Past Exam Questions

(i) The graph is misleading because it uses a false zero, starting the y-axis at 4980 instead of 0, exaggerating the increase in sales.

(ii)(a) To construct a stem-and-leaf diagram, separate each number into a 'stem' (all but the last digit) and a 'leaf' (the last digit). Sort the numbers:

34, 35, 41, 44, 45, 50, 52, 62, 73, 73, 79, 83, 84, 84, 85, 85, 86, 86, 87, 89, 91.

Organize them into a stem-and-leaf format:

Key: \( 3 \mid 4 \) represents 34.

(ii)(b) To find the median, locate the middle value in the ordered list. With 21 numbers, the median is the 11th value: 79.

(a) The back-to-back stem-and-leaf diagram is drawn with the Lions on the left and the Tigers on the right. The stems represent the tens digit, and the leaves represent the units digit:

Key: \( 1 \mid 8 \mid 3 \) means 181 cm (Lions) and 183 cm (Tigers).

(b) Arrange the Lions' heights in ascending order:

169, 178, 179, 180, 181, 186, 187, 189, 190, 190, 196

The median is the 6th value: \( \text{Median} = 186 \) cm.

Lower quartile (LQ) is the median of the first half: \( LQ = 179 \) cm.

Upper quartile (UQ) is the median of the second half: \( UQ = 190 \) cm.

\( IQR = UQ - LQ = 190 - 179 = 11 \) cm.

(c) Comparisons:

• The Tigers are generally taller than the Lions, as their median height (190 cm) is greater than the Lions’ median (186 cm).
• The Tigers have a larger spread, as their upper quartile is higher, indicating more variation in player heights.

(a) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 6 \mid 2 \mid 3 \) means 26 m (Lakeview) and 23 m (Riverside).

(b) Arrange Lakeview’s data in ascending order:

10, 14, 19, 22, 26, 27, 28, 30, 32, 33, 41

Lower quartile (LQ) = 19 (3rd value in lower half).

Upper quartile (UQ) = 32 (3rd value in upper half).

\( IQR = UQ - LQ = 32 - 19 = 13 \) m.

(a) A stem-and-leaf diagram includes all raw data, allowing for further statistical processes such as finding the frequency, mean, mode, or standard deviation, which cannot be found using a box-and-whisker plot.

(b) The back-to-back stem-and-leaf diagram is drawn with Amazons on the left and Giants on the right. The stems are the tens digits, and the leaves are the units digits of the heights. See the mark-scheme for the correct diagram.

(c) To find the interquartile range (IQR) of the Amazons' heights, first order the data: 178, 181, 182, 184, 190, 196, 198, 201, 202, 205, 215. The lower quartile (LQ) is the median of the first half: 182 cm. The upper quartile (UQ) is the median of the second half: 202 cm. Therefore, the IQR is calculated as:

\(\text{IQR} = UQ - LQ = 202 - 182 = 20 \text{ cm}\)

(d) Let the height of the fourth new player be \(h\). The sum of the original 11 players' heights is 2132 cm. The sum of the heights of all 15 players is:

\(191.2 \times 15 = 2868 \text{ cm}\)

The sum of the heights of the four new players is:

\(180 + 185 + 190 + h = 555 + h\)

Equating the total heights:

\(2132 + 555 + h = 2868\)

\(2687 + h = 2868\)

\(h = 2868 - 2687 = 181 \text{ cm}\)

(a) Back-to-back stem-and-leaf diagram (stem = tens, leaf = units; Dados on the left, Linva on the right):

Key: \( 6 \mid 3 \mid 2 \) means \(36\) cm (Dados) and \(32\) cm (Linva).

(b) Dados ordered: 6, 8, 10, 10, 12, 15, 16, 22, 28, 36, 42.

Median \(= 15\ \text{cm}\).

Lower quartile \(Q_1 = 10\ \text{cm}\), upper quartile \(Q_3 = 28\ \text{cm}\).

Interquartile range: \( \text{IQR} = Q_3 - Q_1 = 28 - 10 = 18\ \text{cm}. \)

(c) Linva: median \(=12\ \text{cm}\), \(Q_1=9\ \text{cm}\), \(Q_3=32\ \text{cm}\) so \( \text{IQR}=23\ \text{cm} \).

Dados has higher central tendency (median \(15\ \text{cm}\) vs \(12\ \text{cm}\)) but smaller spread (IQR \(18\ \text{cm}\) vs \(23\ \text{cm}\)).

On average the snowfall in Davos is higher.

The amount of snowfall in Linva varies more than in Davos.

(a)

Key: \( 1 \mid 4 \mid 2 \) means \(\$41{,}000 \)(Company A) and \(\$42{,}000\) (Company B).

(b) Ordered salaries (A): 30, 32, 35, 41, 41, 42, 47, 49, 52, 53, 64.

Median \(=\) 6th value \(= \$42{,}000\).

Lower quartile \(Q_1 = \$35{,}000\); upper quartile \(Q_3 = \$52{,}000\).

Interquartile range \( \text{IQR} = Q_3 - Q_1 = \$52{,}000 - \$35{,}000 = \$17{,}000 \).

(c) Sum of given 11 salaries (B): \(\$433{,}000\).

Total for 12 employees: \(38{,}500 \times 12 = \$462{,}000\).

New employee’s salary: \( \$462{,}000 - \$433{,}000 = \$29{,}000 \).

(i)

Key: \( 8 \mid 4 \mid 5 \) represents \(48\) minutes (Thaters School) and \(45\) minutes (Whitefay Park School).

(ii) Ordered times (Thaters): 38, 43, 48, 52, 54, 56, 57, 58, 58, 61, 62, 66, 75.

Lower quartile \(Q_1 = \frac{48 + 52}{2} = 50\).

Upper quartile \(Q_3 = \frac{61 + 62}{2} = 61.5\).

Interquartile range \( \text{IQR} = Q_3 - Q_1 = 61.5 - 50 = 11.5 \) minutes.

(i)

Key: \( 3 \mid 6 \mid 4 \) means 63 kg for Dolphins and 64 kg for Sharks.

(ii) Ordered weights (Dolphins): 62, 63, 65, 65, 69, 72, 73, 75, 80, 82, 82.

Median = 72 kg (middle value).

Lower quartile \(Q_1 = 65\) kg, upper quartile \(Q_3 = 80\) kg.

Interquartile range \( \text{IQR} = Q_3 - Q_1 = 80 - 65 = 15 \) kg.

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 are: 20, 10, 5, 10, 15.

Calculate frequency densities:

\(\frac{18}{20} = 0.9\)

\(\frac{48}{10} = 4.8\)

\(\frac{34}{5} = 6.8\)

\(\frac{32}{10} = 3.2\)

\(\frac{18}{15} = 1.2\)

Draw bars with these heights on the histogram, ensuring the axes are labeled with 'Frequency Density' and 'Time in minutes'.

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

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

For 10–29: \(\frac{10}{20} = 0.5\)

For 30–39: \(\frac{24}{10} = 2.4\)

For 40–49: \(\frac{30}{10} = 3\)

For 50–59: \(\frac{14}{10} = 1.4\)

For 60–89: \(\frac{12}{30} = 0.4\)

Draw bars with these heights on the graph, with bar ends at 9.5, 29.5, 39.5, 59.5, 89.5. Label axes with 'Frequency density (fd)' and 'Speed/km h^-1'.

(ii) To calculate the mean speed, use the midpoints of each class interval:

Midpoints: 19.5, 34.5, 44.5, 54.5, 74.5

Calculate the weighted mean:

\(\frac{19.5 \times 10 + 34.5 \times 24 + 44.5 \times 30 + 54.5 \times 14 + 74.5 \times 12}{90}\)

\(= \frac{195 + 828 + 1335 + 763 + 894}{90}\)

\(= \frac{4015}{90} \approx 44.6 \text{ km h}^{-1}\)

(i) To find the lower quartile, we need to determine the position of the 25th percentile in the cumulative frequency distribution. The cumulative frequencies are:

- 10–14: 6

\(- 15–19: 6 + 12 = 18\)

\(- 20–24: 18 + 14 = 32\)

\(- 25–34: 32 + 10 = 42\)

\(- 35–59: 42 + 8 = 50\)

The 25th percentile corresponds to the 12.5th value. Since the cumulative frequency reaches 18 in the 15–19 interval, the lower quartile is in the 15–19 kg class interval.

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

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

- 10–14: \(\frac{6}{5} = 1.2\)

- 15–19: \(\frac{12}{5} = 2.4\)

- 20–24: \(\frac{14}{5} = 2.8\)

- 25–34: \(\frac{10}{10} = 1.0\)

- 35–59: \(\frac{8}{25} = 0.32\)

Draw bars with these heights on the histogram, ensuring the horizontal axis ranges from at least 9.5 to 59.5 and the vertical axis is labeled with frequency density.

Solution:

(1) The total frequency is 242, so:

\(14 + 46 + 102 + a + 40 = 242\)

\(a = 40\)

(2) To estimate the mean, take midpoints of each class: 0.5, 1.5, 3.5, 7.5, 20.

Total \(f \times x = 7 + 69 + 357 + 300 + 800 = 1533\).

Mean \(= \dfrac{1533}{242} \approx 6.3\) minutes.

(3) The histogram can be drawn with the following class widths and frequencies per unit:

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

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

For \(0 \leq t < 20\): \(\text{FD} = \frac{320}{20} = 16\)

For \(20 \leq t < 40\): \(\text{FD} = \frac{280}{20} = 14\)

For \(40 \leq t < 60\): \(\text{FD} = \frac{220}{20} = 11\)

For \(60 \leq t < 100\): \(\text{FD} = \frac{220}{40} = 5.5\)

For \(100 \leq t < 140\): \(\text{FD} = \frac{100}{40} = 2.5\)

Draw bars with these heights on the histogram.

(ii) To estimate the mean, use the midpoints of each class interval:

\(\text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\text{Total Frequency}}\)

\(= \frac{(10 \times 320) + (30 \times 280) + (50 \times 220) + (80 \times 220) + (120 \times 100)}{1140}\)

\(= \frac{3200 + 8400 + 11000 + 17600 + 12000}{1140}\)

\(= \frac{52200}{1140}\)

\(= 45.8\)

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

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

Class widths: 5, 5, 10, 20, 40 (in thousands)

Frequency densities: 8, 6, 1.8, 1.7, 0.2

Plot these on the histogram with the horizontal axis labeled 'Capacity (1000s)' and the vertical axis labeled 'Frequency Density'.

(ii) Calculate the mean capacity using the midpoints of each class:

\(\text{Mean} = \frac{(5 \times 40) + (10 \times 30) + (17.5 \times 18) + (32.5 \times 34) + (62.5 \times 8)}{130}\)

\(= \frac{2420}{130} = 18.6 \text{ thousand}\)

(iii) To find the median and lower quartile groups:

Median position: \(\frac{130 + 1}{2} = 65.5\)

Lower quartile position: \(\frac{130 + 1}{4} = 32.75\)

The median group is 8000–12000 and the lower quartile group is 3000–7000.

(i) To find the number of journey times between 25 and 45 minutes, subtract the cumulative frequency at 25 minutes from the cumulative frequency at 45 minutes: \(50 - 18 = 32\).

(ii) To draw the histogram, calculate the frequency for each interval: \([0, 10]: 0\), \((10, 25]: 18\), \((25, 45]: 32\), \((45, 60]: 9\), \((60, 80]: 4\). Then, calculate the frequency density by dividing the frequency by the class width: \(\frac{18}{15} = 1.2\), \(\frac{32}{20} = 1.6\), \(\frac{9}{15} = 0.6\), \(\frac{4}{20} = 0.2\). Draw bars with these heights.

(iii) To estimate the mean journey time, use the midpoints of each interval: \((17.5 \times 18) + (35 \times 32) + (52.5 \times 9) + (70 \times 4)\). Sum these products and divide by the total frequency: \(\frac{2187.5}{63} = 34.7\) minutes.

(i) To draw the histogram, calculate the frequency density for each class interval. Frequency density is given by:

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

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

Frequency densities: \(\frac{18}{20} = 0.9\), \(\frac{15}{5} = 3\), \(\frac{21}{5} = 4.2\), \(\frac{52}{10} = 5.2\), \(\frac{28}{20} = 1.4\)

Draw bars with these heights for each class interval.

(ii) Calculate the mean:

Midpoints: 30.5, 43, 48, 55.5, 70.5

Mean: \(\frac{(30.5 \times 18) + (43 \times 15) + (48 \times 21) + (55.5 \times 52) + (70.5 \times 28)}{134} = \frac{7062}{134} = 52.701 \approx 52.7\)

Calculate the variance:

\(\text{Var} = \frac{(30.5^2 \times 18) + (43^2 \times 15) + (48^2 \times 21) + (55.5^2 \times 52) + (70.5^2 \times 28)}{134} - 52.701^2\)

\(= \frac{392203.5}{134} - 52.701^2 = 149.496\)

Standard deviation: \(\sqrt{149.496} = 12.2\)

To find \(a\) and \(b\), we use the formula for frequency density: \(\text{fd} = \frac{f}{\text{cw}}\), where \(f\) is the frequency and \(\text{cw}\) is the class width.

For the class 63 - 64, \(\text{cw} = 2\) and \(f = 9\). Therefore, \(a = \frac{9}{2} = 4.5\).

For the class 68 - 71, \(\text{cw} = 4\) and \(\text{fd} = 1.5\). Therefore, \(1.5 = \frac{b}{4}\), which gives \(b = 6\).

(i) To find the class containing the upper quartile, calculate the position of the upper quartile in the ordered data set. The upper quartile position is given by \( \frac{3}{4} \times 104 = 78 \). Cumulative frequencies are: 8, 33, 61, 92, 104. The 78th value falls in the class 5.5–7.0 cm.

(ii) To draw the histogram, calculate the frequency density for each class using the formula \( \text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} \).

• Class 2.0–3.5: \( \text{Width} = 1.5,\ \text{Frequency} = 8,\ \text{Density} = \frac{8}{1.5} = 5.33 \)
• Class 3.5–4.5: \( \text{Width} = 1.0,\ \text{Frequency} = 25,\ \text{Density} = 25 \)
• Class 4.5–5.5: \( \text{Width} = 1.0,\ \text{Frequency} = 28,\ \text{Density} = 28 \)
• Class 5.5–7.0: \( \text{Width} = 1.5,\ \text{Frequency} = 31,\ \text{Density} = \frac{31}{1.5} = 20.7 \)
• Class 7.0–9.0: \( \text{Width} = 2.0,\ \text{Frequency} = 12,\ \text{Density} = \frac{12}{2.0} = 6 \)

Draw bars with these heights and ensure correct labels and bar widths with no gaps.

(i) The number of athletes who ran between 10.5 and 11.0 seconds is the difference in cumulative frequencies:

\(10 - 4 = 6.\)

(ii) To draw the histogram, calculate the frequency for each interval:

• 10.0 to 10.5: 4 athletes
• 10.5 to 11.0: 6 athletes
• 11.0 to 12.0: 30 athletes
• 12.0 to 12.5: 9 athletes
• 12.5 to 13.5: 8 athletes

Use these frequencies to draw the histogram with appropriate widths and heights.

(iii) Calculate the mean using midpoints:

\(E(X) = \frac{(10.25 \times 4) + (10.75 \times 6) + (11.5 \times 30) + (12.25 \times 9) + (13 \times 8)}{57} = 11.7\)

Calculate the variance:

\(\text{Var}(X) = \frac{(10.25^2 \times 4) + (10.75^2 \times 6) + (11.5^2 \times 30) + (12.25^2 \times 9) + (13^2 \times 8)}{57} - (11.7)^2 = 0.547\)

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

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

For each interval:

• 100 – 800: \(\frac{8}{800} = 0.01\)
• 900 – 1200: \(\frac{12}{400} = 0.03\)
• 1300 – 2000: \(\frac{50}{800} = 0.0625\)
• 2100 – 3200: \(\frac{48}{1200} = 0.04\)
• 3300 – 4800: \(\frac{32}{1600} = 0.02\)

Draw bars with these heights on the histogram.

(b) The median is the 75th value. Cumulative frequencies are:

• 100 – 800: 8
• 900 – 1200: 20
• 1300 – 2000: 70
• 2100 – 3200: 118
• 3300 – 4800: 150

The median falls in the 2100 – 3200 interval.

(c) To find the interquartile range, calculate the lower quartile (LQ) and upper quartile (UQ):

LQ is the 37.5th value, which is in the 1300 – 2000 interval.

UQ is the 112.5th value, which is in the 2100 – 3200 interval.

Greatest possible value of IQR is \(3200 - 1250 = 1999\).