Answer :
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]
