(i) To draw the cumulative frequency graph, calculate the cumulative frequencies:

• 1 – 20: 10
• 21 – 40: 10 + 32 = 42
• 41 – 50: 42 + 62 = 104
• 51 – 60: 104 + 50 = 154
• 61 – 70: 154 + 28 = 182
• 71 – 90: 182 + 18 = 200

Plot these cumulative frequencies against the upper class boundaries: (20.5, 10), (40.5, 42), (50.5, 104), (60.5, 154), (70.5, 182), (90.5, 200).

(ii) To estimate the number of days when over 30 rooms were occupied, find the cumulative frequency for 30 rooms using linear interpolation between 20 and 40:

Using linear interpolation:

\(10 + \frac{30 - 20.5}{20} \times 32 = 25.2\)

Thus, the cumulative frequency for 30 rooms is approximately 25.2. Therefore, the number of days when over 30 rooms were occupied is \(200 - 25.2 = 174.8\), which rounds to 175 days.

(iii) To find the number of rooms occupied on 75% of the days, calculate 75% of 200 days:

\(0.75 \times 200 = 150\)

Using the cumulative frequency graph, find the number of rooms corresponding to a cumulative frequency of 150. This occurs between 51 and 60 rooms. Using linear interpolation:

\(50.5 + \frac{150 - 104}{50} \times 10 = 59\)

Thus, on 75% of the days, at most 59 rooms were occupied.

(i) To find the median, draw the cumulative frequency graph using the upper class boundaries and cumulative frequencies. The median is the value at the 2500th school (since there are 5000 schools). From the graph, the median number of pupils is approximately 270.

(ii) 80% of 5000 schools is 4000 schools. From the cumulative frequency table, 4000 schools correspond to a number of pupils less than 160. Therefore, \(n = 160\).

(iii) The number of schools with pupils between 201 and 250 is given by the difference in cumulative frequencies: \(2100 - 1600 = 500\).

(iv) To estimate the mean, use the midpoints of the intervals: \((50.5 \times 200 + 125.5 \times 600 + 175.5 \times 800 + 225.5 \times 500 + 300.5 \times 2000 + 400.5 \times 600 + 525.5 \times 300) / 5000 = 268\).

To draw the cumulative frequency graph, we first calculate the cumulative frequencies:

Plot these cumulative frequencies against the upper class boundaries on a graph.

For part (b), to find the 70th percentile, calculate:

\(60 \times 0.7 = 42\)

Locate 42 on the cumulative frequency axis and read across to the curve, then down to the \(x\)-axis to estimate the 70th percentile, which is approximately 126.

To compare the central tendency, we find the medians from the cumulative frequency graphs:

• Median for Country A: approximately 2.0 to 2.1 kg.
• Median for Country B: approximately 3.8 to 3.9 kg.

Alternatively, using means:

• Mean for Country A: approximately 2.0 to 2.1 kg.
• Mean for Country B: approximately 3.4 to 3.5 kg.

Country B has a higher median and mean, indicating heavier babies on average.

To compare the spread, we calculate the interquartile range (IQR) or standard deviation (sd):

• IQR for Country A: 2.4 - 1.5 = 0.9 kg or sd = 0.5 to 0.7 kg.
• IQR for Country B: 4.5 - 2.2 = 2.3 kg or sd = 1.2 to 1.4 kg.

Country B has a greater IQR and sd, indicating a greater spread of weights.

(i) To find \(a\) and \(b\):

\(a = 422 + 72 = 494\)

\(b = 540 - 494 = 46\)

(ii) Draw a cumulative frequency graph using the points: (10, 210), (20, 344), (30, 422), (40, 494), (60, 540).

(iii) The median is found by locating the 270th person on the cumulative frequency graph, which corresponds to approximately 13.5 to 14.6 minutes.

(iv) Calculate the mean \(m\) using midpoints:

\(m = \frac{(5 \times 210 + 15 \times 134 + 25 \times 78 + 35 \times 72 + 50 \times 46)}{540} = \frac{9830}{540} = 18.2\) minutes

Calculate the standard deviation \(s\):

\(s = \sqrt{\frac{(5^2 \times 210 + 15^2 \times 134 + 25^2 \times 78 + 35^2 \times 72 + 50^2 \times 46)}{540} - 18.2^2} = 14.2\) minutes

(v) Calculate the range \((m - \frac{1}{2}s)\) to \((m + \frac{1}{2}s)\):

\(18.2 \pm 7.1 = 11.1, 25.3\)

From the cumulative frequency graph, estimate the number of people between these times: 390 - 225 = 155 to 170 people.