Prepare a cumulative frequency table for the given data.

a) To prepare a cumulative frequency table, we first determine the upper class boundaries and then sum the frequencies.

Step 1: Determine the upper class boundaries.
For a class like $0-9$, the upper class boundary is 9.5. For $10-19$, it's 19.5, and so on.

Step 2: Calculate the cumulative frequency.
The cumulative frequency for a class is the sum of its frequency and the frequencies of all preceding classes.

Here is the cumulative frequency table:

Marks	Frequency (f)	Upper Class Boundary	Cumulative Frequency (cf)
0-9	4	9.5	4
10-19	6	19.5	10
20-29	10	29.5	20
30-39	10	39.5	30
40-49	13	49.5	43
50-59	17	59.5	60
60-69	20	69.5	80
70-79	15	79.5	95
80-89	4	89.5	99
90-99	1	99.5	100

b) To draw a cumulative frequency curve (ogive):

Step 1: Plot the cumulative frequency against the upper class boundaries.
The points to plot are:
$(-0.5, 0)$, $(9.5, 4)$, $(19.5, 10)$, $(29.5, 20)$, $(39.5, 30)$, $(49.5, 43)$, $(59.5, 60)$, $(69.5, 80)$, $(79.5, 95)$, $(89.5, 99)$, $(99.5, 100)$.

Step 2: Draw a smooth curve connecting these points. The x-axis represents the marks (upper class boundaries), and the y-axis represents the cumulative frequency.

c) Since I cannot draw a curve, I will estimate the values using interpolation from the cumulative frequency table, which is equivalent to reading from a correctly drawn curve. The total frequency $N = 100$.

• Median value: The median is the value at the $\frac{N}{2}$ position. $\frac{N}{2} = \frac{100}{2} = 50^{th}$ value. The $50^{th}$ value falls in the $50-59$ class (cumulative frequency goes from 43 to 60). Using the formula for median:

$$\text{Median} = L + \left(\frac{\frac{N}{2} - cf_b}{f_m}\right) \times c$$

Where $L = 49.5$ (lower boundary of median class), $cf_b = 43$ (cumulative frequency before median class), $f_m = 17$ (frequency of median class), $c = 10$ (class width).

$$\text{Median} = 49.5 + \left(\frac{50 - 43}{17}\right) \times 10$$ $$\text{Median} = 49.5 + \left(\frac{7}{17}\right) \times 10$$ $$\text{Median} = 49.5 + 4.1176...$$ $$\text{Median} \approx \boxed{53.62}$$

• Interquartile range (IQR): IQR $= Q_3 - Q_1$.

First Quartile ($Q_1$): The value at the $\frac{N}{4}$ position.
$\frac{N}{4} = \frac{100}{4} = 25^{th}$ value.
The $25^{th}$ value falls in the $30-39$ class (cumulative frequency goes from 20 to 30).

$$Q_1 = L + \left(\frac{\frac{N}{4} - cf_b}{f_{q1}}\right) \times c$$

Where $L = 29.5$, $cf_b = 20$, $f_{q1} = 10$, $c = 10$.

$$Q_1 = 29.5 + \left(\frac{25 - 20}{10}\right) \times 10$$ $$Q_1 = 29.5 + \left(\frac{5}{10}\right) \times 10$$ $$Q_1 = 29.5 + 5 = 34.5$$

Third Quartile ($Q_3$): The value at the $\frac{3N}{4}$ position.
$\frac{3N}{4} = \frac{3 \times 100}{4} = 75^{th}$ value.
The $75^{th}$ value falls in the $60-69$ class (cumulative frequency goes from 60 to 80).

$$Q_3 = L + \left(\frac{\frac{3N}{4} - cf_b}{f_{q3}}\right) \times c$$

Where $L = 59.5$, $cf_b = 60$, $f_{q3} = 20$, $c = 10$.

$$Q_3 = 59.5 + \left(\frac{75 - 60}{20}\right) \times 10$$ $$$$