Measures of Central Tendency — chapter notes

A measure of central tendency is one number that represents a whole data set.

• Mean $\bar{x}$: the sum of the values divided by the number of values (the fair share).
• Median: the middle value of the data arranged in order.
• Mode: the value that occurs most often (a set may have one mode, several modes or none that stands out).

The mean uses every value, so an extreme value pulls it; the median is not affected by extreme values.

Add all the values and divide by how many there are.

Example: for $12, 7, 15, 9, 7, 18, 2$ the sum is $70$ and the mean is $\dfrac{70}{7} = 10$.

Useful facts: if every value is increased by $k$ the mean increases by $k$; if every value is multiplied by $k$ the mean is multiplied by $k$. When the mean is given, total = mean $\times$ number of values finds a missing value or the effect of replacing a value.

Median. Arrange the values in order. If $n$ is odd, the median is the $\dfrac{n + 1}{2}$th value. If $n$ is even, it is the mean of the $\dfrac{n}{2}$th and the next value.

Example: $23, 24, 27, 27, 29, 31, 35, 40$ ($n = 8$): median $= \dfrac{27 + 29}{2} = 28$.

Mode. The value that occurs most often: here $27$. A data set can have two modes (bimodal).

Each value $x$ is listed with its frequency $f$.

• Mean: $\bar{x} = \dfrac{\sum fx}{\sum f}$.
• Median: make a cumulative frequency column; with $n$ observations take the $\dfrac{n + 1}{2}$th (odd $n$) or the mean of the $\dfrac{n}{2}$th and next (even $n$).
• Mode: the value with the greatest frequency.

Example: goals in $40$ matches: $\sum fx = 82$, mean $2.05$, median $2$, mode $1$.

In grouped data each class is represented by its class mark $x = \dfrac{\text{lower limit} + \text{upper limit}}{2}$.

• Continuous classes ($20$–$30$, $30$–$40$): class mark $25$, $35$, $\ldots$
• Discontinuous classes ($20$–$29$, $30$–$39$): class mark $24.5$, $34.5$, $\ldots$

Direct method: find $x$, then $fx$, then $\bar{x} = \dfrac{\sum fx}{\sum f}$.

Example: reading minutes with $\sum f = 50$ and $\sum fx = 2260$ give $\bar{x} = 45.2$.

Choose an assumed mean $a$ (a class mark near the middle).

• Short-cut: $d = x - a$ and $\bar{x} = a + \dfrac{\sum fd}{\sum f}$.
• Step deviation: with equal class width $h$, $u = \dfrac{x - a}{h}$ and $\bar{x} = a + h \times \dfrac{\sum fu}{\sum f}$.

All three methods give the same mean. Example: electricity data, $a = 150$: $150 + \dfrac{-60}{50} = 148.8$; step deviation with $h = 20$: $150 + 20 \times \dfrac{-3}{50} = 148.8$.

Missing frequency. If the mean is given and one frequency $y$ is unknown, write $\sum f$ and $\sum fx$ in terms of $y$ and use $\sum fx = \text{mean} \times \sum f$.

Example: frequencies $6, 8, 12, y, 4$ in classes $10$–$20$ to $50$–$60$ with mean $34.5$: $930 + 45y = 34.5(30 + y)$, so $y = 10$. With two unknowns, use the total frequency as a second equation.

Changed data. Work with totals: replacing $9$ by $21$ in $12$ numbers of mean $18$ gives $\dfrac{216 - 9 + 21}{12} = 19$. Combined mean $= \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$.

• Median class: find $\dfrac{n}{2}$ and the cumulative frequencies; the median class is the first class whose cumulative frequency is at least $\dfrac{n}{2}$.
• Modal class: the class with the highest frequency.

Example: ages of $50$ members, cumulative frequencies $7, 19, 28, 42, 48, 50$: $\dfrac{n}{2} = 25$ lies in $40$–$50$ (median class); the highest frequency $14$ gives the modal class $50$–$60$. The two can be different classes.

Not in the ICSE 2027 syllabus: calculating the median or mode of grouped data by formula; only the median class and the modal class are identified.