Central moments for grouped data
Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. The mean of $X$ is denoted by $\overline{x}$ and is given by
$$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$
Formula
The first four central moments are as follows
$m_1=0$
$m_2 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2$
$m_3 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3$
$m_4 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4$
where,
$N$total number of observations$\overline{x}$sample mean
Example 1
Following tables shows a frequency distribution of daily number of car accidents at a particular cross road during a month of April.
| No.of car accidents ($x$) | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| No. of days ($f$) | 9 | 11 | 6 | 3 | 1 |
Compute first four central moments for the above frequency distribution.
Solution
| $x$ | Freq ($f$) | $f*x$ | |
|---|---|---|---|
| 2 | 9 | 18 | |
| 3 | 11 | 33 | |
| 4 | 6 | 24 | |
| 5 | 3 | 15 | |
| 6 | 1 | 6 | |
| Total | 30 | 96 |
The mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{96}{30}\\ &=3.2 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i*(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i*(x_i-xb)^3$ | $(x_i-xb)^4$ | $f_i*(x_i-xb)^4$ | |
|---|---|---|---|---|---|---|---|---|
| 2 | 9 | 1.44 | 12.96 | -1.728 | -15.552 | 2.0736 | 18.6624 | |
| 3 | 11 | 0.04 | 0.44 | -0.008 | -0.088 | 0.0016 | 0.0176 | |
| 4 | 6 | 0.64 | 3.84 | 0.512 | 3.072 | 0.4096 | 2.4576 | |
| 5 | 3 | 3.24 | 9.72 | 5.832 | 17.496 | 10.4976 | 31.4928 | |
| 6 | 1 | 7.84 | 7.84 | 21.952 | 21.952 | 61.4656 | 61.4656 | |
| Total | 96 | 34.8 | 26.88 | 114.096 |
The first central moment $m_1$ is always zero.
The second central moment is
$$ \begin{aligned} m_2 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2\\ &=\frac{34.8}{30}\\ &=1.16 \end{aligned} $$
The third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{26.88}{30}\\ &=0.896 \end{aligned} $$
The fourth central moment is
$$ \begin{aligned} m_4 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4\\ &=\frac{114.096}{30}\\ &=3.8032 \end{aligned} $$
Example 2
The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute five number summary for the following frequency distribution.
| Time spent on Internet ($x$) | 10-12 | 13-15 | 16-18 | 19-21 | 22-24 |
|---|---|---|---|---|---|
| No. of students ($f$) | 3 | 12 | 15 | 24 | 2 |
Solution
| Class | $x_i$ | $f_i$ | $f_i*x_i$ | |
|---|---|---|---|---|
| 10-12 | 11 | 3 | 33 | |
| 13-15 | 14 | 12 | 168 | |
| 16-18 | 17 | 15 | 255 | |
| 19-21 | 20 | 24 | 480 | |
| 22-24 | 23 | 2 | 46 | |
| Total | 56 | 982 |
The mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357 \end{aligned} $$
| $x_i$ | $f_i$ | $(x_i-xb)^2$ | $f_i(x_i-xb)^2$ | $(x_i-xb)^3$ | $f_i(x_i-xb)^3$ | $(x_i-xb)^4$ | $f_i(x_i-xb)^4$ | |
|---|---|---|---|---|---|---|---|---|
| 11 | 3 | 42.7154 | 128.1462 | -279.1749 | -837.5247 | 1824.6032 | 5473.8096 | |
| 14 | 12 | 12.5012 | 150.0144 | -44.2004 | -530.4048 | 156.2794 | 1875.3528 | |
| 17 | 15 | 0.287 | 4.305 | -0.1537 | -2.3055 | 0.0824 | 1.236 | |
| 20 | 24 | 6.0728 | 145.7472 | 14.9651 | 359.1624 | 36.8786 | 885.0864 | |
| 23 | 2 | 29.8586 | 59.7172 | 163.1562 | 326.3124 | 891.5345 | 1783.069 | |
| Total | 56 | 487.93 | -684.7602 | 10018.5538 |
The first central moment $m_1$ is always zero.
The second central moment is
$$ \begin{aligned} m_2 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2\\ &=\frac{487.93}{56}\\ &=8.713 \end{aligned} $$
The third central moment is
$$ \begin{aligned} m_3 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3\\ &=\frac{-684.7602}{56}\\ &=-12.2279 \end{aligned} $$
The fourth central moment is
$$ \begin{aligned} m_4 &=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4\\ &=\frac{10018.5538}{56}\\ &=178.9027 \end{aligned} $$
