Draw a stem-and-leaf display for this data Organise the data into class intervals. Calculate the actual mean for this data. Calculate an estimated mean for this data.

Question

Asked on August 3, 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 · 2026-08-03T21:07:43.057Z

a) To calculate the mean, mode, median, and range for the appearance ratings of the four additional customers (Customer 3, 4, 5, 6), we first list all 36 ratings: 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5. Step 1: Calculate the mean, mode, median, and range for part (a). • Mean: The sum of all ratings is 1 × 1 + 4 × 2 + 12 × 3 + 12 × 4 + 7 × 5 = 1 + 8 + 36 + 48 + 35 = 128. The total number of ratings is 36. Mean = (128)/(36) ≈ 3.56 • Mode: The values that appear most frequently are 3 and 4, both appearing 12 times. Mode = 3 and 4 • Median: With 36 data points, the median is the average of the 18th and 19th values. After sorting, both the 18th and 19th values are 4. Median = (4+4)/(2) = 4 • Range: The difference between the maximum and minimum values is 5 - 1 = 4. Range = 4 b) Calculate the mean, mode, median, and range for the appearance and service ratings for Store C. Step 2: Calculate for Appearance rating. • Mean: Mean = (Score × Frequency) Frequency = ((1 × 2820) + (2 × 3130) + (3 × 18080) + (4 × 45922) + (5 × 30048))/(100000) Mean = (2820 + 6260 + 54240 + 183688 + 150240)/(100000) = (397248)/(100000) = 3.97248 • Mode: The score with the highest frequency (45 922) is 4. Mode = 4 • Median: The total frequency is 100 000. The median position is the 50 000th value. The cumulative frequency shows that the 50 000th value falls within the score of 4 (cumulative frequency up to score 3 is 24 030, up to score 4 is 69 952). Median = 4 • Range: The difference between the maximum and minimum scores is 5 - 1 = 4. Range = 4 Step 3: Calculate for Service rating. • Mean: Mean = (Score × Frequency) Frequency = ((1 × 3910) + (2 × 10050) + (3 × 19050) + (4 × 45890) + (5 × 21100))/(100000) Mean = (3910 + 20100 + 57150 + 183560 + 105500)/(100000) = (370220)/(100000) = 3.7022 • Mode: The score with the highest frequency (45 890) is 4. Mode = 4 • Median: The total frequency is 100 000. The median position is the 50 000th value. The cumulative frequency shows that the 50 000th value falls within the score of 4 (cumulative frequency up to score 3 is 33 010, up to score 4 is 78 900). Median = 4 • Range: The difference between the maximum and minimum scores is 5 - 1 = 4. Range = 4 c) The data for Station C is grouped in class intervals. Step 4: Answer (c)(1) and (c)(2). (1) Why is it not possible to calculate the actual mean for Station C? It is not possible to calculate the actual mean because the exact values of the data points within each class interval are unknown. Only the frequency of values within a given range is provided. (2) Calculate an estimated mean for Station C. We use the midpoint of each class interval (x) and its frequency (f). Total frequency N = 124+326+402+678+2402+857+211 = 5000. Sum of (f × x) = (124 × 35) + (326 × 45) + (402 × 55) + (678 × 65) + (2402 × 75) + (857 × 85) + (211 × 95) = 4340 + 14670 + 22110 + 44070 + 180150 + 72845 + 20045 = 358230. Estimated Mean = ( fx)/(N) = (358230)/(5000) = 71.646 Step 5: Answer (c)(3) and (c)(4). (3) In what interval is the mode for Station C? The modal class is the class interval with the highest frequency. The highest frequency is 2402, which corresponds to the interval 70 x < 80. Modal Interval = 70 x < 80 (4) In what interval is the median for Station C? The total number of data points is N = 5000. The median position is (N)/(2) = (5000)/(2) = 2500. Cumulative frequencies: 30 x < 40: 124 40 x < 50: 124 + 326 = 450 50 x < 60: 450 + 402 = 852 60 x < 70: 852 + 678 = 1530 70 x < 80: 1530 + 2402 = 3932 The 2500th value falls within the interval 70 x < 80. Median Interval = 70 x < 80 d) The raw data for litres of diesel purchased by 30 truck drivers at Station C. Step 6: Answer (d)(1). (1) Draw a stem-and-leaf display for this data. First, sort the data: 44, 49, 50, 52, 54, 55, 56, 57, 59, 64, 64, 65, 66, 68, 68, 71, 71, 72, 74, 75, 77, 78, 80, 82, 84, 88, 88, 96, 97, 98. c|l Stem & Leaf \\ 4 & 4, 9 \\ 5 & 0, 2, 4, 5, 6, 7, 9 \\ 6 & 4, 4, 5, 6, 8, 8 \\ 7 & 1, 1, 2, 4, 5, 7, 8 \\ 8 & 0, 2, 4, 8, 8 \\ 9 & 6, 7, 8 \\ Key: 4 | 4 = 44 litres Step 7: Answer (d)(2) and (d)(3). (2) Organise the data into class intervals. Using a class width of 10: |c|c| Class Interval & Frequency \\ 40 x < 50 & 2 \\ 50 x < 60 & 7 \\ 60 x < 70 & 6 \\ 70 x < 80 & 7 \\ 80 x < 90 & 5 \\ 90 x < 100 & 3 \\ (3) Calculate the actual mean for this data. Sum of all data points = 82+64+55+50+49+44+52+59+68+74+71+78+88+98+96+77+75+54+57+56+64+66+80+84+88+72+71+65+68+97 = 2009. Total number of data points N = 30. Actual Mean = (2009)/(30) ≈ 66.97 Step 8: Answer (d)(4). (4) Calculate an estimated mean for this data. Using the class intervals from (d)(2) and their midpoints: Sum of (f × x) = (2 × 45) + (7 × 55) + (6 × 65) + (7 × 75) + (5 × 85) + (3 × 95) = 90 + 385 + 390 + 525 + 425 + 285 = 2100. Total frequency N = 30. Estimated Mean = ( fx)/(N) = (2100)/(30) = 70 What's next? 📸

Accepted Answer · 2026-08-03T21:07:43.057Z

a) To calculate the mean, mode, median, and range for the appearance ratings of the four additional customers (Customer 3, 4, 5, 6), we first list all 36 ratings: 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5. Step 1: Calculate the mean, mode, median, and range for part (a). • Mean: The sum of all ratings is 1 × 1 + 4 × 2 + 12 × 3 + 12 × 4 + 7 × 5 = 1 + 8 + 36 + 48 + 35 = 128. The total number of ratings is 36. Mean = (128)/(36) ≈ 3.56 • Mode: The values that appear most frequently are 3 and 4, both appearing 12 times. Mode = 3 and 4 • Median: With 36 data points, the median is the average of the 18th and 19th values. After sorting, both the 18th and 19th values are 4. Median = (4+4)/(2) = 4 • Range: The difference between the maximum and minimum values is 5 - 1 = 4. Range = 4 b) Calculate the mean, mode, median, and range for the appearance and service ratings for Store C. Step 2: Calculate for Appearance rating. • Mean: Mean = (Score × Frequency) Frequency = ((1 × 2820) + (2 × 3130) + (3 × 18080) + (4 × 45922) + (5 × 30048))/(100000) Mean = (2820 + 6260 + 54240 + 183688 + 150240)/(100000) = (397248)/(100000) = 3.97248 • Mode: The score with the highest frequency (45 922) is 4. Mode = 4 • Median: The total frequency is 100 000. The median position is the 50 000th value. The cumulative frequency shows that the 50 000th value falls within the score of 4 (cumulative frequency up to score 3 is 24 030, up to score 4 is 69 952). Median = 4 • Range: The difference between the maximum and minimum scores is 5 - 1 = 4. Range = 4 Step 3: Calculate for Service rating. • Mean: Mean = (Score × Frequency) Frequency = ((1 × 3910) + (2 × 10050) + (3 × 19050) + (4 × 45890) + (5 × 21100))/(100000) Mean = (3910 + 20100 + 57150 + 183560 + 105500)/(100000) = (370220)/(100000) = 3.7022 • Mode: The score with the highest frequency (45 890) is 4. Mode = 4 • Median: The total frequency is 100 000. The median position is the 50 000th value. The cumulative frequency shows that the 50 000th value falls within the score of 4 (cumulative frequency up to score 3 is 33 010, up to score 4 is 78 900). Median = 4 • Range: The difference between the maximum and minimum scores is 5 - 1 = 4. Range = 4 c) The data for Station C is grouped in class intervals. Step 4: Answer (c)(1) and (c)(2). (1) Why is it not possible to calculate the actual mean for Station C? It is not possible to calculate the actual mean because the exact values of the data points within each class interval are unknown. Only the frequency of values within a given range is provided. (2) Calculate an estimated mean for Station C. We use the midpoint of each class interval (x) and its frequency (f). Total frequency N = 124+326+402+678+2402+857+211 = 5000. Sum of (f × x) = (124 × 35) + (326 × 45) + (402 × 55) + (678 × 65) + (2402 × 75) + (857 × 85) + (211 × 95) = 4340 + 14670 + 22110 + 44070 + 180150 + 72845 + 20045 = 358230. Estimated Mean = ( fx)/(N) = (358230)/(5000) = 71.646 Step 5: Answer (c)(3) and (c)(4). (3) In what interval is the mode for Station C? The modal class is the class interval with the highest frequency. The highest frequency is 2402, which corresponds to the interval 70 x < 80. Modal Interval = 70 x < 80 (4) In what interval is the median for Station C? The total number of data points is N = 5000. The median position is (N)/(2) = (5000)/(2) = 2500. Cumulative frequencies: 30 x < 40: 124 40 x < 50: 124 + 326 = 450 50 x < 60: 450 + 402 = 852 60 x < 70: 852 + 678 = 1530 70 x < 80: 1530 + 2402 = 3932 The 2500th value falls within the interval 70 x < 80. Median Interval = 70 x < 80 d) The raw data for litres of diesel purchased by 30 truck drivers at Station C. Step 6: Answer (d)(1). (1) Draw a stem-and-leaf display for this data. First, sort the data: 44, 49, 50, 52, 54, 55, 56, 57, 59, 64, 64, 65, 66, 68, 68, 71, 71, 72, 74, 75, 77, 78, 80, 82, 84, 88, 88, 96, 97, 98. c|l Stem & Leaf \\ 4 & 4, 9 \\ 5 & 0, 2, 4, 5, 6, 7, 9 \\ 6 & 4, 4, 5, 6, 8, 8 \\ 7 & 1, 1, 2, 4, 5, 7, 8 \\ 8 & 0, 2, 4, 8, 8 \\ 9 & 6, 7, 8 \\ Key: 4 | 4 = 44 litres Step 7: Answer (d)(2) and (d)(3). (2) Organise the data into class intervals. Using a class width of 10: |c|c| Class Interval & Frequency \\ 40 x < 50 & 2 \\ 50 x < 60 & 7 \\ 60 x < 70 & 6 \\ 70 x < 80 & 7 \\ 80 x < 90 & 5 \\ 90 x < 100 & 3 \\ (3) Calculate the actual mean for this data. Sum of all data points = 82+64+55+50+49+44+52+59+68+74+71+78+88+98+96+77+75+54+57+56+64+66+80+84+88+72+71+65+68+97 = 2009. Total number of data points N = 30. Actual Mean = (2009)/(30) ≈ 66.97 Step 8: Answer (d)(4). (4) Calculate an estimated mean for this data. Using the class intervals from (d)(2) and their midpoints: Sum of (f × x) = (2 × 45) + (7 × 55) + (6 × 65) + (7 × 75) + (5 × 85) + (3 × 95) = 90 + 385 + 390 + 525 + 425 + 285 = 2100. Total frequency N = 30. Estimated Mean = ( fx)/(N) = (2100)/(30) = 70 What's next? 📸

Draw a stem-and-leaf display for this data Organise the data into class intervals. Calculate the actual mean for this data. Calculate an estimated mean for this data.

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Draw a stem-and-leaf display for this data Organise the data into class intervals. Calculate the actual mean for this data. Calculate an estimated mean for this data.

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