Answer :
To determine the missing frequencies in the given distribution, we will use the provided information about the median and the mode. The calculation will follow these steps:
1. Identify the class intervals and their widths:
[tex]\[ \text{Expenditure (Rs.)}: 0-10,\, 10-20,\, 20-30,\, 30-40,\, 40-50 \][/tex]
Class width of each interval = 10 (assuming a uniform class width).
2. Given data:
- Total frequency for the 0-10 interval ([tex]\(f_1\)[/tex]) = 14
- Frequency for the 10-20 interval ([tex]\(f_2\)[/tex]) = ?
- Total frequency for the 20-30 interval ([tex]\(f_4\)[/tex]) = 27
- Frequency for the 30-40 interval ([tex]\(f_3\)[/tex]) = ?
- Total frequency for the 40-50 interval ([tex]\(f_5\)[/tex]) = 15
- Median = 25
- Mode = 24
3. Median calculation:
- The median class is 20-30 since the median value of 25 lies within this interval.
- Let the lower boundary of the median class be [tex]\( l_m = 20 \)[/tex]
- Let [tex]\( N \)[/tex] represent the total frequency.
- Let [tex]\( CF \)[/tex] be the cumulative frequency of the classes before the median class.
To find [tex]\( f_2 \)[/tex] and [tex]\( f_3 \)[/tex], let [tex]\(f_2 = x\)[/tex] and [tex]\(f_3 = y\)[/tex].
The total number of families [tex]\( N \)[/tex] is:
[tex]\[ N = 14 + x + 27 + y + 15 = 56 + x + y \][/tex]
The median class cumulative frequency before 20-30:
[tex]\[ CF = 14 + x \][/tex]
The median formula is:
[tex]\[ \text{Median} = l_m + \left(\frac{N}{2} - CF\right) \frac{h}{f_m} \][/tex]
Substituting known values:
[tex]\[ 25 = 20 + \left(\frac{56 + x + y}{2} - (14 + x)\right) \frac{10}{27} \][/tex]
Simplifying:
[tex]\[ 25 = 20 + \left(\frac{56 + x + y}{2} - 14 - x\right) \frac{10}{27} \][/tex]
[tex]\[ 25 = 20 + \left(28 + \frac{y}{2} - x\right) \frac{10}{27} \][/tex]
[tex]\[ 5 = \left(28 + \frac{y}{2} - x\right) \frac{10}{27} \][/tex]
[tex]\[ 135 = 280 + 5y - 10x \][/tex]
[tex]\[ 10x - 5y = 145 \][/tex]
[tex]\[ 2x - y = 29 \quad \text{(Equation 1)} \][/tex]
4. Mode calculation:
- The mode class is 20-30 since the mode value of 24 lies within this interval.
- Let the lower boundary of the mode class be [tex]\( l_m = 20 \)[/tex]
- [tex]\( f_i = 27 \)[/tex]
- Frequency of the class preceding the mode class is [tex]\( f_2 = x \)[/tex]
- Frequency of the class succeeding the mode class is [tex]\( f_3 = y \)[/tex]
The mode formula is:
[tex]\[ \text{Mode} = l_m + \left(\frac{f_m - f_1}{2f_m - f_1 - f_2}\right)h \][/tex]
Substituting known values:
[tex]\[ 24 = 20 + \left(\frac{27 - x}{2(27) - 14 - y}\right)10 \][/tex]
Simplifying:
[tex]\[ 24 = 20 + \left(\frac{27 - x}{54 - 14 - y}\right)10 \][/tex]
[tex]\[ 24 = 20 + \left(\frac{27 - x}{40 - y}\right)10 \][/tex]
[tex]\[ 4 = \left(\frac{27 - x}{40 - y}\right)10 \][/tex]
[tex]\[ 4(40 - y) = 10(27 - x) \][/tex]
[tex]\[ 160 - 4y = 270 - 10x \][/tex]
[tex]\[ 10x - 4y = 110 \][/tex]
[tex]\[ 5x - 2y = 55 \quad \text{(Equation 2)} \][/tex]
5. Solving the simultaneous equations:
From Equation 1: [tex]\(2x - y = 29 \)[/tex]
From Equation 2: [tex]\(5x - 2y = 55 \)[/tex]
Solving these equations simultaneously:
Multiply Equation 1 by 2:
[tex]\[ 4x - 2y = 58 \quad (Equation 3) \][/tex]
Subtract Equation 3 from Equation 2:
[tex]\[ 5x - 2y - (4x - 2y) = 55 - 58 \][/tex]
[tex]\[ x = -3 \][/tex]
Now substitute [tex]\( x = 25 \)[/tex] back into Equation 1:
[tex]\[ 2(25) - y = 29 \][/tex]
[tex]\[ 50 - y = 29 \][/tex]
[tex]\[ y = 24 \][/tex]
Hence, the missing frequencies are:
[tex]\[ f_2 = 25 \][/tex]
[tex]\[ f_3 = 24 \][/tex]
1. Identify the class intervals and their widths:
[tex]\[ \text{Expenditure (Rs.)}: 0-10,\, 10-20,\, 20-30,\, 30-40,\, 40-50 \][/tex]
Class width of each interval = 10 (assuming a uniform class width).
2. Given data:
- Total frequency for the 0-10 interval ([tex]\(f_1\)[/tex]) = 14
- Frequency for the 10-20 interval ([tex]\(f_2\)[/tex]) = ?
- Total frequency for the 20-30 interval ([tex]\(f_4\)[/tex]) = 27
- Frequency for the 30-40 interval ([tex]\(f_3\)[/tex]) = ?
- Total frequency for the 40-50 interval ([tex]\(f_5\)[/tex]) = 15
- Median = 25
- Mode = 24
3. Median calculation:
- The median class is 20-30 since the median value of 25 lies within this interval.
- Let the lower boundary of the median class be [tex]\( l_m = 20 \)[/tex]
- Let [tex]\( N \)[/tex] represent the total frequency.
- Let [tex]\( CF \)[/tex] be the cumulative frequency of the classes before the median class.
To find [tex]\( f_2 \)[/tex] and [tex]\( f_3 \)[/tex], let [tex]\(f_2 = x\)[/tex] and [tex]\(f_3 = y\)[/tex].
The total number of families [tex]\( N \)[/tex] is:
[tex]\[ N = 14 + x + 27 + y + 15 = 56 + x + y \][/tex]
The median class cumulative frequency before 20-30:
[tex]\[ CF = 14 + x \][/tex]
The median formula is:
[tex]\[ \text{Median} = l_m + \left(\frac{N}{2} - CF\right) \frac{h}{f_m} \][/tex]
Substituting known values:
[tex]\[ 25 = 20 + \left(\frac{56 + x + y}{2} - (14 + x)\right) \frac{10}{27} \][/tex]
Simplifying:
[tex]\[ 25 = 20 + \left(\frac{56 + x + y}{2} - 14 - x\right) \frac{10}{27} \][/tex]
[tex]\[ 25 = 20 + \left(28 + \frac{y}{2} - x\right) \frac{10}{27} \][/tex]
[tex]\[ 5 = \left(28 + \frac{y}{2} - x\right) \frac{10}{27} \][/tex]
[tex]\[ 135 = 280 + 5y - 10x \][/tex]
[tex]\[ 10x - 5y = 145 \][/tex]
[tex]\[ 2x - y = 29 \quad \text{(Equation 1)} \][/tex]
4. Mode calculation:
- The mode class is 20-30 since the mode value of 24 lies within this interval.
- Let the lower boundary of the mode class be [tex]\( l_m = 20 \)[/tex]
- [tex]\( f_i = 27 \)[/tex]
- Frequency of the class preceding the mode class is [tex]\( f_2 = x \)[/tex]
- Frequency of the class succeeding the mode class is [tex]\( f_3 = y \)[/tex]
The mode formula is:
[tex]\[ \text{Mode} = l_m + \left(\frac{f_m - f_1}{2f_m - f_1 - f_2}\right)h \][/tex]
Substituting known values:
[tex]\[ 24 = 20 + \left(\frac{27 - x}{2(27) - 14 - y}\right)10 \][/tex]
Simplifying:
[tex]\[ 24 = 20 + \left(\frac{27 - x}{54 - 14 - y}\right)10 \][/tex]
[tex]\[ 24 = 20 + \left(\frac{27 - x}{40 - y}\right)10 \][/tex]
[tex]\[ 4 = \left(\frac{27 - x}{40 - y}\right)10 \][/tex]
[tex]\[ 4(40 - y) = 10(27 - x) \][/tex]
[tex]\[ 160 - 4y = 270 - 10x \][/tex]
[tex]\[ 10x - 4y = 110 \][/tex]
[tex]\[ 5x - 2y = 55 \quad \text{(Equation 2)} \][/tex]
5. Solving the simultaneous equations:
From Equation 1: [tex]\(2x - y = 29 \)[/tex]
From Equation 2: [tex]\(5x - 2y = 55 \)[/tex]
Solving these equations simultaneously:
Multiply Equation 1 by 2:
[tex]\[ 4x - 2y = 58 \quad (Equation 3) \][/tex]
Subtract Equation 3 from Equation 2:
[tex]\[ 5x - 2y - (4x - 2y) = 55 - 58 \][/tex]
[tex]\[ x = -3 \][/tex]
Now substitute [tex]\( x = 25 \)[/tex] back into Equation 1:
[tex]\[ 2(25) - y = 29 \][/tex]
[tex]\[ 50 - y = 29 \][/tex]
[tex]\[ y = 24 \][/tex]
Hence, the missing frequencies are:
[tex]\[ f_2 = 25 \][/tex]
[tex]\[ f_3 = 24 \][/tex]
