The following table shows the distribution of the masses of 120 kg of wood, correct to the nearest kg. Draw a histogram, make a cumulative frequency table and curve, use the graph to find the semi interquartile range and 60th percentile.

Question

Asked on July 13, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-13T05:24:30.957Z

Here are the solutions to your questions. Please note that I cannot draw graphs directly, so I will provide instructions on how to draw them and use interpolation for calculations that would typically be read from a graph. Step 1: Prepare data for the Histogram (Question 1) To draw a histogram, we first need to determine the class boundaries. Since the data is given in discrete intervals (e.g., 15-24), we adjust them to be continuous. Original Classes | Class Boundaries (kg) | Frequency -----------------|-----------------------|----------- 15 - 24 | 14.5 - 24.5 | 14 25 - 34 | 24.5 - 34.5 | 54 35 - 44 | 34.5 - 44.5 | 24 45 - 54 | 44.5 - 54.5 | 26 55 - 64 | 54.5 - 64.5 | 2 To draw the histogram: • Draw the Mass (kg) on the x-axis, marking the class boundaries (14.5, 24.5, 34.5, 44.5, 54.5, 64.5). • Draw the Frequency on the y-axis. • Draw rectangular bars for each class interval. The width of each bar should span its class boundaries, and the height should correspond to its frequency. For example, a bar from 14.5 to 24.5 with a height of 14. Step 2: Make a Cumulative Frequency Table (Question 2) We calculate the cumulative frequency by adding the frequency of each class to the sum of frequencies of all preceding classes. Masses (kg) | Frequency (f) | Upper Class Boundary (kg) | Cumulative Frequency (cf) ------------|-----------------|---------------------------|---------------------------- 15 - 24 | 14 | 24.5 | 14 25 - 34 | 54 | 34.5 | 14 + 54 = 68 35 - 44 | 24 | 44.5 | 68 + 24 = 92 45 - 54 | 26 | 54.5 | 92 + 26 = 118 55 - 64 | 2 | 64.5 | 118 + 2 = 120 Step 3: Draw a Cumulative Frequency Curve (Ogive) (Question 3) To draw the cumulative frequency curve: • Draw the Upper Class Boundary (kg) on the x-axis. • Draw the Cumulative Frequency on the y-axis. • Plot the points using the upper class boundaries and their corresponding cumulative frequencies. It's common to start with a point at the lower boundary of the first class with a cumulative frequency of 0. The points to plot are: (14.5, 0), (24.5, 14), (34.5, 68), (44.5, 92), (54.5, 118), (64.5, 120). • Connect these points with a smooth curve. Step 4: Use the graph to find (Question 4) Since a graph cannot be drawn here, we will use linear interpolation from the cumulative frequency table to find the values. The total number of observations is N = 120. a) Semi-interquartile range The semi-interquartile range (SIQR) is given by (Q_3 - Q_1)/(2). First Quartile (Q_1): The position of Q_1 is (N)/(4) = (120)/(4) = 30. The 30th observation falls in the 25-34 class (cumulative frequency 68, previous is 14). Using the formula for Q_1: Q_1 = L + ((N)/(4) - cf_bf) × c where L = 24.5 (lower boundary of Q_1 class), cf_b = 14 (cumulative frequency before Q_1 class), f = 54 (frequency of Q_1 class), c = 10 (class width). Q_1 = 24.5 + ((30 - 14)/(54)) × 10 Q_1 = 24.5 + ((16)/(54)) × 10 Q_1 = 24.5 + 2.96 Q_1 ≈ 27.46 kg Third Quartile (Q_3): The position of Q_3 is (3N)/(4) = (3 × 120)/(4) = 90. The 90th observation falls in the 35-44 class (cumulative frequency 92, previous is 68). Using the formula for Q_3: Q_3 = L + ((3N)/(4) - cf_bf) × c where L = 34.5 (lower boundary of Q_3 class), cf_b = 68 (cumulative frequency before Q_3 class), f = 24 (frequency of Q_3 class), c = 10 (class width). Q_3 = 34.5 + ((90 - 68)/(24)) × 10 Q_3 = 34.5 + ((22)/(24)) × 10 Q_3 = 34.5 + 9.17 Q_3 ≈ 43.67 kg Now, calculate the semi-interquartile range: SIQR = (Q_3 - Q_1)/(2) = (43.67 - 27.46)/(2) = (16.21)/(2) SIQR = 8.11 kg b) 60th percentile The position of the 60th percentile (P_60) is (60)/(100) × N = (60)/(100) × 120 = 72. The 72nd observation falls in the 35-44 class (cumulative frequency 92, previous is 68). Using the formula for P_60: P_60 = L + ((60N)/(100) - cf_bf) × c where L = 34.5 (lower boundary of P_60 class), cf_b = 68 (cumulative frequency before P_60 class), f = 24 (frequency of P_60 class), c = 10 (class width). P_60 = 34.5 + ((72 - 68)/(24)) × 10 P_60 = 34.5 + ((4)/(24)) × 10 P_60 = 34.5 + 1.67 P_60 ≈ 36.17 kg Send me the next one 📸

Accepted Answer · 2026-07-13T05:24:30.957Z

Here are the solutions to your questions. Please note that I cannot draw graphs directly, so I will provide instructions on how to draw them and use interpolation for calculations that would typically be read from a graph. Step 1: Prepare data for the Histogram (Question 1) To draw a histogram, we first need to determine the class boundaries. Since the data is given in discrete intervals (e.g., 15-24), we adjust them to be continuous. Original Classes | Class Boundaries (kg) | Frequency -----------------|-----------------------|----------- 15 - 24 | 14.5 - 24.5 | 14 25 - 34 | 24.5 - 34.5 | 54 35 - 44 | 34.5 - 44.5 | 24 45 - 54 | 44.5 - 54.5 | 26 55 - 64 | 54.5 - 64.5 | 2 To draw the histogram: • Draw the Mass (kg) on the x-axis, marking the class boundaries (14.5, 24.5, 34.5, 44.5, 54.5, 64.5). • Draw the Frequency on the y-axis. • Draw rectangular bars for each class interval. The width of each bar should span its class boundaries, and the height should correspond to its frequency. For example, a bar from 14.5 to 24.5 with a height of 14. Step 2: Make a Cumulative Frequency Table (Question 2) We calculate the cumulative frequency by adding the frequency of each class to the sum of frequencies of all preceding classes. Masses (kg) | Frequency (f) | Upper Class Boundary (kg) | Cumulative Frequency (cf) ------------|-----------------|---------------------------|---------------------------- 15 - 24 | 14 | 24.5 | 14 25 - 34 | 54 | 34.5 | 14 + 54 = 68 35 - 44 | 24 | 44.5 | 68 + 24 = 92 45 - 54 | 26 | 54.5 | 92 + 26 = 118 55 - 64 | 2 | 64.5 | 118 + 2 = 120 Step 3: Draw a Cumulative Frequency Curve (Ogive) (Question 3) To draw the cumulative frequency curve: • Draw the Upper Class Boundary (kg) on the x-axis. • Draw the Cumulative Frequency on the y-axis. • Plot the points using the upper class boundaries and their corresponding cumulative frequencies. It's common to start with a point at the lower boundary of the first class with a cumulative frequency of 0. The points to plot are: (14.5, 0), (24.5, 14), (34.5, 68), (44.5, 92), (54.5, 118), (64.5, 120). • Connect these points with a smooth curve. Step 4: Use the graph to find (Question 4) Since a graph cannot be drawn here, we will use linear interpolation from the cumulative frequency table to find the values. The total number of observations is N = 120. a) Semi-interquartile range The semi-interquartile range (SIQR) is given by (Q_3 - Q_1)/(2). First Quartile (Q_1): The position of Q_1 is (N)/(4) = (120)/(4) = 30. The 30th observation falls in the 25-34 class (cumulative frequency 68, previous is 14). Using the formula for Q_1: Q_1 = L + ((N)/(4) - cf_bf) × c where L = 24.5 (lower boundary of Q_1 class), cf_b = 14 (cumulative frequency before Q_1 class), f = 54 (frequency of Q_1 class), c = 10 (class width). Q_1 = 24.5 + ((30 - 14)/(54)) × 10 Q_1 = 24.5 + ((16)/(54)) × 10 Q_1 = 24.5 + 2.96 Q_1 ≈ 27.46 kg Third Quartile (Q_3): The position of Q_3 is (3N)/(4) = (3 × 120)/(4) = 90. The 90th observation falls in the 35-44 class (cumulative frequency 92, previous is 68). Using the formula for Q_3: Q_3 = L + ((3N)/(4) - cf_bf) × c where L = 34.5 (lower boundary of Q_3 class), cf_b = 68 (cumulative frequency before Q_3 class), f = 24 (frequency of Q_3 class), c = 10 (class width). Q_3 = 34.5 + ((90 - 68)/(24)) × 10 Q_3 = 34.5 + ((22)/(24)) × 10 Q_3 = 34.5 + 9.17 Q_3 ≈ 43.67 kg Now, calculate the semi-interquartile range: SIQR = (Q_3 - Q_1)/(2) = (43.67 - 27.46)/(2) = (16.21)/(2) SIQR = 8.11 kg b) 60th percentile The position of the 60th percentile (P_60) is (60)/(100) × N = (60)/(100) × 120 = 72. The 72nd observation falls in the 35-44 class (cumulative frequency 92, previous is 68). Using the formula for P_60: P_60 = L + ((60N)/(100) - cf_bf) × c where L = 34.5 (lower boundary of P_60 class), cf_b = 68 (cumulative frequency before P_60 class), f = 24 (frequency of P_60 class), c = 10 (class width). P_60 = 34.5 + ((72 - 68)/(24)) × 10 P_60 = 34.5 + ((4)/(24)) × 10 P_60 = 34.5 + 1.67 P_60 ≈ 36.17 kg Send me the next one 📸

The following table shows the distribution of the masses of 120 kg of wood, correct to the nearest kg. Draw a histogram, make a cumulative frequency table and curve, use the graph to find the semi interquartile range and 60th percentile.

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The following table shows the distribution of the masses of 120 kg of wood, correct to the nearest kg. Draw a histogram, make a cumulative frequency table and curve, use the graph to find the semi interquartile range and 60th percentile.

Need help with your own homework?

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