Answer :
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
