Past Exam Questions

(a) To draw the histogram, calculate the frequency density for each class interval using the formula:

\(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}\)

Class widths: 5, 5, 10, 20, 30

Frequency densities: \(\frac{10}{5} = 2\), \(\frac{5}{5} = 1\), \(\frac{26}{10} = 2.6\), \(\frac{32}{20} = 1.6\), \(\frac{18}{30} = 0.6\)

Draw bars with these heights for each class interval.

(b) The lower quartile (LQ) is in the 11 – 20 interval, and the upper quartile (UQ) is in the 21 – 40 interval. The greatest possible interquartile range (IQR) is:

\(40 - 11 = 29\)

(c) To estimate the mean, calculate the midpoint of each class interval:

Midpoints: 3, 8, 15.5, 30.5, 55.5

Calculate the mean using:

\(\text{Mean} = \frac{3 \times 10 + 8 \times 5 + 15.5 \times 26 + 30.5 \times 32 + 55.5 \times 18}{91}\)

\(= \frac{30 + 40 + 403 + 976 + 999}{91}\)

\(= \frac{2448}{91} \approx 26.9\)

(a) To draw the histogram, calculate the class widths: \( 10,\ 5,\ 15,\ 20,\ 10 \). Then calculate the frequency density for each class:

\( \text{Frequency density} = \dfrac{\text{frequency}}{\text{class width}} \)

\( 1.8,\ 4.8,\ 2,\ 1,\ 0.8 \)

Ensure all heights are correct on the diagram using a linear scale. The correct bar ends are: \( 10.5,\ 15.5,\ 30.5,\ 50.5,\ 60.5 \).

(b) The greatest possible value of the interquartile range (IQR) is calculated using the class intervals \( 11\text{–}15 \) and \( 31\text{–}50 \). Thus, the greatest IQR \( = 50 - 11 = 39 \).

(c) To calculate the mean:

\( \text{Mean} = \frac{ 18 \times 5.5 + 24 \times 13 + 30 \times 23 + {}\\ 20 \times 40.5 + 8 \times 55.5 }{100}\) \(= \frac{2355}{100} = 23.6 \)

To calculate the variance:

\( \text{Var} = \frac{ 18 \times 5.5^2 + 24 \times 13^2 + 30 \times 23^2 + {}\\ 20 \times 40.5^2 + 8 \times 55.5^2 }{100}\) \(- \text{mean}^2 \)

\( = \frac{77917.5}{100} - 23.6^2\) \(= 224.57 \)

Standard deviation \( = \sqrt{224.57} = 15.0 \).

(a) To draw the cumulative frequency graph, plot the points (40, 0), (60, 14), (65, 38), (70, 60), (85, 106), and (100, 120) on a graph with the x-axis representing weight (kg) and the y-axis representing cumulative frequency. Connect these points with a smooth curve.

(b) To find the weight W such that 35% of the students weigh more than W kg, calculate 65% of the total cumulative frequency (since 100% - 35% = 65%).

Calculate 65% of 120: \(120 \times 0.65 = 78\).

Using the cumulative frequency graph, find the weight corresponding to a cumulative frequency of 78. This weight is approximately 76 kg.

(a) Calculate cumulative frequencies:

• 0 – 9: 15
• 10 – 14: 15 + 48 = 63
• 15 – 19: 63 + 66 = 129
• 20 – 30: 129 + 21 = 150

Plot these cumulative frequencies against the upper class boundaries on a graph to draw the cumulative frequency curve.

(b) To find the length d such that 40% of the fish have a length of d cm or more, calculate 60% of 150, which is 90. From the cumulative frequency graph, find the length corresponding to a cumulative frequency of 90. This is approximately 16.5 cm.

(c) Calculate the midpoints of each class:

• 0 – 9: 4.75
• 10 – 14: 12
• 15 – 19: 17
• 20 – 30: 25

\(Use the formula for variance:\)

\(\text{Var} = \frac{(4.75^2 \times 15) + (12^2 \times 48) + (17^2 \times 66) + (25^2 \times 21)}{150} - 15.295^2\)

Calculate:

\(\text{Var} = \frac{39449.4375}{150} - 15.295^2 = 29.1\)

(i) From the cumulative frequency graph, the cumulative frequency at 14 cm is approximately 55, and at 24 cm is approximately 156. Therefore, the number of leaves between 14 and 24 cm is \(156 - 55 = 101\).

(ii) 10% of 160 leaves is 16 leaves. Therefore, 144 leaves have a length less than \(L\). From the graph, \(L\) corresponds to a cumulative frequency of 144, which is approximately 22 cm.

(iii) The median is at the 80th leaf (\(\frac{160}{2}\)), which corresponds to approximately 15.6 cm on the graph. The lower quartile \(Q_1\) is at the 40th leaf, approximately 12.7 cm, and the upper quartile \(Q_3\) is at the 120th leaf, approximately 18.8 cm. The interquartile range is \(Q_3 - Q_1 = 18.8 - 12.7 = 6.1\).

(iv) The median for Ransha's data is higher than Sharim's data. The interquartile range for Ransha's data is lower than Sharim's data, indicating less spread.

To find the median, locate the 700th pupil on the cumulative frequency graph (since there are 1400 pupils in total). For Mathematics, the median is at 40 marks, and for English, it is at 55 marks.

To find the interquartile range (IQR), locate the 350th and 1050th pupils on the cumulative frequency graph. For Mathematics, the IQR is from 12 to 80, giving an IQR of 68. For English, the IQR is from 42 to 62, giving an IQR of 20.

Thus, the median of English is larger than the median of Mathematics, indicating a higher central tendency for English marks. The IQR for Mathematics is larger than for English, indicating a greater spread in Mathematics marks.

(i) Plot the cumulative frequency points: (0, 0), (10, 16), (20, 50), (30, 106), (50, 146), (70, 176), (90, 200). Join these points with a smooth curve or straight lines.

(ii) The median is the value at the \(\frac{200}{2} = 100\)th position. From the graph, this corresponds to approximately 29 minutes.

(iii) For 80 drivers, the time is at least \(T\) minutes, meaning \(200 - 80 = 120\) drivers took less than \(T\) minutes. From the graph, \(T\) corresponds to approximately 37 minutes.

(iv) Calculate the estimated mean using midpoints: Frequencies are 16, 34, 56, 40, 30, 24. Midpoints are 5, 15, 25, 40, 60, 80. Estimated mean \(= \frac{5 \times 16 + 15 \times 34 + 25 \times 56 + 40 \times 40 + 60 \times 30 + 80 \times 24}{200} = \frac{7310}{200} = 36.55\) minutes.

(i) Plot the cumulative frequency points at the upper class boundaries: (20, 52), (30, 94), (40, 142), (50, 172), (70, 222), (100, 250). Draw a smooth curve or straight lines connecting these points.

(ii) To find \(k\), locate the cumulative frequency of 150 on the graph and read the corresponding rainfall value. From the graph, \(k \approx 42\) mm.

(iii) Calculate the frequencies for each class interval: \(52, 42, 48, 30, 50, 28\).

Find the midpoints of each class: \(10, 25, 35, 45, 60, 85\).

Calculate the mean: \(\bar{x} = \frac{(10 \times 52) + (25 \times 42) + (35 \times 48) + (45 \times 30) + (60 \times 50) + (85 \times 28)}{250} = \frac{9980}{250} = 39.9 \text{ mm}\)

Calculate the variance: \(s^2 = \frac{(10^2 \times 52) + (25^2 \times 42) + (35^2 \times 48) + (45^2 \times 30) + (60^2 \times 50) + (85^2 \times 28)}{250} - \bar{x}^2 = 539.59\)

Standard deviation: \(s = \sqrt{539.59} = 23.2 \text{ mm}\)

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

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

To find the median, locate the point where the cumulative frequency is 450 (half of 900) on the graph. From the graph, the median time \(t\) is approximately 4.8 minutes.

To solve this problem, we first need to construct the cumulative frequency table from the given frequency distribution:

(i) Plot these cumulative frequencies against the upper class boundaries on the grid to draw the cumulative frequency graph.

(ii) To estimate the percentage of trees with circumference larger than 75 cm, we need to find the cumulative frequency at 75 cm. Using linear interpolation between 50 cm and 80 cm:

The total number of trees is 140. The number of trees with circumference larger than 75 cm is:

The percentage is then:

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

• 3 – 5 seconds: Cumulative frequency = 10
• 6 – 8 seconds: Cumulative frequency = 10 + 15 = 25
• 9 – 11 seconds: Cumulative frequency = 25 + 17 = 42
• 12 – 16 seconds: Cumulative frequency = 42 + 4 = 46
• 17 – 25 seconds: Cumulative frequency = 46 + 2 = 48

Plot these cumulative frequencies against the upper class boundaries: (5.5, 10), (8.5, 25), (11.5, 42), (16.5, 46), (25.5, 48).

(ii) To find \(t\), note that 35 cars accelerated in more than \(t\) seconds, so 13 cars accelerated in \(t\) seconds or less. From the cumulative frequency graph, find the time corresponding to a cumulative frequency of 13. This gives \(t \approx 6.5\) seconds, so \(6 < t < 7\).

(i) The frequency for each class is calculated by multiplying the frequency density by the class width:

• For length 0-5 cm: Frequency density = 2, Class width = 5, Frequency = 2 × 5 = 10
• For length 5-10 cm: Frequency density = 8, Class width = 5, Frequency = 8 × 5 = 40
• For length 10-20 cm: Frequency density = 12, Class width = 10, Frequency = 12 × 10 = 120
• For length 20-25 cm: Frequency density = 6, Class width = 5, Frequency = 6 × 5 = 30

(ii) Cumulative frequencies are calculated as follows:

• Length < 5 cm: Cumulative frequency = 10
• Length < 10 cm: Cumulative frequency = 10 + 40 = 50
• Length < 20 cm: Cumulative frequency = 50 + 120 = 170
• Length < 25 cm: Cumulative frequency = 170 + 30 = 200

Plot these cumulative frequencies against the upper class boundaries to draw the cumulative frequency graph.

(iii) From the cumulative frequency graph:

• Median is the value at the 100th position (half of 200), which is approximately 14.2 cm.
• Lower quartile (LQ) is the value at the 50th position, approximately 10 cm.
• Upper quartile (UQ) is the value at the 150th position, approximately 18.5 cm.
• Interquartile range (IQR) = UQ - LQ = 18.5 - 10 = 8.5 cm.

(iv) Estimate the mean using midpoints and frequencies:

• Mean = \(\frac{(2.5 \times 10) + (7.5 \times 40) + (15 \times 120) + (22.5 \times 30)}{200}\)
• Mean = \(\frac{25 + 300 + 1800 + 675}{200} = \frac{2800}{200} = 14\)

(i) To find the median, locate the 100th caterpillar on the cumulative frequency graph (since there are 200 caterpillars). The median length is approximately 3.2 cm. For the interquartile range, find the lower quartile (50th caterpillar) and upper quartile (150th caterpillar). The lower quartile is approximately 2.6 cm and the upper quartile is approximately 3.7 cm. Thus, the interquartile range (IQR) is \(3.7 - 2.6 = 1.1\) cm.

(ii) To estimate the number of caterpillars with lengths between 2 and 3.5 cm, find the cumulative frequencies at these lengths. At 2 cm, the cumulative frequency is approximately 24, and at 3.5 cm, it is approximately 134. Therefore, the number of caterpillars is \(134 - 24 = 110\).

(iii) 6% of 200 caterpillars is 12 caterpillars. Therefore, 188 caterpillars have lengths less than \(l\). Locate the 188th caterpillar on the graph to estimate \(l\). The length \(l\) is approximately 4.5 cm.

(a) To find the interquartile range (IQR), we need to determine the values of the lower quartile (\(Q_1\)) and the upper quartile (\(Q_3\)).

\(Q_1\) is at the 25th percentile, which corresponds to \(0.25 \times 120 = 30\) on the cumulative frequency axis. From the graph, \(Q_1 \approx 23.7\) seconds.

\(Q_3\) is at the 75th percentile, which corresponds to \(0.75 \times 120 = 90\) on the cumulative frequency axis. From the graph, \(Q_3 \approx 31\) seconds.

Thus, the interquartile range is \(Q_3 - Q_1 = 31 - 23.7 = 7.3\) seconds.

(b) To find \(T\), note that 35% of the children took longer than \(T\) seconds. This means 65% took \(T\) seconds or less.

65% of 120 children is \(0.65 \times 120 = 78\).

From the graph, the time corresponding to a cumulative frequency of 78 is approximately \(28.5\) seconds.

(i) Draw cumulative frequency graphs for both girls and boys using the given data:

• Girls: Cumulative frequencies are 0, 12, 33, 50, 60.
• Boys: Cumulative frequencies are 0, 20, 43, 55, 60.

Plot these points on the graph with height on the horizontal axis and cumulative frequency on the vertical axis.

(ii) To find the percentage of girls shorter than 165 cm:

• From the cumulative frequency graph for girls, find the cumulative frequency at 165 cm, which is approximately 18.
• Calculate the percentage: \(\frac{18}{60} \times 100 = 30\%\).

(iii) One advantage of using box-and-whisker plots is that they clearly show the median, quartiles, and range, making it easier to compare the spread and central tendency of the data for boys and girls.

(i) From the cumulative frequency diagram, determine the quartiles: \(Q_1 = 2.6\), \( median= 3.8 \text{ to } 3.85 \), \(Q_3 = 6.4 \text{ to } 6.6\). Draw a box-and-whisker plot using these values. Ensure the axis is labeled with time in seconds and is linear from 2 to 10.

(ii) Calculate the interquartile range (IQR): \(IQR = Q_3 - Q_1 = 6.5 - 2.6 = 3.9\).Calculate \(1.5 \times IQR = 1.5 \times 3.9 = 5.85\).Check for outliers: \(Q_1 - 5.85 = -3.25\) (negative, so no lower outliers) and \(Q_3 + 5.85 = 12.35\) (greater than 10, so no upper outliers).

Therefore, there are no outliers.

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

Plot these points and join them with a smooth curve.

(ii) 28% of the daffodils are of height \(h\) cm or more, which means 72% are less than \(h\). 72% of 200 is 144, so \(h\) corresponds to the cumulative frequency of 144. From the graph, \(h \approx 22.5 \text{ cm}\).

(iii) To calculate the standard deviation:

Use the midpoints of the intervals: \(7, 13, 18, 23, 28\).

Calculate the variance:

\(\text{var} = \frac{(7^2 \times 22 + 13^2 \times 32 + 18^2 \times 78 + 23^2 \times 40 + 28^2 \times 28)}{200} - 18.39^2\)

\(= \frac{74870}{200} - 18.39^2\)

\(= 374.35 - 18.39^2\)

\(= 36.1579\)

Standard deviation \(\text{sd} = \sqrt{36.1579} = 6.01\).

(i) Plot the cumulative frequency points: (40, 0), (50, 12), (60, 34), (65, 64), (70, 92), (90, 144). Draw a smooth curve through these points.

(ii) To find \(c\), note that 64 people weigh more than \(c\), so 80 people weigh less than \(c\). From the graph, this corresponds to a weight of 67.2 kg. Thus, \(c = 67.2\) kg.

(iii) Calculate the frequencies: 12, 22, 30, 28, 52. Use midpoints: 45, 55, 62.5, 67.5, 80. Calculate the mean:

\(\text{Mean} = \frac{(45 \times 12) + (55 \times 22) + (62.5 \times 30) + (67.5 \times 28) + (80 \times 52)}{144} = \frac{9675}{144} = 67.2 \text{ kg}\)

\(Calculate the variance:\)

\(\text{Variance} = \frac{(45^2 \times 12) + (55^2 \times 22) + (62.5^2 \times 30) + (67.5^2 \times 28) + (80^2 \times 52)}{144} - (67.2)^2 = 127.59\)

Standard deviation:

\(\text{Standard deviation} = \sqrt{127.59} = 11.3 \text{ kg}\)