Past Exam Questions

(i) To find the median and quartiles for females:

There are 39 females, so the median is the 20th value.

Counting from the smallest, the 20th value is $22 700.

The lower quartile (Q1) is the 10th value, which is $21 700.

The upper quartile (Q3) is the 30th value, which is $24 000.

(ii) For the box-and-whisker plots:

Use the given quartiles for males: Median = \($24 000, LQ = $22 600, UQ = $25 300.\)

Draw the box-and-whisker plots on a uniform scale with labels for salary in $000.

(i) To find the median for A, list the values in ascending order: 0.110, 0.111, 0.120, 0.130, 0.131, 0.133, 0.141, 0.150, 0.154, 0.155, 0.160, 0.164, 0.165, 0.166, 0.168, 0.169, 0.170, 0.171, 0.180, 0.183, 0.183, 0.186, 0.188, 0.188, 0.189, 0.198, 0.199, 0.200.
The median is the middle value: 0.186.
To find the interquartile range (IQR), calculate the lower quartile (Q1) and upper quartile (Q3).
Q1 is the 7th value: 0.179.
Q3 is the 21st value: 0.198.
IQR = Q3 - Q1 = 0.198 - 0.179 = 0.019.

(ii) For variable B, the given median is 0.171, Q1 is 0.164, and Q3 is 0.179.
Draw box-and-whisker plots for A and B using these values, ensuring a uniform scale from at least 0.15 to 0.21.

To draw the box-and-whisker plots:

History:

• Order the marks: 27, 28, 33, 33, 37, 38, 39, 42, 48, 49, 51, 55, 57
• Lowest mark: 27
• Highest mark: 57
• Median: 39 (7th value)
• Lower quartile (LQ): 33 (average of 3rd and 4th values)
• Upper quartile (UQ): 50 (average of 10th and 11th values)

Physics:

• Lowest mark: 17
• Highest mark: 74
• Lower quartile (LQ): 28
• Median: 39
• Upper quartile (UQ): 67

Draw the box-and-whisker plots on a uniform scale with appropriate labels.

Difference: The Physics marks are more spread out than the History marks, as indicated by the wider range and interquartile range in the Physics data.

(i) Count the total number of leaves in the first stem-and-leaf diagram:

\(4 + 2 + 4 + 20 + 12 + 7 + 18 = 67\)

Thus, there are 67 shelves in this section of the library.

(ii) To draw a box-and-whisker plot, we need the quartiles:

• Lower Quartile (LQ) = 64
• Median = 73
• Upper Quartile (UQ) = 90

Draw the box-and-whisker plot using these quartiles and the minimum and maximum values from the data.

(iii) The books in the second section have a smaller interquartile range (IQR) or standard deviation, indicating less variability in the number of books per shelf compared to the first section.

To find the median and quartiles, first arrange the data in ascending order: 104, 104, 109, 115, 117, 120, 124, 125, 132, 134, 142, 145, 158, 160, 162.

The median is the middle value. Since there are 15 data points, the median is the 8th value: 125.

The lower quartile (LQ) is the median of the first half of the data (excluding the overall median if the number of data points is odd). The first half is: 104, 104, 109, 115, 117, 120, 124. The median of this set is the 4th value: 115.

The upper quartile (UQ) is the median of the second half of the data. The second half is: 132, 134, 142, 145, 158, 160, 162. The median of this set is the 4th value: 145.

For the box-and-whisker plot, draw a number line from at least 110 to 160. Plot the LQ at 115, the median at 125, and the UQ at 145. Draw a box from 115 to 145 with a line at 125. Extend whiskers to the minimum (104) and maximum (162) values.

(a) To find the median for Company A, list the diameters in order: 0.330, 0.331, 0.332, 0.333, 0.334, 0.335, 0.336, 0.337, 0.338, 0.339, 0.340, 0.341, 0.342, 0.343, 0.344, 0.345, 0.346, 0.347, 0.348. The median is the 10th value, which is 0.355 cm.

To find the interquartile range (IQR), calculate the lower quartile (Q1) and upper quartile (Q3). Q1 is the 5th value, 0.348 cm, and Q3 is the 15th value, 0.366 cm. Thus, IQR = Q3 - Q1 = 0.366 - 0.348 = 0.018 cm.

(b) Draw box-and-whisker plots using the given quartiles for Company B and calculated quartiles for Company A. Ensure the plots are accurately labeled and scaled.

(c) One comparison could be that Company B has a wider interquartile range than Company A, indicating more variability in the diameters of pipes produced by Company B.

(i) To find the lower quartile (LQ), subtract the interquartile range (IQR) from the upper quartile (UQ):

Given: UQ = 4 hours 42 minutes, IQR = 3 hours 48 minutes.

Convert times to minutes: UQ = 282 minutes, IQR = 228 minutes.

LQ = UQ - IQR = 282 - 228 = 54 minutes.

Convert back to hours and minutes: 54 minutes = 0 hours 54 minutes.

Therefore, the lower quartile is 2 hours 48 minutes.

(ii) To draw the box-and-whisker plot:

• Minimum value: 0.5 hours (30 minutes).
• Lower quartile: 2.8 hours (2 hours 48 minutes).
• Median: 3.6 hours (3 hours 36 minutes).
• Upper quartile: 4.7 hours (4 hours 42 minutes).
• Maximum value: 5.2 hours (5 hours 12 minutes).

Draw a number line from 0 to 6 hours. Use a scale where 2 cm represents 60 minutes (1 hour). Plot the minimum, lower quartile, median, upper quartile, and maximum values accordingly.

(i) A stem-and-leaf diagram shows all the data, allowing for easy identification of the shape, mode, and other characteristics of the distribution.

(ii) To find the median and quartiles for the group who do not exercise:

• Arrange the data in ascending order.
• There are 63 data points, so the median is the 32nd value: 6.5.
• The lower quartile (Q1) is the 16th value: 5.4.
• The upper quartile (Q3) is the 48th value: 8.3.

(iii) Draw two box-and-whisker plots on a single diagram with a linear scale from 3 to 10. Label the plots for 'exercise' and 'not exercise'.

(i) To find the median for country Q, list the weights in order: 51, 55, 62, 63, 64, 68, 71, 73, 74, 75, 76, 77, 77, 78, 78, 79, 82, 83, 86, 87, 87, 88, 88, 90, 92, 92, 94, 104, 105. There are 29 values, so the median is the 15th value: 78.

For the lower quartile (Q1), find the 7.5th value, which is between the 7th and 8th values:\( (71 + 73)/2 = 72.\)

For the upper quartile (Q3), find the 22.5th value, which is between the 22nd and 23rd values: \((87 + 88)/2 = 87.5.\)

(ii) Draw box-and-whisker plots for both countries P and Q using the given quartiles.

(iii) From the box plots, it is evident that people in country P are generally heavier than those in country Q. Additionally, the weights in country Q are more spread out, indicating greater variability.

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

Key: 8 | 7 | 2 means 78 kg for Rebels and 72 kg for Sharks.

(b) To find the median for the Rebels, arrange the data in order: 75, 78, 79, 80, 82, 82, 83, 84, 85, 86, 89, 93, 95, 99, 102. The median is the 8th value, which is 84 kg.

To find the interquartile range (IQR), calculate the lower quartile (Q1) and upper quartile (Q3). Q1 is the 4th value: 80 kg. Q3 is the 12th value: 93 kg. Thus, IQR = Q3 - Q1 = 93 - 80 = 13 kg.

(c) Draw a box-and-whisker plot for the Rebels with endpoints at 75 and 102, and median and quartiles as found in part (b).

(d) One comparison is that the average weight of the Rebels is higher than the average weight of the Sharks.

(a) To find the median for machine A, arrange the data in ascending order: 0.220, 0.231, 0.231, 0.232, 0.233, 0.234, 0.235, 0.236, 0.237, 0.238, 0.239, 0.240, 0.241, 0.242, 0.243, 0.244, 0.245, 0.246, 0.247. The median is the 10th value, which is 0.238 m.

To find the interquartile range (IQR), calculate the lower quartile (LQ) and upper quartile (UQ). LQ is the 5th value: 0.231 m. UQ is the 15th value: 0.245 m. Thus, IQR = UQ - LQ = 0.245 - 0.231 = 0.014 m.

(b) For the box-and-whisker plot, use the following values for machine A: LQ = 0.231, Median = 0.238, UQ = 0.245, minimum = 0.220, maximum = 0.254. For machine B: LQ = 0.224, Median = 0.232, UQ = 0.243, minimum = 0.211, maximum = 0.256.

(c) Comparison 1: The median length of rods produced by machine A is greater than that of machine B (0.238 m vs. 0.232 m).

Comparison 2: The interquartile range for machine A is smaller than that for machine B (0.014 m vs. 0.019 m), indicating that the lengths of rods produced by machine A are less spread out.

(i) Advantage: A box-and-whisker plot provides a visual summary of data distribution, showing the median, quartiles, and potential outliers, which helps in understanding the spread and shape of the data.

Disadvantage: It does not provide detailed information about individual data points or the exact distribution of data within the quartiles.

(iii)(a) To draw the box-and-whisker plot:

• Identify the minimum value: 190
• Identify the maximum value: 375
• Calculate the median: 280
• Calculate the lower quartile (LQ): 256 or 256.5
• Calculate the upper quartile (UQ): 329

Draw the box from LQ to UQ with a line at the median. Extend whiskers from the minimum to LQ and from UQ to the maximum.

(iii)(b) The interquartile range (IQR) is calculated as:

\(IQR = UQ - LQ = 329 - 256 = 73\) or \(329 - 256.5 = 72.5\)

To draw the box-and-whisker plot, we first find the quartiles:

Lower quartile \(( Q_1 ) = 18\)

Median\( = 25\)

Upper quartile \(( Q_3 ) = 50\)

Interquartile range\( (IQR) = Q_3 - Q_1 = 50 - 18 = 32\)

For outliers, calculate 1.5 times the IQR:

\(1.5 × IQR = 1.5 × 32 = 48\)

Check for outliers:

Lower bound\( = Q_1 - 1.5 × IQR = 18 - 48 = -30 \)(no values below this)

Upper bound \(= Q_3 + 1.5 × IQR = 50 + 48 = 98 \)(no values above this)

Since all data values are within the bounds, there are no outliers.

First, arrange the data in ascending order:

2, 2, 5, 6, 9, 10, 10, 10, 12, 12, 13, 13, 14, 17, 17, 18, 18, 28, 28, 34, 35, 38, 44, 65, 82, 88, 104

There are \(27\) data points, so the median is the \(14^\text{th}\) value:

\(\text{Median} = 17\)

To find the lower quartile (LQ), take the median of the first \(13\) values (the \(7^\text{th}\) value):

\(\text{LQ} = 10\)

To find the upper quartile (UQ), take the median of the last \(13\) values (the \(7^\text{th}\) of that half, overall \(21^\text{st}\) value):

\(\text{UQ} = 35\)

Draw a box-and-whisker plot with these values:

• Minimum = 2
• LQ = 10
• Median = 17
• UQ = 35
• Maximum = 104

First, calculate the frequencies for each class interval:

• 40 < x ≤ 60: 14
• 60 < x ≤ 65: 24
• 65 < x ≤ 70: 22
• 70 < x ≤ 85: 46
• 85 < x ≤ 100: 14

Next, determine the midpoints for each class interval:

• 40 < x ≤ 60: 50
• 60 < x ≤ 65: 62.5
• 65 < x ≤ 70: 67.5
• 70 < x ≤ 85: 77.5
• 85 < x ≤ 100: 92.5

Calculate the mean:

\(\text{Mean} = \frac{0 \times 20 + 14 \times 50 + 24 \times 62.5 + 22 \times 67.5 + 46 \times 77.5 + 14 \times 92.5}{120} = \frac{8545}{120} = 71.2 \text{ kg}\)

\(Calculate the variance:\)

\(\text{Variance} = \frac{0 \times 20^2 + 14 \times 50^2 + 24 \times 62.5^2 + 22 \times 67.5^2 + 46 \times 77.5^2 + 14 \times 92.5^2}{120} - 71.2^2\)

\(= \frac{625062.5}{120} - 71.2^2 = 138.23\)

Calculate the standard deviation:

\(\text{Standard deviation} = \sqrt{138.23} = 11.8 \text{ kg}\)

(i) To find the mean age of all 32 members, calculate the total age of all members and divide by the total number of members:

\(\text{Total age of Juniors} = 15.5 \times 12 = 186\)

\(\text{Total age of Seniors} = 910\)

\(\text{Total age of all members} = 186 + 910 = 1096\)

\(\text{Mean age} = \frac{1096}{32} = 34.25 \text{ years}\)

\((ii) To find the standard deviation of all 32 members, first find the variance:\)

For Juniors, \(\text{variance} = \frac{\Sigma x^2}{12} - 15.5^2 = 1.2^2\)

\(\Sigma x^2 = 2900.28\)

For the whole group:

\(\Sigma z^2 = \Sigma x^2 + \Sigma y^2 = 2900.28 + 42850 = 45750\)

\(\text{Variance} = \frac{45750}{32} - (34.25)^2 = 256.63\)

\(\text{Standard deviation} = \sqrt{256.63} = 16.0 \text{ years}\)

(i) To find the mean price of all 51 holidays, use the formula for the combined mean:

\(\text{Mean} = \frac{30 \times 1500 + 21 \times 2400}{51} = \frac{45000 + 50400}{51} = 1870.59\)

Rounding to 3 significant figures, the mean is 1870.

(ii) First, find \(\sum x_F^2\) and \(\sum x_L^2\) using the given standard deviations and means:

For Farfield Travel:

\(230^2 = \frac{\sum x_F^2}{30} - 1500^2\)

\(\sum x_F^2 = 30 \times (230^2 + 1500^2) = 69\,087\,000\)

For Lacket Travel:

\(160^2 = \frac{\sum x_L^2}{21} - 2400^2\)

\(\sum x_L^2 = 21 \times (160^2 + 2400^2) = 121\,497\,600\)

\(Now, calculate the combined variance:\)

\(\text{New variance} = \frac{69\,087\,000 + 121\,497\,600}{51} - 1870.59^2 = 237\,853\)

\(The standard deviation is the square root of the variance:\)

\(\text{New standard deviation} = \sqrt{237\,853} \approx 488\)