The table shows the distribution of the lengths of 20 iron rods measured in metres :
| Length (m) | 1.0 - 1.1 | 1.2 - 1.3 | 1.4 - 1.5 | 1.6 - 1.7 | 1.8 - 1.9 |
| Frequency | 2 | 3 | 8 | 5 | 2 |
Using an assumed mean of 1.45, calculate the mean of the distribution.
Assumed mean = 1.45
| Length (m) |
Freq (f) |
Mid-length (x) |
\(d = x - A\) | \(fd\) |
| 1.0 - 1.1 | 2 | 1.05 | -0.4 | -0.8 |
| 1.2 - 1.3 | 3 | 1.25 | -0.2 | -0.6 |
| 1.4 - 1.5 | 8 | 1.45 | 0 | 0 |
| 1.6 - 1.7 | 5 | 1.65 | 0.2 | 1.0 |
| 1.8 - 1.9 | 2 | 1.85 | 0.4 | 0.8 |
| 20 | 0.4 |
Mean \(\bar{x} = A + \frac{\sum fd}{\sum f}\)
= \(1.45 + \frac{0.4}{20}\)
= \(1.45 + 0.02\)
= \(1.47\)
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