Past Exam Questions

(i) From the cumulative frequency graph, determine the lower quartile (LQ), median, and upper quartile (UQ):

• LQ = 15 (thousands of euros)
• Median = 18 (thousands of euros)
• UQ = 26 (thousands of euros)

Draw a box-and-whisker plot with these values, using a linear scale from 0 to 80 (thousands of euros). The whiskers extend from the minimum (5) to LQ and from UQ to the maximum (80).

(ii) Comment: Most (3/4) of the sample earn less than 26,000 euros, indicating not many earn high salaries.

(iii) Calculate the interquartile range (IQR):

\(\text{IQR} = UQ - LQ = 26 - 15 = 11\)

(a) A high outlier is any salary above:

\(UQ + 1.5 \times \text{IQR} = 26 + 1.5 \times 11 = 42.5 \text{ (thousands of euros)} = 42500 \text{ euros}\)

(b) A low outlier is any salary below:

\(LQ - 1.5 \times \text{IQR} = 15 - 1.5 \times 11 = -1.5 \text{ (thousands of euros)}\)

Since the lowest salary is 5000 euros, no salaries are low enough to be classified as outliers.

(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.

(i) The interval \(-2 \leq t < 0\) indicates that some trains arrived up to 2 minutes early.

(ii) To draw the cumulative frequency graph, first construct the cumulative frequency table:

Plot these points on a graph with the x-axis representing the number of minutes late and the y-axis representing the cumulative frequency. Connect the points with a smooth curve or straight lines.

To estimate the median, find the value corresponding to the cumulative frequency of 102 (half of 204). This is approximately between 2.1 and 2.4 minutes.

To estimate the interquartile range, find the values corresponding to the cumulative frequencies of 51 (quarter of 204) and 153 (three-quarters of 204). The interquartile range is approximately between 3.2 and 3.6 minutes.

To construct the cumulative frequency graph, plot the following points based on the given data:

• (3, 0) - The least amount of time spent is 3 hours, so cumulative frequency starts at 0.
• (15, 160) - The lower quartile is 15 hours, so 25% of 640 people is 160.
• (20, 320) - The median is 20 hours, so 50% of 640 people is 320.
• (35, 480) - The upper quartile is 35 hours, so 75% of 640 people is 480.
• (60, 640) - The greatest amount of time spent is 60 hours, so cumulative frequency is 640.

Draw a smooth curve or straight lines connecting these points to form the cumulative frequency graph.

To estimate how many people watched more than 50 hours of television, find the cumulative frequency at 50 hours on the graph and subtract from 640. The mark scheme suggests an estimate of 60 to 70 people.

(a) Plot the cumulative frequency graph using the points: (20, 32), (30, 66), (35, 112), (40, 178), (50, 228), (60, 250). Ensure the graph is smooth and passes through these points.

(b) To find the 60th percentile, locate 150 on the cumulative frequency axis and find the corresponding time on the graph. The estimated time is approximately 38 minutes.

(c) Calculate the standard deviation using the formula for variance:

First, find the midpoints of each class interval: 10, 25, 32.5, 37.5, 45, 55.

Use the formula for variance:

\(\text{Variance} = \frac{1}{250} \left( 32 \times 10^2 + 34 \times 25^2 + 46 \times 32.5^2 + 66 \times 37.5^2 + 50 \times 45^2 + 22 \times 55^2 \right) - 34.4^2\)

\(= \frac{333650}{250} - 34.4^2 = 151.24\)

The standard deviation is \(\sqrt{151.24} \approx 12.3\) minutes.

(a) Draw a cumulative frequency graph with the given cumulative frequencies at the upper class boundaries: (200, 16), (300, 46), (500, 88), (900, 122), (1200, 134), (1600, 140).

(b) To find the interquartile range (IQR), use the cumulative frequency graph:

• Lower quartile (LQ) at 25% of 140: 35
• Upper quartile (UQ) at 75% of 140: 105
• IQR = UQ - LQ = 105 - 35 = 70

(c) Calculate the mean and standard deviation:

• Class midpoints: 100, 250, 400, 700, 1050, 1400
• Frequencies: 16, 30, 42, 34, 12, 6
• Mean: \(\frac{16 \times 100 + 30 \times 250 + 42 \times 400 + 34 \times 700 + 12 \times 1050 + 6 \times 1400}{140} = 505\)
• Variance: \(\frac{16 \times 100^2 + 30 \times 250^2 + 42 \times 400^2 + 34 \times 700^2 + 12 \times 1050^2 + 6 \times 1400^2}{140} - 505^2 = 324^2\)
• Standard deviation: \(\sqrt{324^2} = 324\)

(a) To estimate the number of plants with heights less than 100 cm, locate 100 cm on the x-axis and find the corresponding cumulative frequency on the y-axis. The graph shows approximately 60 plants.

(b) The 65th percentile corresponds to 65% of 160 plants, which is 104 plants. Locate 104 on the y-axis and find the corresponding height on the x-axis. The graph shows this height to be approximately 136 cm.

(c) The interquartile range (IQR) is the difference between the upper quartile (Q3) and the lower quartile (Q1). For 160 plants, Q1 is at the 40th plant (25% of 160) and Q3 is at the 120th plant (75% of 160). From the graph, Q1 is approximately 76 cm and Q3 is approximately 150 cm. Therefore, the IQR is 150 - 76 = 74 cm.

(a) To draw the cumulative frequency graph:

• Calculate the cumulative frequencies: 12, 28, 60, 126, 146, 150.
• Plot these cumulative frequencies against the upper boundaries of the intervals: 4.5, 10.5, 20.5, 30.5, 40.5, 60.5.
• Draw a smooth curve through the points.

(b) To find \(d\) for 30% of journeys:

• Calculate 70% of 150: \(0.7 \times 150 = 105\).
• Find the distance corresponding to a cumulative frequency of 105 on the graph. \(d \approx 27 \text{ km}\).

(c) To calculate the mean distance:

• Find midpoints of each interval: 2.25, 7.5, 15.5, 25.5, 35.5, 50.5.
• Calculate the mean: \(\bar{x} = \frac{(2.25 \times 12) + (7.5 \times 16) + (15.5 \times 32) + (25.5 \times 66) + (35.5 \times 20) + (50.5 \times 4)}{150}\)
• \(\bar{x} = \frac{3238}{150} \approx 21.6 \text{ km}\).

(a) Plot the cumulative frequency points: (20, 12), (30, 48), (40, 106), (60, 134), (100, 150). Draw a smooth curve through these points.

(b) To find \(k\), calculate 24% of 150: \(150 \times 0.76 = 114\). From the graph, \(k = 45\) minutes corresponds to a cumulative frequency of 114.

(c) Calculate frequencies: 12, 36, 58, 28, 16. Use midpoints: 10, 25, 35, 50, 80.

Mean: \(\frac{10 \times 12 + 25 \times 36 + 35 \times 58 + 50 \times 28 + 80 \times 16}{150} = \frac{120 + 900 + 2030 + 1400 + 1280}{150} = 38.2\).

Variance: \(\frac{12 \times 10^2 + 36 \times 25^2 + 58 \times 35^2 + 28 \times 50^2 + 16 \times 80^2}{150} - \text{mean}^2 = \frac{1200 + 22500 + 71050 + 70000 + 102400}{150} - 38.2^2 = 321.76\).

Standard deviation: \(\sqrt{321.76} = 17.9\).

(i) To draw the box-and-whisker plots:

• For girls: Minimum = 5, Lower quartile = 5.5, Median = 7, Upper quartile = 9, Maximum = 13.
• For boys: Minimum = 4, Lower quartile = 6, Median = 8.5, Upper quartile = 11, Maximum = 16.

Draw the plots on the same scale, labeling the axes with 'Time in seconds'.

(ii) Comparisons:

• The range for girls (13 - 5 = 8) is smaller than for boys (16 - 4 = 12).
• The interquartile range for girls (9 - 5.5 = 3.5) is smaller than for boys (11 - 6 = 5).
• The median time for girls (7) is less than the median time for boys (8.5), indicating girls are generally quicker.
• The distribution for boys is almost symmetrical, while for girls it is positively skewed.

To draw a box-and-whisker plot, follow these steps:

1. Arrange the data in ascending order: 14, 15, 17, 18, 19, 19, 21, 22, 22, 23, 24, 25, 26, 26, 28, 28, 30, 32, 36, 38, 41, 42, 45, 46, 47.
2. Find the median (Q2): The median is the 13th value, which is 26.
3. Find the lower quartile (Q1): The lower quartile is the median of the first 12 values, which is the 6.5th value, calculated as \\((19 + 21) / 2 = 20\\).
4. Find the upper quartile (Q3): The upper quartile is the median of the last 12 values, which is the 18.5th value, calculated as \\((36 + 38) / 2 = 37\\).
5. Draw the box from Q1 to Q3, with a line at the median (Q2).
6. Draw whiskers from the minimum value (14) to Q1 and from Q3 to the maximum value (47).

The box-and-whisker plot should have a linear scale with the axis labeled as 'Glasses of water'.

(i) To find the median and quartiles for type A:

• Arrange the data in ascending order.
• Median position = \(\frac{61 + 1}{2} = 31\), so the median is the 31st value, which is 0.52 seconds.
• Lower quartile (Q1) position = \(\frac{61 + 1}{4} = 15.5\), so Q1 is the average of the 15th and 16th values, which is 0.41 seconds.
• Upper quartile (Q3) position = \(\frac{3(61 + 1)}{4} = 46.5\), so Q3 is the average of the 46th and 47th values, which is 0.79 seconds.

(ii) Draw box-and-whisker plots for both types A and B using the given quartiles and medians.

(iii) Compare the loading times:

• Smartphone B is quicker as its median is lower.
• Smartphone B is slightly less variable as its interquartile range is smaller.