The times taken, \(t\) seconds, by 1140 people to solve a puzzle are summarised in the table.
\(\begin{array}{|c|c|} \hline \text{Time (} t \text{ seconds)} & \text{Number of people} \\ \hline 0 \leq t < 20 & 320 \\ 20 \leq t < 40 & 280 \\ 40 \leq t < 60 & 220 \\ 60 \leq t < 100 & 220 \\ 100 \leq t < 140 & 100 \\ \hline \end{array}\)
(i) On the grid, draw a histogram to illustrate this information.
(ii) Calculate an estimate of the mean of \(t\).
Solution
(i) 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 \(0 \leq t < 20\): \(\text{FD} = \frac{320}{20} = 16\)
For \(20 \leq t < 40\): \(\text{FD} = \frac{280}{20} = 14\)
For \(40 \leq t < 60\): \(\text{FD} = \frac{220}{20} = 11\)
For \(60 \leq t < 100\): \(\text{FD} = \frac{220}{40} = 5.5\)
For \(100 \leq t < 140\): \(\text{FD} = \frac{100}{40} = 2.5\)
Draw bars with these heights on the histogram.
(ii) To estimate the mean, use the midpoints of each class interval:
\(\text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\text{Total Frequency}}\)
\(= \frac{(10 \times 320) + (30 \times 280) + (50 \times 220) + (80 \times 220) + (120 \times 100)}{1140}\)
\(= \frac{3200 + 8400 + 11000 + 17600 + 12000}{1140}\)
\(= \frac{52200}{1140}\)
\(= 45.8\)
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