Representation of data — practice questions

(i) First, we need to calculate the class widths and frequency densities for the unequal class intervals.

The histogram is drawn with 'Time (t mins)' on the horizontal axis and 'Frequency Density' on the vertical axis. The bars have widths 5, 10, 10, 15 and heights 2.0, 3.0, 2.5, 1.0 respectively. There are no gaps between the bars.

(ii) To estimate the mean, we need the mid-point ($x$) of each class.

Total frequency, $\sum f = 10+30+25+15 = 80$. Total $fx$, $\sum fx = 125 + 600 + 750 + 637.5 = 2112.5$.

Estimated Mean, $\bar{x} = \frac{\sum fx}{\sum f} = \frac{2112.5}{80} = 26.40625$.

So, the estimated mean time is 26.4 minutes (to 3 s.f.).

(i) The total number of plants is $n=120$. The median is the value of the $(n/2)$th plant, which is the 60th plant. Reading from 60 on the cumulative frequency axis across to the curve and down to the height axis gives an estimated median. Median ($Q_2$) ≈ 18 cm.

(ii) The lower quartile ($Q_1$) is at the $(n/4)$th position = 120/4 = 30th plant. Reading from 30 on the y-axis gives $Q_1 \approx$ 12 cm. The upper quartile ($Q_3$) is at the $(3n/4)$th position = 3*120/4 = 90th plant. Reading from 90 on the y-axis gives $Q_3 \approx$ 25 cm.

Interquartile Range (IQR) = $Q_3 - Q_1 \approx 25 - 12 = 13$ cm. So, the estimated IQR is 13 cm.

(iii) We have the five-number summary: Minimum = 5 cm Lower Quartile ($Q_1$) = 12 cm Median ($Q_2$) = 18 cm Upper Quartile ($Q_3$) = 25 cm Maximum = 38 cm

A box-and-whisker plot is drawn with a horizontal scale from 0 to 40. A box is drawn from $Q_1=12$ to $Q_3=25$. A vertical line is drawn inside the box at the median, $M=18$. Whiskers are drawn from the box to the minimum value (5) and the maximum value (38).