(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:

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

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