The following table shows the distribution of the masses of 120 kg of wood, correct to the nearest kg. Draw a histogram, make a cumulative frequency table and curve, use the graph to find the semi interquartile range and 60th percentile.

To draw a histogram, we first need to determine the class boundaries. Since the data is given in discrete intervals (e.g., 15-24), we adjust them to be continuous.

Original Classes	Class Boundaries (kg)	Frequency
15 - 24	14.5 - 24.5	14
25 - 34	24.5 - 34.5	54
35 - 44	34.5 - 44.5	24
45 - 54	44.5 - 54.5	26
55 - 64	54.5 - 64.5	2

To draw the histogram:

• Draw the Mass (kg) on the x-axis, marking the class boundaries (14.5, 24.5, 34.5, 44.5, 54.5, 64.5).
• Draw the Frequency on the y-axis.
• Draw rectangular bars for each class interval. The width of each bar should span its class boundaries, and the height should correspond to its frequency. For example, a bar from 14.5 to 24.5 with a height of 14.

We calculate the cumulative frequency by adding the frequency of each class to the sum of frequencies of all preceding classes.

Masses (kg)	Frequency ($f$)	Upper Class Boundary (kg)	Cumulative Frequency ($cf$)
15 - 24	14	24.5	14
25 - 34	54	34.5	$14 + 54 = 68$
35 - 44	24	44.5	$68 + 24 = 92$
45 - 54	26	54.5	$92 + 26 = 118$
55 - 64	2	64.5	$118 + 2 = 120$

To draw the cumulative frequency curve:

• Draw the Upper Class Boundary (kg) on the x-axis.
• Draw the Cumulative Frequency on the y-axis.
• Plot the points using the upper class boundaries and their corresponding cumulative frequencies. It's common to start with a point at the lower boundary of the first class with a cumulative frequency of 0. The points to plot are: $(14.5, 0)$, $(24.5, 14)$, $(34.5, 68)$, $(44.5, 92)$, $(54.5, 118)$, $(64.5, 120)$.
• Connect these points with a smooth curve.

Since a graph cannot be drawn here, we will use linear interpolation from the cumulative frequency table to find the values. The total number of observations is $N = 120$.

a) Semi-interquartile range
The semi-interquartile range (SIQR) is given by $\frac{Q_3 - Q_1}{2}$.

First Quartile ($Q_1$):
The position of $Q_1$ is $\frac{N}{4} = \frac{120}{4} = 30$.
The 30th observation falls in the 25-34 class (cumulative frequency 68, previous is 14).
Using the formula for $Q_1$:

$$Q_1 = L + \left(\frac{\frac{N}{4} - cf_b}{f}\right) \times c$$

where $L = 24.5$ (lower boundary of $Q_1$ class), $cf_b = 14$ (cumulative frequency before $Q_1$ class), $f = 54$ (frequency of $Q_1$ class), $c = 10$ (class width).

$$Q_1 = 24.5 + \left(\frac{30 - 14}{54}\right) \times 10$$ $$Q_1 = 24.5 + \left(\frac{16}{54}\right) \times 10$$ $$Q_1 = 24.5 + 2.96$$ $$Q_1 \approx 27.46 \text{ kg}$$

Third Quartile ($Q_3$):
The position of $Q_3$ is $\frac{3N}{4} = \frac{3 \times 120}{4} = 90$.
The 90th observation falls in the 35-44 class (cumulative frequency 92, previous is 68).
Using the formula for $Q_3$:

$$Q_3 = L + \left(\frac{\frac{3N}{4} - cf_b}{f}\right) \times c$$

where $L = 34.5$ (lower boundary of $Q_3$ class), $cf_b = 68$ (cumulative frequency before $Q_3$ class), $f = 24$ (frequency of $Q_3$ class), $c = 10$ (class width).

$$Q_3 = 34.5 + \left(\frac{90 - 68}{24}\right) \times 10$$ $$Q_3 = 34.5 + \left(\frac{22}{24}\right) \times 10$$ $$Q_3 = 34.5 + 9.17$$ $$Q_3 \approx 43.67 \text{ kg}$$

Now, calculate the semi-interquartile range:

$$\text{SIQR} = \frac{Q_3 - Q_1}{2} = \frac{43.67 - 27.46}{2} = \frac{16.21}{2}$$ $$\text{SIQR} = \boxed{\text{8.11 kg}}$$

b) 60th percentile
The position of the 60th percentile ($P_{60}$) is $\frac{60}{100} \times N = \frac{60}{100} \times 120 = 72$.
The 72nd observation falls in the 35-44 class (cumulative frequency 92, previous is 68).
Using the formula for $P_{60}$:

$$P_{60} = L + \left(\frac{\frac{60N}{100} - cf_b}{f}\right) \times c$$

where $L = 34.5$ (lower boundary of $P_{60}$ class), $cf_b = 68$ (cumulative frequency before $P_{60}$ class), $f = 24$ (frequency of $P_{60}$ class), $c = 10$ (class width).

$$P_{60} = 34.5 + \left(\frac{72 - 68}{24}\right) \times 10$$ $$P_{60} = 34.5 + \left(\frac{4}{24}\right) \times 10$$ $$P_{60} = 34.5 + 1.67$$ $$P_{60} \approx \boxed{\text{36.17 kg}}$$