Table 2.2: Scores of candidates. Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution. Use your curve to estimate the: (i) Median mark (ii) Upper and lower quartiles (iii) Semi-inter quartile range.

First, determine the class boundaries for the given marks. Since the classes are 21-30, 31-40, etc., the class boundaries are found by taking the midpoint between the upper limit of one class and the lower limit of the next. For example, between 30 and 31, the boundary is 30.5.

Marks (%)	Class Boundaries	No. of students (Frequency, $f$)	Cumulative Frequency (CF)
21-30	20.5 - 30.5	9	9
31-40	30.5 - 40.5	20	9 + 20 = 29
41-50	40.5 - 50.5	31	29 + 31 = 60
51-60	50.5 - 60.5	42	60 + 42 = 102
61-70	60.5 - 70.5	50	102 + 50 = 152
71-80	70.5 - 80.5	32	152 + 32 = 184
81-90	80.5 - 90.5	16	184 + 16 = 200

The total number of students, $N = 200$.

To draw the cumulative frequency curve, plot the cumulative frequency against the upper class boundaries.

• Plot the points: $(30.5, 9)$, $(40.5, 29)$, $(50.5, 60)$, $(60.5, 102)$, $(70.5, 152)$, $(80.5, 184)$, $(90.5, 200)$.
• Start the curve from the lower boundary of the first class with a cumulative frequency of 0, i.e., $(20.5, 0)$.
• Join these points with a smooth curve. The x-axis represents "Marks (%)" and the y-axis represents "Cumulative Frequency".

The median mark is the value corresponding to the $\frac{N}{2}$-th position.

$$\text{Median position} = \frac{200}{2} = 100$$

On your cumulative frequency curve, locate 100 on the y-axis (cumulative frequency). Draw a horizontal line from this point to intersect the curve. From the intersection point, draw a vertical line down to the x-axis (marks). Read the value on the x-axis.
Using interpolation (which approximates reading from a smooth curve):

$$Q_2 = L + \left(\frac{\frac{N}{2} - CF_b}{f_m}\right) \times c$$

Where $L = 50.5$, $CF_b = 60$, $f_m = 42$, $c = 10$.

$$Q_2 = 50.5 + \left(\frac{100 - 60}{42}\right) \times 10 = 50.5 + \left(\frac{40}{42}\right) \times 10 \approx 50.5 + 9.52 = 60.02$$

The estimated median mark is $\boxed{\text{60.0 marks}}$.

The lower quartile is the value corresponding to the $\frac{N}{4}$-th position.

$$\text{Lower quartile position} = \frac{200}{4} = 50$$

On your cumulative frequency curve, locate 50 on the y-axis. Draw a horizontal line to the curve, then a vertical line down to the x-axis. Read the value.
Using interpolation:

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

Where $L = 40.5$, $CF_b = 29$, $f_q = 31$, $c = 10$.

$$Q_1 = 40.5 + \left(\frac{50 - 29}{31}\right) \times 10 = 40.5 + \left(\frac{21}{31}\right) \times 10 \approx 40.5 + 6.77 = 47.27$$

The estimated lower quartile is $\boxed{\text{47.3 marks}}$.

The upper quartile is the value corresponding to the $\frac{3N}{4}$-th position.

$$\text{Upper quartile position} = \frac{3 \times 200}{4} = 150$$

On your cumulative frequency curve, locate 150 on the y-axis. Draw a horizontal line to the curve, then a vertical line down to the x-axis. Read the value.
Using interpolation:

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

Where $L = 60.5$, $CF_b = 102$, $f_q = 50$, $c = 10$.

$$Q_3 = 60.5 + \left(\frac{150 - 102}{50}\right) \times 10 = 60.5 + \left(\frac{48}{50}\right) \times 10 \approx 60.5 + 9.6 = 70.1$$

The estimated upper quartile is $\boxed{\text{70.1 marks}}$.

The semi-interquartile range is half the difference between the upper and lower quartiles.

$$\text{SIQR} = \frac{Q_3 - Q_1}{2}$$ $$\text{SIQR} = \frac{70.1 - 47.3}{2} = \frac{22.8}{2} = 11.4$$

The semi-interquartile range is $\boxed{\text{11.4 marks}}$.