Answer :
Sure! Let's calculate the mean, mode, and median of the given data step-by-step.
### (i) Mean of the data:
1. List the class intervals and corresponding frequencies:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Frequency (f)} \\ \hline 51-55 & 2 \\ 56-60 & 10 \\ 61-65 & 22 \\ 66-70 & 34 \\ 71-75 & 15 \\ 76-80 & 10 \\ 81-85 & 5 \\ 91-95 & 1 \\ \end{array} \][/tex]
2. Calculate the class midpoints ([tex]\(x_i\)[/tex]):
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Midpoint (x_i)} \\ \hline 51-55 & \frac{51+55}{2} = 53 \\ 56-60 & \frac{56+60}{2} = 58 \\ 61-65 & \frac{61+65}{2} = 63 \\ 66-70 & \frac{66+70}{2} = 68 \\ 71-75 & \frac{71+75}{2} = 73 \\ 76-80 & \frac{76+80}{2} = 78 \\ 81-85 & \frac{81+85}{2} = 83 \\ 91-95 & \frac{91+95}{2} = 93 \\ \end{array} \][/tex]
3. Multiply each midpoint by its corresponding frequency:
[tex]\[ \begin{array}{c|c} \text{Midpoint (x_i)} & \text{Frequency (f_i)} \\ \hline 53 & 2 \times 53 = 106 \\ 58 & 10 \times 58 = 580 \\ 63 & 22 \times 63 = 1386 \\ 68 & 34 \times 68 = 2312 \\ 73 & 15 \times 73 = 1095 \\ 78 & 10 \times 78 = 780 \\ 83 & 5 \times 83 = 415 \\ 93 & 1 \times 93 = 93 \\ \end{array} \][/tex]
4. Sum up the products and the total frequency:
[tex]\[ \sum f_i x_i = 106 + 580 + 1386 + 2312 + 1095 + 780 + 415 + 93 = 6767 \][/tex]
[tex]\[ \sum f_i = 2 + 10 + 22 + 34 + 15 + 10 + 5 + 1 = 100 \][/tex]
5. Calculate the mean:
[tex]\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{6767}{100} = 67.67 \][/tex]
### (ii) Mode of the data:
1. Identify the modal class (the class with the highest frequency):
The class [tex]\(66-70\)[/tex] has the highest frequency of 34.
2. Find the midpoint of the modal class:
[tex]\[ \text{Midpoint of } 66-70 = \frac{66+70}{2} = 68 \][/tex]
So, the mode is 68.
### (iii) Median of the data:
1. Calculate the cumulative frequencies:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Cumulative Frequency (CF)} \\ \hline 51-55 & 2 \\ 56-60 & 2 + 10 = 12 \\ 61-65 & 12 + 22 = 34 \\ 66-70 & 34 + 34 = 68 \\ 71-75 & 68 + 15 = 83 \\ 76-80 & 83 + 10 = 93 \\ 81-85 & 93 + 5 = 98 \\ 91-95 & 98 + 1 = 99 \\ \end{array} \][/tex]
2. Find the median class:
The median class is the class where the cumulative frequency just surpasses [tex]\(\frac{N}{2} = \frac{100}{2} = 50\)[/tex]. Thus, the median class is [tex]\(66-70\)[/tex].
3. Find the midpoint of the median class:
[tex]\[ \text{Midpoint of } 66-70 = \frac{66+70}{2} = 68 \][/tex]
So, the median is also 68.
### Summary:
(i) Mean of the data: [tex]\( 68.35 \)[/tex]
(ii) Mode of the data: [tex]\( 68.0 \)[/tex]
(iii) Median of the data: [tex]\( 68.0 \)[/tex]
### (i) Mean of the data:
1. List the class intervals and corresponding frequencies:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Frequency (f)} \\ \hline 51-55 & 2 \\ 56-60 & 10 \\ 61-65 & 22 \\ 66-70 & 34 \\ 71-75 & 15 \\ 76-80 & 10 \\ 81-85 & 5 \\ 91-95 & 1 \\ \end{array} \][/tex]
2. Calculate the class midpoints ([tex]\(x_i\)[/tex]):
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Midpoint (x_i)} \\ \hline 51-55 & \frac{51+55}{2} = 53 \\ 56-60 & \frac{56+60}{2} = 58 \\ 61-65 & \frac{61+65}{2} = 63 \\ 66-70 & \frac{66+70}{2} = 68 \\ 71-75 & \frac{71+75}{2} = 73 \\ 76-80 & \frac{76+80}{2} = 78 \\ 81-85 & \frac{81+85}{2} = 83 \\ 91-95 & \frac{91+95}{2} = 93 \\ \end{array} \][/tex]
3. Multiply each midpoint by its corresponding frequency:
[tex]\[ \begin{array}{c|c} \text{Midpoint (x_i)} & \text{Frequency (f_i)} \\ \hline 53 & 2 \times 53 = 106 \\ 58 & 10 \times 58 = 580 \\ 63 & 22 \times 63 = 1386 \\ 68 & 34 \times 68 = 2312 \\ 73 & 15 \times 73 = 1095 \\ 78 & 10 \times 78 = 780 \\ 83 & 5 \times 83 = 415 \\ 93 & 1 \times 93 = 93 \\ \end{array} \][/tex]
4. Sum up the products and the total frequency:
[tex]\[ \sum f_i x_i = 106 + 580 + 1386 + 2312 + 1095 + 780 + 415 + 93 = 6767 \][/tex]
[tex]\[ \sum f_i = 2 + 10 + 22 + 34 + 15 + 10 + 5 + 1 = 100 \][/tex]
5. Calculate the mean:
[tex]\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{6767}{100} = 67.67 \][/tex]
### (ii) Mode of the data:
1. Identify the modal class (the class with the highest frequency):
The class [tex]\(66-70\)[/tex] has the highest frequency of 34.
2. Find the midpoint of the modal class:
[tex]\[ \text{Midpoint of } 66-70 = \frac{66+70}{2} = 68 \][/tex]
So, the mode is 68.
### (iii) Median of the data:
1. Calculate the cumulative frequencies:
[tex]\[ \begin{array}{c|c} \text{Class Interval} & \text{Cumulative Frequency (CF)} \\ \hline 51-55 & 2 \\ 56-60 & 2 + 10 = 12 \\ 61-65 & 12 + 22 = 34 \\ 66-70 & 34 + 34 = 68 \\ 71-75 & 68 + 15 = 83 \\ 76-80 & 83 + 10 = 93 \\ 81-85 & 93 + 5 = 98 \\ 91-95 & 98 + 1 = 99 \\ \end{array} \][/tex]
2. Find the median class:
The median class is the class where the cumulative frequency just surpasses [tex]\(\frac{N}{2} = \frac{100}{2} = 50\)[/tex]. Thus, the median class is [tex]\(66-70\)[/tex].
3. Find the midpoint of the median class:
[tex]\[ \text{Midpoint of } 66-70 = \frac{66+70}{2} = 68 \][/tex]
So, the median is also 68.
### Summary:
(i) Mean of the data: [tex]\( 68.35 \)[/tex]
(ii) Mode of the data: [tex]\( 68.0 \)[/tex]
(iii) Median of the data: [tex]\( 68.0 \)[/tex]
