Mean of Data

The mean (also called the arithmetic mean or average) is the most familiar way to summarise a data set with a single number. Add all the values and divide by how many there are. Simple to compute, but the result captures the "centre of mass" of the data , a balance point that distils the entire distribution into one number.

Definitions

For a data set $x_1, x_2, \ldots, x_n$, the mean is $\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{1}{n} \sum_{i=1}^{n} x_i.$

If the data is given with frequencies $f_1, f_2, \ldots, f_n$ for distinct values $x_1, x_2, \ldots, x_n$ (or class marks), the mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}.$

For grouped data, $x_i$ is the class mark (midpoint) of each class interval.

Computing the mean , ungrouped data

Example. Heights of $5$ students: $145, 150, 152, 155, 160$ cm.

$\bar{x} = \dfrac{145 + 150 + 152 + 155 + 160}{5} = \dfrac{762}{5} = 152.4$ cm.

Computing the mean , ungrouped frequency

Example. Number of children per family in 20 families:

$\sum f_i x_i = 0 \cdot 3 + 1 \cdot 7 + 2 \cdot 7 + 3 \cdot 2 + 4 \cdot 1 = 31$. $\sum f_i = 20$. Mean: $\bar{x} = \dfrac{31}{20} = 1.55$ children per family.

Computing the mean , grouped data

Example. Marks in classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 4, 8, 12, 10, 6.

Class marks: 5, 15, 25, 35, 45. Multiply: $5 \cdot 4 + 15 \cdot 8 + 25 \cdot 12 + 35 \cdot 10 + 45 \cdot 6 = 20 + 120 + 300 + 350 + 270 = 1060$. Total frequency: $40$. Mean: $\dfrac{1060}{40} = 26.5$.

Properties of the mean

Linearity. If you add a constant $c$ to every observation, the mean shifts by $c$. If you multiply every observation by $c$, the mean scales by $c$.

Sum of deviations. $\sum (x_i - \bar{x}) = 0$. The mean is the unique value for which the sum of "deviations" is zero. This is the "balance point" interpretation.

Affected by extremes. A single very large or very small value can shift the mean significantly. For data with outliers, the median (next lesson) is often more representative.

Worked examples

Example 1. Find the mean of $4, 7, 9, 12, 18$.

$\bar{x} = \dfrac{4 + 7 + 9 + 12 + 18}{5} = \dfrac{50}{5} = 10$.

Example 2. The mean of $5$ values is $20$. If one value, $40$, is added, find the new mean.

Sum was $5 \cdot 20 = 100$. New sum: $140$. New count: $6$. New mean: $\dfrac{140}{6} \approx 23.33$.

Example 3. Find the mean of:

$\sum f_i x_i = 20 + 80 + 90 + 40 = 230$. $\sum f_i = 20$. Mean: $11.5$.

Example 4. Find the mean of grouped data: classes 0-20, 20-40, 40-60, 60-80 with frequencies 5, 8, 4, 3.

Class marks: 10, 30, 50, 70. $\sum f_i x_i = 50 + 240 + 200 + 210 = 700$. $\sum f_i = 20$. Mean: $35$.

Example 5. The mean of $10$ observations is $15$. If each observation is increased by $3$, find the new mean.

New mean = $15 + 3 = 18$ (by linearity).

Try it yourself

1. Mean of $3, 5, 7, 9, 11$.
2. Mean of $2, 4, 6, 8, 10, 12$.
3. The mean of $25$ values is $40$. Find the total.
4. Mean of the data: $x$: 2, 5, 10, 15. $f$: 3, 4, 2, 1.
5. Mean of grouped data: classes 0-10, 10-20, 20-30, 30-40, 40-50. $f$: 2, 6, 10, 8, 4.
6. The mean of $6$ values is $20$. The values are $14, 18, 22, 26, x, 24$. Find $x$.
7. If a constant $5$ is subtracted from every observation, what happens to the mean?
8. If each observation is doubled, what happens to the mean?
9. The mean of $40$ students' marks is $58$. The mean of $30$ girls is $60$ and that of $10$ boys is found incorrectly. Find the boys' mean.
10. Mean of $3, 6, 9, 12, x$ is $9$. Find $x$.

Pitfalls / Insight

• The mean has units. Always state them.
• The mean is affected by outliers. A salary list with one very high salary has a high mean but a possibly low median.
• For grouped data, use class marks (midpoints), not class limits.

Insight. The mean is the balance point of the data , the single number that minimises total squared deviation. It is the default summary statistic in nearly every scientific discipline, from physics to economics to medicine.