTotal of values=2+8+6+9=25add the productsMean=255=5divide total by count\begin{aligned} \text{Total frequency}&=1+2+1+1=5 && \text{count observations}\\ \text{Total of values}&=2+8+6+9=25 && \text{add the products}\\ \text{Mean}&=\frac{25}{5}=5 && \text{divide total by count} \end{aligned}

Answer. The mean is 55.

Check. Expand the table back to 2,4,4,6,92,4,4,6,9. It is exactly Example 1's list, so it must give the same mean.

Watch out

(2+4+6+9)÷4(2+4+6+9)\div4 ignores the repeated four. A frequency is a repetition count: multiply by it in the numerator and add it in the denominator.

The next table gives exact goal counts, so there is no estimation of the individual values.

Median and mode

The median finds the centre by position after ordering the data. The mode finds the most frequent value. Neither starts by adding all the values.

Median and mode

The median is the middle value of ordered data when there is an odd number of observations. With an even number, it is the mean of the two middle values.

The mode is the value occurring most often. There can be more than one mode if several values share the greatest frequency. If all values occur once, there is no mode.

For nn observations, the middle position is (n+1)/2(n+1)/2 when nn is odd. When nn is even, use positions n/2n/2 and n/2+1n/2+1, then average the values at those positions. A position is a counting instruction, rather than the answer itself.

Worked example 3: an odd number of observations

Question. Find the median and mode of 9,4,2,6,49,4,2,6,4.

Plan. Order the five observations, find the middle one and identify the most repeated value.

Ordered data=2,4,4,6,9sort before finding the middleMiddle position=5+12=3two values on each sideThird value=4read the value in that position\begin{aligned} \text{Ordered data}&=2,4,4,6,9 && \text{sort before finding the middle}\\ \text{Middle position}&=\frac{5+1}{2}=3 && \text{two values on each side}\\ \text{Third value}&=4 && \text{read the value in that position} \end{aligned}

Answer. Median 44; mode 44, because four occurs twice and each other value occurs once.

Check. Cross off the smallest and largest, then the next smallest and largest. One four remains in the centre. The mode comes from repetition, which is a different reason even though the answers coincide here.

Now add one observation, eight. There are two middle positions. A frequency table helps locate them without writing out a long repeated list.

Worked example 4: an even number in a frequency table

Question. Find the median and mode after adding 88 to Example 3.

ValueFrequencyPositions in the ordered data
221111
44222,32,3
661144
881155
991166

Plan. Count six observations, locate positions three and four, then average their values.

Middle values=4 and 6positions 3 and 4Median=4+62=5average both middle valuesMode=4greatest frequency is 2\begin{aligned} \text{Middle values}&=4\text{ and }6 && \text{positions 3 and 4}\\ \text{Median}&=\frac{4+6}{2}=5 && \text{average both middle values}\\ \text{Mode}&=4 && \text{greatest frequency is 2} \end{aligned}

Answer. Median 55; mode 44.

Check. The full list is 2,4,4,6,8,92,4,4,6,8,9. Pairing off the extremes leaves 44 and 66, confirming a midpoint of five. The median need not be one of the observations.

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