Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Let's go through the steps to estimate the mean, median, and mode for the given data.
### a) Estimate the Mean
1. Identify the class intervals and frequencies:
- Class Intervals: [tex]\[(65, 99.99), (100, 134.99), (135, 169.99), (170, 204.99), (205, 239.99)\][/tex]
- Frequencies: [tex]\[3, 2, 5, 5, 8\][/tex]
2. Calculate the midpoints for each interval:
[tex]\[ \text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \][/tex]
- For (65, 99.99): [tex]\(\frac{65 + 99.99}{2} = 82.495\)[/tex]
- For (100, 134.99): [tex]\(\frac{100 + 134.99}{2} = 117.495\)[/tex]
- For (135, 169.99): [tex]\(\frac{135 + 169.99}{2} = 152.495\)[/tex]
- For (170, 204.99): [tex]\(\frac{170 + 204.99}{2} = 187.495\)[/tex]
- For (205, 239.99): [tex]\(\frac{205 + 239.99}{2} = 222.495\)[/tex]
3. Calculate the mean using the formula:
[tex]\[ \text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\sum \text{Frequencies}} \][/tex]
- Sum of (Midpoint × Frequency):
[tex]\[ (82.495 \times 3) + (117.495 \times 2) + (152.495 \times 5) + (187.495 \times 5) + (222.495 \times 8) \][/tex]
- Compute individually:
[tex]\[ 247.485 + 234.99 + 762.475 + 937.475 + 1779.96 = 3962.385 \][/tex]
- Total frequency: [tex]\(3 + 2 + 5 + 5 + 8 = 23\)[/tex]
- Mean:
[tex]\[ \text{Mean} = \frac{3962.385}{23} \approx 172.278 \][/tex]
### b) Estimate the Median
1. Calculate the cumulative frequency:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Cumulative Frequency} \\ \hline (65, 99.99) & 3 \\ (100, 134.99) & 3 + 2 = 5 \\ (135, 169.99) & 5 + 5 = 10 \\ (170, 204.99) & 10 + 5 = 15 \\ (205, 239.99) & 15 + 8 = 23 \\ \end{array} \][/tex]
2. Locate the median class (where cumulative frequency ≥ total frequency / 2):
- Total frequency [tex]\(\sum \text{Frequencies} = 23\)[/tex]
- Median location [tex]\( = \frac{23}{2} = 11.5\)[/tex] (Look for cumulative frequency ≥ 11.5)
- The median class is (170, 204.99).
3. Calculate the lower boundary, cumulative frequency before the median class, and interval width:
- Lower Boundary (L) = 170
- Cumulative frequency before the median class (F) = 10
- Interval frequency (f) = 5
- Interval width (h) = 204.99 - 170 = 34.99
4. Calculate the median using the formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \][/tex]
- Median:
[tex]\[ \text{Median} = 170 + \left( \frac{11.5 - 10}{5} \right) \times 34.99 = 170 + \left(\frac{1.5}{5}\right) \times 34.99 = 170 + 10.497 = 180.497 \][/tex]
### c) Estimate the Mode
1. Locate the modal class (class with the highest frequency):
- The highest frequency is 8. The modal class is (205, 239.99).
2. Calculate the lower boundary (L) and the frequency differences:
- Lower Boundary (L) = 205
- Frequency of modal class ([tex]\(f_m\)[/tex]) = 8
- Frequency of class before modal class ([tex]\(f_1\)[/tex]) = 5
- Frequency of class after modal class ([tex]\(f_2\)[/tex]) = 0
- Interval width (h) = 239.99 - 205 = 34.99
3. Calculate the mode using the formula:
[tex]\[ \text{Mode} = L + \left( \frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)} \right) \times h \][/tex]
- Mode:
[tex]\[ \text{Mode} = 205 + \left( \frac{8 - 5}{(8 - 5) + (8 - 0)} \right) \times 34.99 = 205 + \left( \frac{3}{3 + 8} \right) \times 34.99 = 205 + \left( \frac{3}{11} \right) \times 34.99 = 205 + 9.543 = 214.543 \][/tex]
### Summary
- Mean: [tex]\(172.278\)[/tex]
- Median: [tex]\(180.497\)[/tex]
- Mode: [tex]\(214.543\)[/tex]
### a) Estimate the Mean
1. Identify the class intervals and frequencies:
- Class Intervals: [tex]\[(65, 99.99), (100, 134.99), (135, 169.99), (170, 204.99), (205, 239.99)\][/tex]
- Frequencies: [tex]\[3, 2, 5, 5, 8\][/tex]
2. Calculate the midpoints for each interval:
[tex]\[ \text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \][/tex]
- For (65, 99.99): [tex]\(\frac{65 + 99.99}{2} = 82.495\)[/tex]
- For (100, 134.99): [tex]\(\frac{100 + 134.99}{2} = 117.495\)[/tex]
- For (135, 169.99): [tex]\(\frac{135 + 169.99}{2} = 152.495\)[/tex]
- For (170, 204.99): [tex]\(\frac{170 + 204.99}{2} = 187.495\)[/tex]
- For (205, 239.99): [tex]\(\frac{205 + 239.99}{2} = 222.495\)[/tex]
3. Calculate the mean using the formula:
[tex]\[ \text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\sum \text{Frequencies}} \][/tex]
- Sum of (Midpoint × Frequency):
[tex]\[ (82.495 \times 3) + (117.495 \times 2) + (152.495 \times 5) + (187.495 \times 5) + (222.495 \times 8) \][/tex]
- Compute individually:
[tex]\[ 247.485 + 234.99 + 762.475 + 937.475 + 1779.96 = 3962.385 \][/tex]
- Total frequency: [tex]\(3 + 2 + 5 + 5 + 8 = 23\)[/tex]
- Mean:
[tex]\[ \text{Mean} = \frac{3962.385}{23} \approx 172.278 \][/tex]
### b) Estimate the Median
1. Calculate the cumulative frequency:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Cumulative Frequency} \\ \hline (65, 99.99) & 3 \\ (100, 134.99) & 3 + 2 = 5 \\ (135, 169.99) & 5 + 5 = 10 \\ (170, 204.99) & 10 + 5 = 15 \\ (205, 239.99) & 15 + 8 = 23 \\ \end{array} \][/tex]
2. Locate the median class (where cumulative frequency ≥ total frequency / 2):
- Total frequency [tex]\(\sum \text{Frequencies} = 23\)[/tex]
- Median location [tex]\( = \frac{23}{2} = 11.5\)[/tex] (Look for cumulative frequency ≥ 11.5)
- The median class is (170, 204.99).
3. Calculate the lower boundary, cumulative frequency before the median class, and interval width:
- Lower Boundary (L) = 170
- Cumulative frequency before the median class (F) = 10
- Interval frequency (f) = 5
- Interval width (h) = 204.99 - 170 = 34.99
4. Calculate the median using the formula:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \][/tex]
- Median:
[tex]\[ \text{Median} = 170 + \left( \frac{11.5 - 10}{5} \right) \times 34.99 = 170 + \left(\frac{1.5}{5}\right) \times 34.99 = 170 + 10.497 = 180.497 \][/tex]
### c) Estimate the Mode
1. Locate the modal class (class with the highest frequency):
- The highest frequency is 8. The modal class is (205, 239.99).
2. Calculate the lower boundary (L) and the frequency differences:
- Lower Boundary (L) = 205
- Frequency of modal class ([tex]\(f_m\)[/tex]) = 8
- Frequency of class before modal class ([tex]\(f_1\)[/tex]) = 5
- Frequency of class after modal class ([tex]\(f_2\)[/tex]) = 0
- Interval width (h) = 239.99 - 205 = 34.99
3. Calculate the mode using the formula:
[tex]\[ \text{Mode} = L + \left( \frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)} \right) \times h \][/tex]
- Mode:
[tex]\[ \text{Mode} = 205 + \left( \frac{8 - 5}{(8 - 5) + (8 - 0)} \right) \times 34.99 = 205 + \left( \frac{3}{3 + 8} \right) \times 34.99 = 205 + \left( \frac{3}{11} \right) \times 34.99 = 205 + 9.543 = 214.543 \][/tex]
### Summary
- Mean: [tex]\(172.278\)[/tex]
- Median: [tex]\(180.497\)[/tex]
- Mode: [tex]\(214.543\)[/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
