Past Exam Questions

Solution

(i) First, find the class widths from the table:

• \( 1–5 \): \( 5 - 1 = 4 \), but since both ends are included, width \( = 5 \)
• \( 6–20 \): \( 20 - 6 + 1 = 15 \)
• \( 21–35 \): \( 35 - 21 + 1 = 15 \)
• \( 36–60 \): \( 60 - 36 + 1 = 25 \)
• \( 61–80 \): \( 80 - 61 + 1 = 20 \)

Then calculate frequency density using

\( \displaystyle \text{fd} = \frac{\text{Frequency}}{\text{Class Width}} \)

\( \displaystyle \text{fd} = \frac{24}{5},\ \frac{9}{15},\ \frac{21}{15},\ \frac{15}{25},\ \frac{42}{20} = 4.8,\ 0.6,\ 1.4,\ 0.6,\ 2.1 \).

Draw bars with these heights on the histogram; align class boundaries accurately and ensure correct widths with no gaps.

(ii) To estimate the mean, use class midpoints: \( 3,\ 13,\ 28,\ 48,\ 70.5 \).

\( \displaystyle \text{mean} = \frac{ (3 \times 24) + (13 \times 9) + (28 \times 21) + (48 \times 15) + (70.5 \times 42) }{111} = \frac{4458}{111} \approx 40.2\ \text{errors}. \)

(iii) Cumulative frequencies are \( 24,\ 33,\ 54,\ 69,\ 111 \).

Lower quartile position: \( \displaystyle 0.25 \times 111 = 27.75 \) → in class \( 6–20 \).
Upper quartile position: \( \displaystyle 0.75 \times 111 = 83.25 \) → in class \( 61–80 \).

For least possible IQR, take maximum possible LQ at \( 20 \) and minimum possible UQ at \( 61 \):

\( \displaystyle \text{IQR}_{\min} = 61 - 20 = 41 \).

1. In a correct histogram, the bars should be adjacent to each other without gaps, as the data is continuous. The diagram shows gaps between the bars, which is incorrect.

2. In a histogram, the area of each bar should represent the frequency of the data. This requires using frequency density, calculated as \(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}\). The diagram does not use frequency density, leading to incorrect bar heights.

(i) The frequency density for the interval 200 to 250 is 1.5. The width of the interval is 50. Therefore, the number of people is given by:

\(1.5 \times 50 = 75\)

(ii) The frequencies for each interval are 10, 25, 50, 75, 30. The mid-points of the intervals are 125, 162.5, 187.5, 225, 300.

Mean calculation:

\(\text{Mean} = \frac{(10 \times 125) + (25 \times 162.5) + (50 \times 187.5) + (75 \times 225) + (30 \times 300)}{190}\)

\(= \frac{40562.5}{190} = 213\)

Standard deviation calculation:

\(sd^2 = \frac{10 \times 125^2 + 25 \times 162.5^2 + 50 \times 187.5^2 + 75 \times 225^2 + 30 \times 300^2}{190} - (213)^2\)

\(sd = 46.5 \text{ or } 46.6\)

(iii) The answers are estimates because the mid-points of each interval were used instead of the raw data.

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

Midpoints: 3, 8, 15.5, 25.5, 40.5

Mean = \(\frac{(3 \times 59) + (8 \times 67) + (15.5 \times 38) + (25.5 \times 18) + (40.5 \times 11)}{193} = 11.4\)

To calculate the standard deviation, use:

\(\sigma^2 = \frac{(3^2 \times 59) + (8^2 \times 67) + (15.5^2 \times 38) + (25.5^2 \times 18) + (40.5^2 \times 11)}{193} - (11.4)^2\)

\(\sigma = 9.78 \text{ or } 9.79\)

(ii) For the histogram, calculate frequency densities:

Frequency density = \(\frac{\text{Frequency}}{\text{Class width}}\)

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

Frequency densities: \(\frac{59}{5} = 11.8, \frac{67}{5} = 13.4, \frac{38}{10} = 3.8, \frac{18}{10} = 1.8, \frac{11}{20} = 0.55\)

Draw bars with these heights at midpoints 5.5, 10.5, 20.5, 30.5, and 40.5.

(i) To find the median, calculate the cumulative frequency to find the middle value. The cumulative frequencies are: 19, 31, 59, 81, 99, 112. The median is the 56th value, which falls in the 15 < t ≤ 20 interval. The upper quartile (UQ) is the 84th value, which falls in the 25 < t ≤ 40 interval.

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

• 0 < t ≤ 10: fd = 19/10 = 1.9
• 10 < t ≤ 15: fd = 12/5 = 2.4
• 15 < t ≤ 20: fd = 28/5 = 5.6
• 20 < t ≤ 25: fd = 22/5 = 4.4
• 25 < t ≤ 40: fd = 18/15 = 1.2
• 40 < t ≤ 60: fd = 13/20 = 0.65

Draw bars with these heights and corresponding widths.

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

\(\frac{(5 \times 19) + (12.5 \times 12) + (17.5 \times 28) + (22.5 \times 22) + (32.5 \times 18) + (50 \times 13)}{112}\) =\( \frac{2465}{112} = 22.0 \text{ minutes}\)

Solution

(i) The median is the 110th value (since \( \dfrac{220}{2} = 110 \)). From the cumulative frequency table, the median lies in the interval \(45\text{–}50\ \text{g}\).

(ii) The lower quartile (LQ) is the 55th value, which lies in the interval \(40\text{–}45\ \text{g}\). The upper quartile (UQ) is the 165th value, which lies in the interval \(50\text{–}60\ \text{g}\). Therefore, the smallest interquartile range (IQR) could be \(5\ \text{g}\) (if \( \text{LQ} = 45 \) and \( \text{UQ} = 50 \)), and the largest could be \(20\ \text{g}\) (if \( \text{LQ} = 40 \) and \( \text{UQ} = 60 \)).

(iii) The number of sausages weighing between \(50\ \text{g}\) and \(60\ \text{g}\) is the difference in cumulative frequencies: \( 210 - 160 = 50 \).

(iv) To draw the histogram, first find class frequencies from cumulative frequencies, then compute frequency density \( \big( \text{fd} = \dfrac{\text{frequency}}{\text{class width}} \big) \):

\[ \begin{aligned} &<20:\ \text{fd} = \dfrac{0}{10} = 0,\\[4pt] &20\text{–}30:\ \text{fd} = \dfrac{20}{10} = 2,\\[4pt] &30\text{–}40:\ \text{fd} = \dfrac{30}{10} = 3,\\[4pt] &40\text{–}45:\ \text{fd} = \dfrac{50}{5} = 10,\\[4pt] &45\text{–}50:\ \text{fd} = \dfrac{60}{5} = 12,\\[4pt] &50\text{–}60:\ \text{fd} = \dfrac{50}{10} = 5,\\[4pt] &60\text{–}70:\ \text{fd} = \dfrac{10}{10} = 1. \end{aligned} \]

Plot these frequency densities against the corresponding weight intervals, with bar widths equal to class widths and no gaps.

(i) To find the frequencies, use the frequency density from the histogram. The frequency is calculated by multiplying the frequency density by the class width.

• For \( 2 < t \le 4 \): Frequency density \(= 10\), Width \(= 2\), Frequency \(= 10 \times 2 = 20\)
• For \( 4 < t \le 6 \): Frequency density \(= 22\), Width \(= 2\), Frequency \(= 22 \times 2 = 44\)
• For \( 6 < t \le 7 \): Frequency density \(= 34\), Width \(= 1\), Frequency \(= 34 \times 1 = 34\)
• For \( 7 < t \le 8 \): Frequency density \(= 30\), Width \(= 1\), Frequency \(= 30 \times 1 = 30\)
• For \( 8 < t \le 10 \): Frequency density \(= 15\), Width \(= 2\), Frequency \(= 15 \times 2 = 30\)
• For \( 10 < t \le 16 \): Frequency density \(= 6\), Width \(= 6\), Frequency \(= 6 \times 6 = 36\)

(ii) To estimate the mean time, calculate the midpoints of each class interval and use them to find the mean:

• Midpoints: \( 3,\ 5,\ 6.5,\ 7.5,\ 9,\ 13 \)
• \( \text{Mean} = \frac{ (3 \times 20) + (5 \times 44) + (6.5 \times 34) + {}\\ (7.5 \times 30) + (9 \times 30) + (13 \times 36) }{194}\) \(= \frac{1464}{194} = 7.55\ \text{minutes} \)

(i) The frequency density for the 1 – 10 interval is given by:

\(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{2x}{10}\)

We know the height is 3 cm, so:

\(3 = \frac{2x}{10}\)

Solving for \(x\):

\(3 \times 10 = 2x\)

\(30 = 2x\)

\(x = 15\)

For the 51 – 70 interval, the class width is 20, and the frequency is \(x = 15\).

\(\text{Frequency Density} = \frac{15}{20} = 0.75\)

Thus, the height of the 51 – 70 rectangle is 0.75 cm.

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

\(\text{Mean} = \frac{(5.5 \times 30) + (15.5 \times 60) + (23 \times 45) + (28 \times 75) + (40.5 \times 60) + (60.5 \times 15)}{285}\)

\(= \frac{165 + 930 + 1035 + 2100 + 2430 + 907.5}{285}\)

\(= \frac{7567.5}{285}\)

\(\approx 26.6 \text{ grams}\)

(i) To draw the histogram, calculate the frequency density for each class interval. The class widths are 20, 10, 10, 10, 10, 15. The frequency densities are calculated as follows:

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

For 1–20: \(\frac{40}{20} = 2.0\)

For 21–30: \(\frac{34}{10} = 3.4\)

For 31–40: \(\frac{56}{10} = 5.6\)

For 41–50: \(\frac{54}{10} = 5.4\)

For 51–60: \(\frac{29}{10} = 2.9\)

For 61–75: \(\frac{21}{15} = 1.4\)

Draw the histogram with these frequency densities.

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

Midpoints: 10.5, 25.5, 35.5, 45.5, 55.5, 68

Mean = \(\frac{\sum xf}{234} = \frac{8769.5}{234} = 37.5\)

To calculate the standard deviation, first find the variance:

Variance = \(\frac{\sum x^2f}{234} - \text{mean}^2\)

Standard Deviation = \(\sqrt{\text{Variance}} = 16.9\)

(i) To draw the histogram, calculate the frequency density for each class interval. Frequency density is given by \(\text{Frequency} / \text{Class Width}\).

For \(0.1 \leq t \leq 0.5\), Frequency = 11, Class Width = 0.4, Frequency Density = \(11 / 0.4 = 27.5\).

For \(0.6 \leq t \leq 1.0\), Frequency = 15, Class Width = 0.4, Frequency Density = \(15 / 0.4 = 37.5\).

For \(1.1 \leq t \leq 2.0\), Frequency = 18, Class Width = 0.9, Frequency Density = \(18 / 0.9 = 20\).

For \(2.1 \leq t \leq 3.0\), Frequency = 30, Class Width = 0.9, Frequency Density = \(30 / 0.9 = 33.33\).

For \(3.1 \leq t \leq 4.5\), Frequency = 21, Class Width = 1.4, Frequency Density = \(21 / 1.4 = 15\).

Draw bars with these heights on the histogram.

(ii) Calculate the mean time spent:

Find the midpoints of each interval: \(0.3, 0.8, 1.55, 2.55, 3.8\).

Multiply each midpoint by its frequency and sum these products: \((0.3 \times 11) + (0.8 \times 15) + (1.55 \times 18) + (2.55 \times 30) + (3.8 \times 21)\) \(= 199.5\).

Divide by the total frequency: \(199.5 / 95 = 2.1\).

Thus, the estimated mean time spent is 2.1 hours.

To calculate the standard deviation, we first find the midpoints of each class interval:

Midpoints: 10, 30, 45, 55, 80

Using the formula for variance:

\(\text{Var} = \frac{32 \times 10^2 + 46 \times 30^2 + 96 \times 45^2 + 52 \times 55^2 + 24 \times 80^2}{250} - 43.2^2\)

\(= \frac{549900}{250} - 43.2^2\)

\(= 2199.6 - 1866.24\)

\(= 333.36\)

Standard deviation \(\text{Sd} = \sqrt{333.36} = 18.3\)

To construct the grouped frequency table, we need to divide the range from \( 41.5 \) kg to \( 61.5 \) kg into five equal intervals.

\( \text{Total range} = 61.5 - 41.5 = 20 \) kg

\( \text{Class width} = \dfrac{20}{5} = 4 \) kg

The intervals are:

• 41.5–45.5
• 45.5–49.5
• 49.5–53.5
• 53.5–57.5
• 57.5–61.5

Next, count the number of weights in each interval:

• 41.5–45.5: 45, 44, 45 (4 weights)
• 45.5–49.5: 47, 49, 46, 48, 49 (7 weights)
• 49.5–53.5: 50, 51, 52, 50, 51, 52, 51, 53, 52, 53 (10 weights)
• 53.5–57.5: 55, 55, 54, 56, 57 (5 weights)
• 57.5–61.5: 58, 59, 60, 61 (4 weights)

Thus, the frequencies for each interval are \( 4,\ 7,\ 10,\ 5,\ 4 \) respectively.

(i) The modal age group is the one with the highest frequency density. From the histogram, the age group 30-35 years has the highest frequency density.

(ii) To find the number of fathers between 25 and 30 years old, multiply the frequency density by the class width. The frequency density for 25-30 years is 4.8, and the class width is 5 years. Therefore, the number of fathers is:

\(4.8 \times 5 = 24\)

(iii) To find the total number of fathers in the sample, sum the frequencies for all age groups. The frequencies are calculated by multiplying the frequency density by the class width for each group:

\(4 \times 1 + 3.6 \times 5 + 4.8 \times 5 + 5.6 \times 5 + 5.2 \times 5 + 2 \times 5 = 4 + 18 + 24 + 28 + 26 + 10 = 110\)

(i) To find the frequency density, use the formula:

\(\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}\)

For the class 3.20 ≤ x < 3.40, the class width is \(3.40 - 3.20 = 0.20\).

Given frequency is 8, so:

\(a = \frac{8}{0.20} = 40\)

(ii) To draw the histogram:

- Use uniform linear scales from at least 2.8 to 3.4 on the x-axis and 0 to 240 on the y-axis.

- Label both axes, with 'length (m)' on the x-axis and 'frequency density' on the y-axis.

- Draw bars with the following heights based on frequency density:

• 2.80 ≤ x < 3.00: height = 85
• 3.00 ≤ x < 3.10: height = 240
• 3.10 ≤ x < 3.20: height = 190
• 3.20 ≤ x < 3.40: height = 40

- Ensure correct widths with no gaps between bars.

To construct the histogram, we first determine the frequency of factory floor areas within each interval:

- Interval \(0 \leq x < 500\): 150, 350, 450 (3 factories)

- Interval \(500 \leq x < 1000\): 578, 595, 644, 722, 798, 802, 904 (7 factories)

- Interval \(1000 \leq x < 2000\): 1000, 1330, 1533, 1561, 1778, 1960 (6 factories)

- Interval \(2000 \leq x < 3000\): 2167, 2330, 2433 (3 factories)

- Interval \(3000 \leq x < 4000\): 3231 (1 factory)

The frequencies are therefore 3, 7, 6, 3, and 1 for each respective interval.

According to the mark scheme, the scaled frequencies are the same as the frequencies: 3, 7, 3, 1.5, 0.5.

The histogram should have bars with these heights, and the axes should be labeled appropriately with 'scaled f' and 'area, m\(^2\)'.

(i) To calculate the mean, use the midpoints of each interval:

Mean = \(\frac{(2.5 \times 11 + 7.5 \times 20 + 15 \times 32 + 25 \times 18 + 35 \times 10 + 55 \times 6)}{97} = 18.4\) minutes.

To calculate the standard deviation, use:

\(\text{sd} = \sqrt{\frac{(2.5^2 \times 11 + 7.5^2 \times 20 + 15^2 \times 32 + 25^2 \times 18 + 35^2 \times 10 + 55^2 \times 6)}{97} - \text{mean}^2}\) = \(13.3\) minutes.

(ii) To draw the histogram, calculate the frequency densities:

• 0 ≤ t < 5: \(\frac{11}{5} = 2.2\)
• 5 ≤ t < 10: \(\frac{20}{5} = 4.0\)
• 10 ≤ t < 20: \(\frac{32}{10} = 3.2\)
• 20 ≤ t < 30: \(\frac{18}{10} = 1.8\)
• 30 ≤ t < 40: \(\frac{10}{10} = 1.0\)
• 40 ≤ t < 70: \(\frac{6}{30} = 0.2\)

Plot these frequency densities against the time intervals on the histogram.

(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: 20, 10, 10, 20, 30

Frequency densities: \(\frac{440}{20} = 22\), \(\frac{720}{10} = 72\), \(\frac{920}{10} = 92\), \(\frac{300}{20} = 15\), \(\frac{120}{30} = 4\)

Plot these on the histogram with time on the x-axis and frequency density on the y-axis.

(b) Calculate the standard deviation using the midpoints of each class interval:

Midpoints: 10, 25, 35, 50, 75

Variance formula:

\(\frac{440(10^2) + 720(25^2) + 920(35^2) + 300(50^2) + 120(75^2)}{2500} - 31.44^2\)

\(= \frac{3046000}{2500} - 31.44^2 = 229.9264\)

Standard deviation \(= \sqrt{229.9264} = 15.2\)

(c) The upper quartile lies in the class interval 30–40.

(d) The standard deviation stays the same because the data remains in the same intervals, so the overall distribution is unchanged.

(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: 30, 15, 20, 10, 25

Frequency Densities: \(\frac{21}{30} = 0.7\), \(\frac{30}{15} = 2\), \(\frac{68}{20} = 3.4\), \(\frac{86}{10} = 8.6\), \(\frac{45}{25} = 1.8\)

Draw bars with these heights for each class interval.

(b) The median is the middle value of the ordered data set. Since there are 250 children, the median is the 125.5th value. Cumulative frequencies: 21, 51, 119, 205, 250. The median falls in the 66 – 75 interval.

(c) The mean is 61.05 minutes, which is different from the median interval. This indicates the distribution is not symmetrical, possibly skewed.

(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 the intervals:

• \(0 \leq t < 5\): \(\frac{23}{5} = 4.6\)
• \(5 \leq t < 10\): \(\frac{102}{5} = 20.4\)
• \(10 \leq t < 20\): \(\frac{135}{10} = 13.5\)
• \(20 \leq t < 30\): \(\frac{76}{10} = 7.6\)
• \(30 \leq t < 50\): \(\frac{24}{20} = 1.2\)

Draw bars with these heights, ensuring axes are labeled correctly.

(b) To estimate the mean time, use the midpoints of each interval:

• \(0 \leq t < 5\): midpoint = 2.5
• \(5 \leq t < 10\): midpoint = 7.5
• \(10 \leq t < 20\): midpoint = 15
• \(20 \leq t < 30\): midpoint = 25
• \(30 \leq t < 50\): midpoint = 40

Calculate the mean:

\(\frac{(2.5 \times 23) + (7.5 \times 102) + (15 \times 135) + (25 \times 76) + (40 \times 24)}{360}\)

\(= \frac{5707.5}{360} \approx 15.9\) minutes

(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: 10, 10, 20, 20, 40

Frequency densities: \(\frac{16}{10} = 1.6\), \(\frac{54}{10} = 5.4\), \(\frac{78}{20} = 3.9\), \(\frac{32}{20} = 1.6\), \(\frac{20}{40} = 0.5\)

Draw bars with these heights on the histogram.

(b) To estimate the mean time, use the midpoints of each interval:

Midpoints: 5, 15, 30, 50, 80

Mean = \(\frac{16 \times 5 + 54 \times 15 + 78 \times 30 + 32 \times 50 + 20 \times 80}{200}\)

\(= \frac{80 + 810 + 2340 + 1600 + 1600}{200}\)

\(= \frac{6430}{200} = 32.15\)

Accept 32.2 as the answer.

(c) The greatest possible value of the interquartile range is calculated by taking a value in the upper quartile (40-60) and subtracting a value in the lower quartile (10-20):

\(Greatest possible value = 60 - 10 = 50\)