The problem provides a cumulative frequency diagram representing the floor area (in $m^2$) of 80 houses. We need to use the diagram to estimate: (i) the median, (ii) the lower quartile, (iii) the interquartile range, (iv) the number of houses with a floor area greater than $120 m^2$.

1. Problem Description

The problem provides a cumulative frequency diagram representing the floor area (in $m^2$) of 80 houses. We need to use the diagram to estimate:

(i) the median,

(ii) the lower quartile,

(iii) the interquartile range,

(iv) the number of houses with a floor area greater than $120 m^2$.

2. Solution Steps

(i) The median corresponds to the value at the cumulative frequency of $\frac{1}{2} \times 80 = 40$. From the graph, the corresponding floor area is approximately $90 m^2$.

(ii) The lower quartile corresponds to the value at the cumulative frequency of $\frac{1}{4} \times 80 = 20$. From the graph, the corresponding floor area is approximately $68 m^2$.

(iii) The interquartile range is the difference between the upper quartile and the lower quartile.

The upper quartile corresponds to the value at the cumulative frequency of $\frac{3}{4} \times 80 = 60$. From the graph, the corresponding floor area (upper quartile) is approximately $104 m^2$.

Interquartile Range = Upper Quartile - Lower Quartile

Interquartile Range = $104 - 68 = 36 m^2$.

(iv) To find the number of houses with a floor area greater than $120 m^2$, we first find the cumulative frequency corresponding to $120 m^2$. From the graph, it is approximately

5

2. This means 52 houses have a floor area less than or equal to $120 m^2$.

The total number of houses is

8

0. Therefore, the number of houses with a floor area greater than $120 m^2$ is $80 - 52 = 28$.

3. Final Answer

(i) 90 $m^2$

(ii) 68 $m^2$

(iii) 36 $m^2$

(iv) 28