Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To find the missing frequencies [tex]\( f_1 \)[/tex] and [tex]\( f_2 \)[/tex], follow these steps:
1. Calculate the class midpoints (also called class marks) for each interval:
The midpoint (class mark) is calculated as the average of the lower and upper bounds of each class interval.
[tex]\[ \text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \][/tex]
Given class intervals: [tex]\( 0-49, 50-99, 100-149, 150-199, 200-249, 250-299, 300-349 \)[/tex].
[tex]\[ \begin{aligned} &0-49 \rightarrow 24.5, \\ &50-99 \rightarrow 74.5, \\ &100-149 \rightarrow 124.5, \\ &150-199 \rightarrow 174.5, \\ &200-249 \rightarrow 224.5, \\ &250-299 \rightarrow 274.5, \\ &300-349 \rightarrow 324.5 \\ \end{aligned} \][/tex]
2. Set up the equation for the mean:
The mean of a frequency distribution is given by:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the mean,
- [tex]\( f_i \)[/tex] is the frequency of the [tex]\( i \)[/tex]-th class,
- [tex]\( x_i \)[/tex] is the midpoint of the [tex]\( i \)[/tex]-th class.
Given [tex]\( \bar{x} = 148 \)[/tex], [tex]\( \sum f_i = 100 \)[/tex], and the midpoints [tex]\( x_i \)[/tex] calculated above:
[tex]\[ \frac{24.5 \times 10 + 74.5 \times 15 + 124.5 \times f_1 + 174.5 \times 20 + 224.5 \times 15 + 274.5 \times f_2 + 324.5 \times 2}{100} = 148 \][/tex]
3. Form the equation for the sum of the frequencies:
[tex]\[ 10 + 15 + f_1 + 20 + 15 + f_2 + 2 = 100 \][/tex]
Simplifying this:
[tex]\[ 62 + f_1 + f_2 = 100 \Rightarrow f_1 + f_2 = 38 \][/tex]
4. Substitute [tex]\( f_1 + f_2 = 38 \)[/tex] into the mean equation:
[tex]\[ 24.5 \times 10 + 74.5 \times 15 + 124.5 \times f_1 + 174.5 \times 20 + 224.5 \times 15 + 274.5 \times f_2 + 324.5 \times 2 = 14800 \][/tex]
Calculate the known sums first:
[tex]\[ \begin{aligned} &24.5 \times 10 = 245,\\ &74.5 \times 15 = 1117.5,\\ &174.5 \times 20 = 3490,\\ &224.5 \times 15 = 3367.5,\\ &324.5 \times 2 = 649 \end{aligned} \][/tex]
Summing these up:
[tex]\[ 245 + 1117.5 + 3490 + 3367.5 + 649 = 8869 \][/tex]
So the equation becomes:
[tex]\[ 8869 + 124.5 f_1 + 274.5 f_2 = 14800 \][/tex]
Simplify and isolate the terms involving [tex]\( f_1 \)[/tex] and [tex]\( f_2 \)[/tex]:
[tex]\[ 124.5 f_1 + 274.5 f_2 = 5931 \][/tex]
5. Solve the system of linear equations:
We have two equations:
[tex]\[ \begin{cases} f_1 + f_2 = 38 \\ 124.5 f_1 + 274.5 f_2 = 5931 \end{cases} \][/tex]
Substitute [tex]\( f_2 = 38 - f_1 \)[/tex] into the second equation:
[tex]\[ 124.5 f_1 + 274.5 (38 - f_1) = 5931 \][/tex]
Simplify and solve for [tex]\( f_1 \)[/tex]:
[tex]\[ 124.5 f_1 + 10431 - 274.5 f_1 = 5931 \][/tex]
[tex]\[ -150 f_1 = -4500 \][/tex]
[tex]\[ f_1 = 30 \][/tex]
Substitute [tex]\( f_1 = 30 \)[/tex] back into [tex]\( f_2 = 38 - f_1 \)[/tex]:
[tex]\[ f_2 = 38 - 30 = 8 \][/tex]
Therefore, the missing frequencies are:
[tex]\[ f_1 = 30 \quad \text{and} \quad f_2 = 8 \][/tex]
1. Calculate the class midpoints (also called class marks) for each interval:
The midpoint (class mark) is calculated as the average of the lower and upper bounds of each class interval.
[tex]\[ \text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \][/tex]
Given class intervals: [tex]\( 0-49, 50-99, 100-149, 150-199, 200-249, 250-299, 300-349 \)[/tex].
[tex]\[ \begin{aligned} &0-49 \rightarrow 24.5, \\ &50-99 \rightarrow 74.5, \\ &100-149 \rightarrow 124.5, \\ &150-199 \rightarrow 174.5, \\ &200-249 \rightarrow 224.5, \\ &250-299 \rightarrow 274.5, \\ &300-349 \rightarrow 324.5 \\ \end{aligned} \][/tex]
2. Set up the equation for the mean:
The mean of a frequency distribution is given by:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the mean,
- [tex]\( f_i \)[/tex] is the frequency of the [tex]\( i \)[/tex]-th class,
- [tex]\( x_i \)[/tex] is the midpoint of the [tex]\( i \)[/tex]-th class.
Given [tex]\( \bar{x} = 148 \)[/tex], [tex]\( \sum f_i = 100 \)[/tex], and the midpoints [tex]\( x_i \)[/tex] calculated above:
[tex]\[ \frac{24.5 \times 10 + 74.5 \times 15 + 124.5 \times f_1 + 174.5 \times 20 + 224.5 \times 15 + 274.5 \times f_2 + 324.5 \times 2}{100} = 148 \][/tex]
3. Form the equation for the sum of the frequencies:
[tex]\[ 10 + 15 + f_1 + 20 + 15 + f_2 + 2 = 100 \][/tex]
Simplifying this:
[tex]\[ 62 + f_1 + f_2 = 100 \Rightarrow f_1 + f_2 = 38 \][/tex]
4. Substitute [tex]\( f_1 + f_2 = 38 \)[/tex] into the mean equation:
[tex]\[ 24.5 \times 10 + 74.5 \times 15 + 124.5 \times f_1 + 174.5 \times 20 + 224.5 \times 15 + 274.5 \times f_2 + 324.5 \times 2 = 14800 \][/tex]
Calculate the known sums first:
[tex]\[ \begin{aligned} &24.5 \times 10 = 245,\\ &74.5 \times 15 = 1117.5,\\ &174.5 \times 20 = 3490,\\ &224.5 \times 15 = 3367.5,\\ &324.5 \times 2 = 649 \end{aligned} \][/tex]
Summing these up:
[tex]\[ 245 + 1117.5 + 3490 + 3367.5 + 649 = 8869 \][/tex]
So the equation becomes:
[tex]\[ 8869 + 124.5 f_1 + 274.5 f_2 = 14800 \][/tex]
Simplify and isolate the terms involving [tex]\( f_1 \)[/tex] and [tex]\( f_2 \)[/tex]:
[tex]\[ 124.5 f_1 + 274.5 f_2 = 5931 \][/tex]
5. Solve the system of linear equations:
We have two equations:
[tex]\[ \begin{cases} f_1 + f_2 = 38 \\ 124.5 f_1 + 274.5 f_2 = 5931 \end{cases} \][/tex]
Substitute [tex]\( f_2 = 38 - f_1 \)[/tex] into the second equation:
[tex]\[ 124.5 f_1 + 274.5 (38 - f_1) = 5931 \][/tex]
Simplify and solve for [tex]\( f_1 \)[/tex]:
[tex]\[ 124.5 f_1 + 10431 - 274.5 f_1 = 5931 \][/tex]
[tex]\[ -150 f_1 = -4500 \][/tex]
[tex]\[ f_1 = 30 \][/tex]
Substitute [tex]\( f_1 = 30 \)[/tex] back into [tex]\( f_2 = 38 - f_1 \)[/tex]:
[tex]\[ f_2 = 38 - 30 = 8 \][/tex]
Therefore, the missing frequencies are:
[tex]\[ f_1 = 30 \quad \text{and} \quad f_2 = 8 \][/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
