Question 3.32: Calculate the mean deviation from the arithmetic mean for th...
Calculate the mean deviation from the arithmetic mean for the data shown in the following table.
| Length of rivet (mm) | Frequency |
| 9.8 | 3 |
| 9.9 | 18 |
| 9.95 | 36 |
| 10.0 | 62 |
| 10.05 | 56 |
| 10.1 | 20 |
| 10.2 | 5 |
The easiest way to tackle this problem is to set up a table of values in a similar manner to the table we produced for Example 3.29, with the headings for such a table being taken from the above formula for finding the mean deviation for a frequency distribution.
| Rivet length (x) |
f | fx | |x-\bar{x}| | f|x-\bar{x}| |
| 9.8 | 3 | 29.4 | 0.208 | 0.624 |
| 9.9 | 18 | 178.2 | 0.108 | 1.944 |
| 9.95 | 36 | 358.2 | 0.058 | 2.088 |
| 10.0 | 62 | 620.0 | 0.008 | 0.496 |
| 10.05 | 56 | 562.8 | 0.042 | 2.352 |
| 10.1 | 20 | 202 | 0.092 | 1.84 |
| 10.2 | 5 | 51 | 0.192 | 0.96 |
| Total | ∑f = 200 | ∑fx = 2001.6 | \sum f|x-\bar{x}| = 10.304 |
Arithmetic mean:
\bar{x}=\frac{\Sigma f x}{\sum f}=\frac{2001.6}{200}=10.008
\bar{x} = 10.008 was required to complete the last two columns in the table.
Then the mean deviation from the mean of the rivet lengths is:
\begin{aligned}&=\frac{\Sigma f|x-\bar{x}|}{\Sigma f}=\frac{10.304}{200}=0.05152 \ mm \\&\simeq 0.05 \ mm \end{aligned}
This small average deviation from the arithmetic mean for rivet length is what we would expect in this case, with the deviation being due to very small manufacturing errors. This is, therefore, an example of a frequency distribution tightly packed around the average for the distribution.