The times taken, in minutes, by 360 employees at a large company to travel from home to work are summarised in the following table.
| Time \( t \) (minutes) |
\( 0 \le t < 5 \) |
\( 5 \le t < 10 \) |
\( 10 \le t < 20 \) |
\( 20 \le t < 30 \) |
\( 30 \le t < 50 \) |
| Frequency |
23 |
102 |
135 |
76 |
24 |
(a) Draw a histogram to represent this information.
(b) Calculate an estimate of the mean time taken by an employee to travel to work.
Solution
(a) To draw the histogram, calculate the frequency density for each class interval using the formula:
\(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}\)
For the intervals:
- \(0 \leq t < 5\): \(\frac{23}{5} = 4.6\)
- \(5 \leq t < 10\): \(\frac{102}{5} = 20.4\)
- \(10 \leq t < 20\): \(\frac{135}{10} = 13.5\)
- \(20 \leq t < 30\): \(\frac{76}{10} = 7.6\)
- \(30 \leq t < 50\): \(\frac{24}{20} = 1.2\)
Draw bars with these heights, ensuring axes are labeled correctly.
(b) To estimate the mean time, use the midpoints of each interval:
- \(0 \leq t < 5\): midpoint = 2.5
- \(5 \leq t < 10\): midpoint = 7.5
- \(10 \leq t < 20\): midpoint = 15
- \(20 \leq t < 30\): midpoint = 25
- \(30 \leq t < 50\): midpoint = 40
Calculate the mean:
\(\frac{(2.5 \times 23) + (7.5 \times 102) + (15 \times 135) + (25 \times 76) + (40 \times 24)}{360}\)
\(= \frac{5707.5}{360} \approx 15.9\) minutes
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