The problem provides a frequency table showing the distribution of floor areas for 80 houses. Part (i) requires us to calculate the estimated mean floor area. Part (iii) asks us to find the probability that one house has a floor area greater than $130 m^2$ and the other has a floor area less than or equal to $60 m^2$, when two houses are randomly picked.

1. Problem Description

The problem provides a frequency table showing the distribution of floor areas for 80 houses. Part (i) requires us to calculate the estimated mean floor area. Part (iii) asks us to find the probability that one house has a floor area greater than $130 m^2$ and the other has a floor area less than or equal to $60 m^2$, when two houses are randomly picked.

2. Solution Steps

(i) To calculate the estimated mean floor area, we use the midpoints of each interval and the corresponding frequencies.

The intervals and frequencies are:

- 40 < a <= 60: Frequency = 14, Midpoint = (40+60)/2 = 50

- 60 < a <= 80: Frequency = 17, Midpoint = (60+80)/2 = 70

- 80 < a <= 100: Frequency = 18, Midpoint = (80+100)/2 = 90

- 100 < a <= 130: Frequency = 15, Midpoint = (100+130)/2 = 115

- 130 < a <= 160: Frequency = 9, Midpoint = (130+160)/2 = 145

- 160 < a <= 200: Frequency = 7, Midpoint = (160+200)/2 = 180

The estimated mean is given by:

$Mean = \frac{\sum (Midpoint * Frequency)}{\sum Frequency}$

$Mean = \frac{(50*14) + (70*17) + (90*18) + (115*15) + (145*9) + (180*7)}{14+17+18+15+9+7}$

$Mean = \frac{700 + 1190 + 1620 + 1725 + 1305 + 1260}{80}$

$Mean = \frac{7800}{80}$

$Mean = 97.5$

(iii)

We need to find the probability that one house has floor area greater than $130 m^2$ and the other has floor area $60 m^2$ or less.

Houses with floor area greater than $130 m^2$: $130 < a <= 160$ and $160 < a <= 200$. Total = 9 + 7 =

1

6. Houses with floor area less than or equal to $60 m^2$: $40 < a <= 60$. Total =

1

4. Total houses =

8

0.

We are picking two houses at random. There are two possible scenarios:

1. First house has area > 130 and second house has area <=

6

0. Probability = $(16/80) * (14/79) = 224/6320$

2. First house has area <= 60 and second house has area >

1

3

0. Probability = $(14/80) * (16/79) = 224/6320$

Total probability = $(224/6320) + (224/6320) = 448/6320 = 28/395$

3. Final Answer

(i) 97.5

(iii) 28/395