Get expert insights and community-driven knowledge on IDNLearn.com. Discover reliable answers to your questions with our extensive database of expert knowledge.
Sagot :
Let's analyze the data provided and compute the required statistical measures step-by-step.
### Table Completion
First, let's complete the table by adding the midpoints (x_i) of each class interval:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{Class Interval} & \text{Frequency (f_i)} & \text{Midpoint (x_i)} \\ \hline 3<9 & 2 & \frac{3+9}{2} = 6 \\ \hline 9<15 & 5 & \frac{9+15}{2} = 12 \\ \hline 15<21 & 10 & \frac{15+21}{2} = 18 \\ \hline 21<27 & 5 & \frac{21+27}{2} = 24 \\ \hline 27<33 & 3 & \frac{27+33}{2} = 30 \\ \hline \end{array} \][/tex]
### Total Frequency (N)
Sum of the frequencies:
[tex]\[ N = 2 + 5 + 10 + 5 + 3 = 25 \][/tex]
### (i) Mean
To find the mean, we use the formula:
[tex]\[ \text{Mean} = \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
Calculating [tex]\( f_i x_i \)[/tex]:
[tex]\[ \begin{array}{|l|l|l|l|} \hline \text{Class Interval} & \text{Frequency (f_i)} & \text{Midpoint (x_i)} & \text{f_i x_i} \\ \hline 3<9 & 2 & 6 & 12 \\ \hline 9<15 & 5 & 12 & 60 \\ \hline 15<21 & 10 & 18 & 180 \\ \hline 21<27 & 5 & 24 & 120 \\ \hline 27<33 & 3 & 30 & 90 \\ \hline \end{array} \][/tex]
Sum of [tex]\( f_i x_i \)[/tex]:
[tex]\[ \sum f_i x_i = 12 + 60 + 180 + 120 + 90 = 462 \][/tex]
Mean:
[tex]\[ \bar{x} = \frac{462}{25} = 18.48 \][/tex]
### (ii) Median
To find the median, we need the class containing the median. The median is at the [tex]\( \frac{N}{2} \)[/tex]-th position, so here it is:
[tex]\[ \frac{25}{2} = 12.5 \][/tex]
Cumulative frequencies:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{Class Interval} & \text{Frequency (f_i)} & \text{Cumulative Frequency} \\ \hline 3<9 & 2 & 2 \\ \hline 9<15 & 5 & 7 \\ \hline 15<21 & 10 & 17 \\ \hline 21<27 & 5 & 22 \\ \hline 27<33 & 3 & 25 \\ \hline \end{array} \][/tex]
The median class is [tex]\( 15 < 21 \)[/tex] (since 12.5 falls into the cumulative frequency range 7 to 17).
Using the median formula for grouped data:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f_m}\right) \times h \][/tex]
Where:
- [tex]\( L \)[/tex] = lower boundary of the median class interval = 15
- [tex]\( CF \)[/tex] = cumulative frequency before the median class = 7
- [tex]\( f_m \)[/tex] = frequency of the median class = 10
- [tex]\( h \)[/tex] = class width = 6 (from 15 to 21)
[tex]\[ \text{Median} = 15 + \left(\frac{12.5 - 7}{10}\right) \times 6 = 15 + \left(\frac{5.5}{10}\right) \times 6 = 15 + 3.3 = 18.3 \][/tex]
### (iii) Mode
The mode is the value that appears most frequently. The class with the highest frequency is [tex]\( 15 < 21 \)[/tex] with a frequency of 10.
Using the mode formula for grouped data:
[tex]\[ \text{Mode} = L + \left(\frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}}\right) \times h \][/tex]
Where:
- [tex]\( L \)[/tex] = lower boundary of the modal class = 15
- [tex]\( f_m \)[/tex] = frequency of the modal class = 10
- [tex]\( f_{m-1} \)[/tex] = frequency of the class before the modal class = 5
- [tex]\( f_{m+1} \)[/tex] = frequency of the class after the modal class = 5
- [tex]\( h \)[/tex] = class width = 6
[tex]\[ \text{Mode} = 15 + \left(\frac{10 - 5}{2 \times 10 - 5 - 5}\right) \times 6 = 15 + \left(\frac{5}{10}\right) \times 6 = 15 + 3 = 18.0 \][/tex]
### (iv) Variance
To calculate the variance:
[tex]\[ \sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} \][/tex]
Calculating [tex]\( f_i (x_i - \bar{x})^2 \)[/tex]:
[tex]\[ \begin{array}{|l|l|l|l|l|} \hline \text{Class Interval} & \text{Midpoint (x_i)} & x_i - \bar{x} & (x_i - \bar{x})^2 & f_i (x_i - \bar{x})^2 \\ \hline 3<9 & 6 & 6 - 18.48 = -12.48 & 155.2704 & 2 \times 155.2704 = 310.5408 \\ \hline 9<15 & 12 & 12 - 18.48 = -6.48 & 41.9904 & 5 \times 41.9904 = 209.952 \\ \hline 15<21 & 18 & 18 - 18.48 = -0.48 & 0.2304 & 10 \times 0.2304 = 2.304 \\ \hline 21<27 & 24 & 24 - 18.48 = 5.52 & 30.4704 & 5 \times 30.4704 = 152.352 \\ \hline 27<33 & 30 & 30 - 18.48 = 11.52 & 132.7104 & 3 \times 132.7104 = 398.1312 \\ \hline \end{array} \][/tex]
Sum of [tex]\( f_i (x_i - \bar{x})^2 \)[/tex]:
[tex]\[ \sum f_i (x_i - \bar{x})^2 = 310.5408 + 209.952 + 2.304 + 152.352 + 398.1312 = 1073.28 \][/tex]
Variance:
[tex]\[ \sigma^2 = \frac{1073.28}{25} = 42.9696 \][/tex]
### (v) Standard Deviation
Standard deviation is the square root of variance:
[tex]\[ \sigma = \sqrt{42.9696} \approx 6.5551 \][/tex]
### Summary
Therefore, the statistical measures are:
1. Mean = 18.48
2. Median = 18.3
3. Mode = 18.0
4. Variance = 42.9696
5. Standard deviation = 6.5551
### Table Completion
First, let's complete the table by adding the midpoints (x_i) of each class interval:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{Class Interval} & \text{Frequency (f_i)} & \text{Midpoint (x_i)} \\ \hline 3<9 & 2 & \frac{3+9}{2} = 6 \\ \hline 9<15 & 5 & \frac{9+15}{2} = 12 \\ \hline 15<21 & 10 & \frac{15+21}{2} = 18 \\ \hline 21<27 & 5 & \frac{21+27}{2} = 24 \\ \hline 27<33 & 3 & \frac{27+33}{2} = 30 \\ \hline \end{array} \][/tex]
### Total Frequency (N)
Sum of the frequencies:
[tex]\[ N = 2 + 5 + 10 + 5 + 3 = 25 \][/tex]
### (i) Mean
To find the mean, we use the formula:
[tex]\[ \text{Mean} = \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
Calculating [tex]\( f_i x_i \)[/tex]:
[tex]\[ \begin{array}{|l|l|l|l|} \hline \text{Class Interval} & \text{Frequency (f_i)} & \text{Midpoint (x_i)} & \text{f_i x_i} \\ \hline 3<9 & 2 & 6 & 12 \\ \hline 9<15 & 5 & 12 & 60 \\ \hline 15<21 & 10 & 18 & 180 \\ \hline 21<27 & 5 & 24 & 120 \\ \hline 27<33 & 3 & 30 & 90 \\ \hline \end{array} \][/tex]
Sum of [tex]\( f_i x_i \)[/tex]:
[tex]\[ \sum f_i x_i = 12 + 60 + 180 + 120 + 90 = 462 \][/tex]
Mean:
[tex]\[ \bar{x} = \frac{462}{25} = 18.48 \][/tex]
### (ii) Median
To find the median, we need the class containing the median. The median is at the [tex]\( \frac{N}{2} \)[/tex]-th position, so here it is:
[tex]\[ \frac{25}{2} = 12.5 \][/tex]
Cumulative frequencies:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{Class Interval} & \text{Frequency (f_i)} & \text{Cumulative Frequency} \\ \hline 3<9 & 2 & 2 \\ \hline 9<15 & 5 & 7 \\ \hline 15<21 & 10 & 17 \\ \hline 21<27 & 5 & 22 \\ \hline 27<33 & 3 & 25 \\ \hline \end{array} \][/tex]
The median class is [tex]\( 15 < 21 \)[/tex] (since 12.5 falls into the cumulative frequency range 7 to 17).
Using the median formula for grouped data:
[tex]\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f_m}\right) \times h \][/tex]
Where:
- [tex]\( L \)[/tex] = lower boundary of the median class interval = 15
- [tex]\( CF \)[/tex] = cumulative frequency before the median class = 7
- [tex]\( f_m \)[/tex] = frequency of the median class = 10
- [tex]\( h \)[/tex] = class width = 6 (from 15 to 21)
[tex]\[ \text{Median} = 15 + \left(\frac{12.5 - 7}{10}\right) \times 6 = 15 + \left(\frac{5.5}{10}\right) \times 6 = 15 + 3.3 = 18.3 \][/tex]
### (iii) Mode
The mode is the value that appears most frequently. The class with the highest frequency is [tex]\( 15 < 21 \)[/tex] with a frequency of 10.
Using the mode formula for grouped data:
[tex]\[ \text{Mode} = L + \left(\frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}}\right) \times h \][/tex]
Where:
- [tex]\( L \)[/tex] = lower boundary of the modal class = 15
- [tex]\( f_m \)[/tex] = frequency of the modal class = 10
- [tex]\( f_{m-1} \)[/tex] = frequency of the class before the modal class = 5
- [tex]\( f_{m+1} \)[/tex] = frequency of the class after the modal class = 5
- [tex]\( h \)[/tex] = class width = 6
[tex]\[ \text{Mode} = 15 + \left(\frac{10 - 5}{2 \times 10 - 5 - 5}\right) \times 6 = 15 + \left(\frac{5}{10}\right) \times 6 = 15 + 3 = 18.0 \][/tex]
### (iv) Variance
To calculate the variance:
[tex]\[ \sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} \][/tex]
Calculating [tex]\( f_i (x_i - \bar{x})^2 \)[/tex]:
[tex]\[ \begin{array}{|l|l|l|l|l|} \hline \text{Class Interval} & \text{Midpoint (x_i)} & x_i - \bar{x} & (x_i - \bar{x})^2 & f_i (x_i - \bar{x})^2 \\ \hline 3<9 & 6 & 6 - 18.48 = -12.48 & 155.2704 & 2 \times 155.2704 = 310.5408 \\ \hline 9<15 & 12 & 12 - 18.48 = -6.48 & 41.9904 & 5 \times 41.9904 = 209.952 \\ \hline 15<21 & 18 & 18 - 18.48 = -0.48 & 0.2304 & 10 \times 0.2304 = 2.304 \\ \hline 21<27 & 24 & 24 - 18.48 = 5.52 & 30.4704 & 5 \times 30.4704 = 152.352 \\ \hline 27<33 & 30 & 30 - 18.48 = 11.52 & 132.7104 & 3 \times 132.7104 = 398.1312 \\ \hline \end{array} \][/tex]
Sum of [tex]\( f_i (x_i - \bar{x})^2 \)[/tex]:
[tex]\[ \sum f_i (x_i - \bar{x})^2 = 310.5408 + 209.952 + 2.304 + 152.352 + 398.1312 = 1073.28 \][/tex]
Variance:
[tex]\[ \sigma^2 = \frac{1073.28}{25} = 42.9696 \][/tex]
### (v) Standard Deviation
Standard deviation is the square root of variance:
[tex]\[ \sigma = \sqrt{42.9696} \approx 6.5551 \][/tex]
### Summary
Therefore, the statistical measures are:
1. Mean = 18.48
2. Median = 18.3
3. Mode = 18.0
4. Variance = 42.9696
5. Standard deviation = 6.5551
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
