Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
