1.1 Determine the missing values in the table above 1.2 Using the table above, calculate the estimated mean rainfall. 1.3 Draw an Ogive (cumulative frequency graph) of the data on the GRID PROVIDED 1.4 Write the following: Modal class Estimated median rainfall

For the first row, "MIDPOINT X FREQUENCY" is 20 and "MIDPOINT" is 10.

For the third row, the rainfall interval is $40 < x \le 60$. The midpoint is the average of the lower and upper bounds.

Question 4.1.2: Using the table above, calculate the estimated mean rainfall.

Using $A=2$ and $B=50$:
Total Frequency ($\sum f$):

Total (Midpoint $\times$ Frequency) ($\sum (x \cdot f)$):

The formula for the estimated mean is:

Question 4.1.3: Draw an Ogive (cumulative frequency graph) of the data on the GRID PROVIDED.
I cannot draw a graph directly. You would plot the cumulative frequency against the upper class boundaries.

Question 4.1.4: Write the following:

a) Modal class

The frequencies are 2, 4, 5, 6, 8, 3, 2. The highest frequency is 8.

The class interval with a frequency of 8 is $80 < x \le 100$.
The modal class is $\boxed{\textbf{80 < x \le 100}}$.

b) Estimated median rainfall

Total frequency ($N$) = 30.
The median position is $\frac{N}{2} = \frac{30}{2} = 15$. The median is the 15th value.

Calculate cumulative frequencies:
$0 < x \le 20$: Cum Freq = 2
$20 < x \le 40$: Cum Freq = $2 + 4 = 6$
$40 < x \le 60$: Cum Freq = $6 + 5 = 11$
$60 < x \le 80$: Cum Freq = $11 + 6 = 17$
The 15th value falls into the class $60 < x \le 80$, as its cumulative frequency (17) is the first to exceed the median position (15). So, the median class is $60 < x \le 80$.

Median $= L + \left( \frac{\frac{N}{2} - CF_b}{f_m} \right) \times w$
Where:
$L = 60$ (lower boundary of the median class)
$N = 30$ (total frequency)
$CF_b = 11$ (cumulative frequency of the class before the median class)
$f_m = 6$ (frequency of the median class)
$w = 20$ (class width, $80 - 60$)