Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Certainly! Let's go step by step to find the mean, median, and mode of the given data.
### i) Mean
To find the mean, we use the formula:
[tex]\[ \text{Mean} (\bar{X}) = \frac{\sum(f * x)}{\sum f} \][/tex]
where \( f \) is the frequency and \( x \) is the midpoint of each class interval.
1. First, find the midpoints of each class:
- For \( 30-39 \), midpoint = \( \frac{30 + 39}{2} = 34.5 \)
- For \( 40-49 \), midpoint = \( \frac{40 + 49}{2} = 44.5 \)
- For \( 50-59 \), midpoint = \( \frac{50 + 59}{2} = 54.5 \)
- For \( 60-69 \), midpoint = \( \frac{60 + 69}{2} = 64.5 \)
- For \( 70-79 \), midpoint = \( \frac{70 + 79}{2} = 74.5 \)
- For \( 80-89 \), midpoint = \( \frac{80 + 89}{2} = 84.5 \)
2. Multiply each midpoint by the corresponding frequency and add the products:
[tex]\[ \sum(f x) = (34.5 11) + (44.5 25) + (54.5 45) + (64.5 15) + (74.5 10) + (84.5 * 14) \][/tex]
3. Add the frequencies:
[tex]\[ \sum f = 11 + 25 + 45 + 15 + 10 + 14 = 120 \][/tex]
4. Calculate the mean:
[tex]\[ \bar{X} = \frac{580.5 + 1112.5 + 2452.5 + 967.5 + 745 + 1183} {120} = \frac{7041}{120} = 52.5 \][/tex]
[tex]\[ \bar{X} = 52.5 \][/tex]
### ii) Median
To find the median, we identify the median class and use interpolation.
1. The total number of students is \( 120 \).
2. Half of the total number of students, \( n/2 \), is \( 60 \).
3. Cumulative frequencies:
[tex]\[ 11, 11 + 25 = 36 \quad 36 + 45 = 81 \quad 81 + 15 = 96 \quad 96 + 10 = 106 \quad 106 + 14 = 120 \][/tex]
4. The median class is the class where the cumulative frequency first exceeds \( 60 \), which is \( 50-59 \). The class boundaries are \( 50 \) and \( 60 \).
5. Using the formula for median:
[tex]\[ \text{Median} = L + \left(\frac{(n/2 - CF)}{f}\right) * h \][/tex]
where:
- \( L \) is the lower class boundary of the median class (\(50\)),
- \( CF \) is the cumulative frequency of the class before the median class (\(36\)),
- \( f \) is the frequency of the median class (\(45\)),
- \( h \) is the class interval (\(10\)).
6. Substitute the values into the formula:
[tex]\[ \text{Median } = 50 + \left(\frac{(60 - 36)}{45}\right) * 10 \][/tex]
[tex]\[ = 50 + \left(\frac{24}{45}\right) * 10 \][/tex]
[tex]\[ = 50 + 5.3333 \][/tex]
[tex]\[ = 55.3333 \][/tex]
### iii) Mode
To find the mode, we identify the modal class, which is the class with the highest frequency.
1. The highest frequency is \( 45 \), which is in the \( 50-59 \) class interval.
2. Using the mode formula:
[tex]\[ \text{Mode} = L + \left(\frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)}\right) * h \][/tex]
where:
- \( L \) is the lower boundary of the modal class (\(50\)),
- \( f_m \) is the frequency of the modal class (\(45\)),
- \( f_1 \) is the frequency of the class before the modal class (\(25\)),
- \( f_2 \) is the frequency of the class after the modal class (\(15\)),
- \( h \) is the class interval (\(10\)).
3. Substitute the values into the formula:
[tex]\[ \text{Mode} = 50 + \left(\frac{45 - 25}{(45 - 25) + (45 - 15)}\right) * 10 \][/tex]
[tex]\[ = 50 + \left(\frac{20}{20 + 30}\right) * 10 \][/tex]
[tex]\[ = 50 + \left(\frac{20}{50}\right) * 10 \][/tex]
[tex]\[ = 50 + 4 \][/tex]
[tex]\[ = 54 \][/tex]
### Conclusion
Therefore, the mean, median, and mode of the distribution are:
- Mean: \( 52.5 \)
- Median: \( 55.3333 \)
- Mode: [tex]\( 54 \)[/tex]
### i) Mean
To find the mean, we use the formula:
[tex]\[ \text{Mean} (\bar{X}) = \frac{\sum(f * x)}{\sum f} \][/tex]
where \( f \) is the frequency and \( x \) is the midpoint of each class interval.
1. First, find the midpoints of each class:
- For \( 30-39 \), midpoint = \( \frac{30 + 39}{2} = 34.5 \)
- For \( 40-49 \), midpoint = \( \frac{40 + 49}{2} = 44.5 \)
- For \( 50-59 \), midpoint = \( \frac{50 + 59}{2} = 54.5 \)
- For \( 60-69 \), midpoint = \( \frac{60 + 69}{2} = 64.5 \)
- For \( 70-79 \), midpoint = \( \frac{70 + 79}{2} = 74.5 \)
- For \( 80-89 \), midpoint = \( \frac{80 + 89}{2} = 84.5 \)
2. Multiply each midpoint by the corresponding frequency and add the products:
[tex]\[ \sum(f x) = (34.5 11) + (44.5 25) + (54.5 45) + (64.5 15) + (74.5 10) + (84.5 * 14) \][/tex]
3. Add the frequencies:
[tex]\[ \sum f = 11 + 25 + 45 + 15 + 10 + 14 = 120 \][/tex]
4. Calculate the mean:
[tex]\[ \bar{X} = \frac{580.5 + 1112.5 + 2452.5 + 967.5 + 745 + 1183} {120} = \frac{7041}{120} = 52.5 \][/tex]
[tex]\[ \bar{X} = 52.5 \][/tex]
### ii) Median
To find the median, we identify the median class and use interpolation.
1. The total number of students is \( 120 \).
2. Half of the total number of students, \( n/2 \), is \( 60 \).
3. Cumulative frequencies:
[tex]\[ 11, 11 + 25 = 36 \quad 36 + 45 = 81 \quad 81 + 15 = 96 \quad 96 + 10 = 106 \quad 106 + 14 = 120 \][/tex]
4. The median class is the class where the cumulative frequency first exceeds \( 60 \), which is \( 50-59 \). The class boundaries are \( 50 \) and \( 60 \).
5. Using the formula for median:
[tex]\[ \text{Median} = L + \left(\frac{(n/2 - CF)}{f}\right) * h \][/tex]
where:
- \( L \) is the lower class boundary of the median class (\(50\)),
- \( CF \) is the cumulative frequency of the class before the median class (\(36\)),
- \( f \) is the frequency of the median class (\(45\)),
- \( h \) is the class interval (\(10\)).
6. Substitute the values into the formula:
[tex]\[ \text{Median } = 50 + \left(\frac{(60 - 36)}{45}\right) * 10 \][/tex]
[tex]\[ = 50 + \left(\frac{24}{45}\right) * 10 \][/tex]
[tex]\[ = 50 + 5.3333 \][/tex]
[tex]\[ = 55.3333 \][/tex]
### iii) Mode
To find the mode, we identify the modal class, which is the class with the highest frequency.
1. The highest frequency is \( 45 \), which is in the \( 50-59 \) class interval.
2. Using the mode formula:
[tex]\[ \text{Mode} = L + \left(\frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)}\right) * h \][/tex]
where:
- \( L \) is the lower boundary of the modal class (\(50\)),
- \( f_m \) is the frequency of the modal class (\(45\)),
- \( f_1 \) is the frequency of the class before the modal class (\(25\)),
- \( f_2 \) is the frequency of the class after the modal class (\(15\)),
- \( h \) is the class interval (\(10\)).
3. Substitute the values into the formula:
[tex]\[ \text{Mode} = 50 + \left(\frac{45 - 25}{(45 - 25) + (45 - 15)}\right) * 10 \][/tex]
[tex]\[ = 50 + \left(\frac{20}{20 + 30}\right) * 10 \][/tex]
[tex]\[ = 50 + \left(\frac{20}{50}\right) * 10 \][/tex]
[tex]\[ = 50 + 4 \][/tex]
[tex]\[ = 54 \][/tex]
### Conclusion
Therefore, the mean, median, and mode of the distribution are:
- Mean: \( 52.5 \)
- Median: \( 55.3333 \)
- Mode: [tex]\( 54 \)[/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
