Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
