The table shows the distribution of monthly income (in thousands of naira) of workers in a factory Monthly Income (N'1000) 135-139 140-149 150-154 155-164 1...
Question 1 Report
The table shows the distribution of monthly income (in thousands of naira) of workers in a factory
| Monthly Income (N'1000) | 135-139 | 140-149 | 150-154 | 155-164 | 165-169 |
| Number of workers | 20 | 42 | 28 | 38 | 22 |
(a) Draw a histogram for the distribution.
(b) Use your graph to estimate the mode of the distribution.
Answer Details
(a) Histogram
Since the class intervals have unequal widths, the heights of the rectangles are the frequency densities:
\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}.\]
| Monthly income (₦'000) | Class boundaries (₦'000) | Frequency | Class width | Frequency density |
|---|---|---|---|---|
| 135–139 | 134.5–139.5 | 20 | 5 | 4.0 |
| 140–149 | 139.5–149.5 | 42 | 10 | 4.2 |
| 150–154 | 149.5–154.5 | 28 | 5 | 5.6 |
| 155–164 | 154.5–164.5 | 38 | 10 | 3.8 |
| 165–169 | 164.5–169.5 | 22 | 5 | 4.4 |
Plot the class boundaries on the horizontal axis and frequency density on the vertical axis. The resulting histogram is:
(b) Estimated mode
The modal class is \(149.5\text{–}154.5\), since it has the greatest frequency density, \(5.6\).
Using the standard intersecting-diagonals construction on the modal rectangle gives
\[\begin{aligned} \text{Mode} &=149.5+\frac{5.6-4.2}{(5.6-4.2)+(5.6-3.8)}\times 5\\ &=149.5+\frac{1.4}{3.2}\times5\\ &=151.6875\approx151.7. \end{aligned}\]
Therefore, the estimated modal monthly income is \(151.7\) thousand naira, that is, approximately ₦151,700.
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