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 the median, lower quartile, interquartile range, and 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 the median, lower quartile, interquartile range, and the number of houses with a floor area greater than $120 m^2$.

2. Solution Steps

(i) The median is the value corresponding to the cumulative frequency of $\frac{1}{2} \times 80 = 40$. From the graph, the floor area corresponding to a cumulative frequency of 40 is approximately $88 m^2$.

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

(iii) The upper quartile is the value corresponding to the cumulative frequency of $\frac{3}{4} \times 80 = 60$. From the graph, the floor area corresponding to a cumulative frequency of 60 is approximately $112 m^2$.

The interquartile range is the difference between the upper and lower quartiles:

$IQR = Upper \ Quatile - Lower \ Quartile$

$IQR = 112 - 68 = 44 m^2$

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

5

2. Since there are 80 houses in total, the number of houses with a floor area greater than $120 m^2$ is $80 - 52 = 28$.

3. Final Answer

(i) Median: $88 m^2$

(ii) Lower quartile: $68 m^2$

(iii) Interquartile range: $44 m^2$

(iv) Number of houses with a floor area greater than $120 m^2$: $28$