Answer :
To find the median of the given data, we follow a systematic approach for grouped data. Here's a detailed step-by-step solution:
1. Sum of all consumers:
- First, we calculate the total number of consumers:
[tex]\[ \text{Total Number of Consumers} = 4 + 9 + 22 + 42 + 56 + 64 + 68 = 265 \][/tex]
2. Cumulative Frequencies:
- Next, we determine the cumulative frequency for each class. This helps in locating the median position.
[tex]\[ \begin{align*} \text{Cumulative Frequency up to 85} &= 4 \\ \text{Cumulative Frequency up to 105} &= 4 + 9 = 13 \\ \text{Cumulative Frequency up to 125} &= 4 + 9 + 22 = 35 \\ \text{Cumulative Frequency up to 145} &= 4 + 9 + 22 + 42 = 77 \\ \text{Cumulative Frequency up to 165} &= 4 + 9 + 22 + 42 + 56 = 133 \\ \text{Cumulative Frequency up to 185} &= 4 + 9 + 22 + 42 + 56 + 64 = 197 \\ \text{Cumulative Frequency up to 205} &= 4 + 9 + 22 + 42 + 56 + 64 + 68 = 265 \\ \end{align*} \][/tex]
3. Finding the Median Position:
- The median position is found using the formula:
[tex]\[ \text{Median Position} = \frac{N}{2} = \frac{265}{2} = 132.5 \][/tex]
4. Identifying the Median Class:
- We look for the class where the cumulative frequency just exceeds the median position.
- Here, the median lies between 145 and 165, because [tex]\(132.5\)[/tex] (median position) falls in the cumulative frequency [tex]\(77\)[/tex] up until [tex]\(145\)[/tex] and [tex]\(133\)[/tex] up until [tex]\(165\)[/tex].
5. Class Interval Parameters:
[tex]\[ \text{Lower Class Boundary, } L = 145 \][/tex]
[tex]\[ \text{Frequency of Median Class, } f = 56 \][/tex]
[tex]\[ \text{Cumulative Frequency of the class before Median Class, } cf = 77 \][/tex]
[tex]\[ \text{Class Interval Width, } h = 165 - 145 = 20 \][/tex]
6. Calculating the Median:
- We use the formula for median in grouped data:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \][/tex]
Substituting the values:
[tex]\[ \text{Median} = 145 + \left( \frac{132.5 - 77}{56} \right) \times 20 \][/tex]
[tex]\[ \text{Median} = 145 + \left( \frac{55.5}{56} \right) \times 20 \][/tex]
[tex]\[ \text{Median} = 145 + 19.82142857142857 \][/tex]
[tex]\[ \text{Median} = 164.82142857142858 \][/tex]
Therefore, the median monthly consumption is approximately [tex]\(164.82\)[/tex].
1. Sum of all consumers:
- First, we calculate the total number of consumers:
[tex]\[ \text{Total Number of Consumers} = 4 + 9 + 22 + 42 + 56 + 64 + 68 = 265 \][/tex]
2. Cumulative Frequencies:
- Next, we determine the cumulative frequency for each class. This helps in locating the median position.
[tex]\[ \begin{align*} \text{Cumulative Frequency up to 85} &= 4 \\ \text{Cumulative Frequency up to 105} &= 4 + 9 = 13 \\ \text{Cumulative Frequency up to 125} &= 4 + 9 + 22 = 35 \\ \text{Cumulative Frequency up to 145} &= 4 + 9 + 22 + 42 = 77 \\ \text{Cumulative Frequency up to 165} &= 4 + 9 + 22 + 42 + 56 = 133 \\ \text{Cumulative Frequency up to 185} &= 4 + 9 + 22 + 42 + 56 + 64 = 197 \\ \text{Cumulative Frequency up to 205} &= 4 + 9 + 22 + 42 + 56 + 64 + 68 = 265 \\ \end{align*} \][/tex]
3. Finding the Median Position:
- The median position is found using the formula:
[tex]\[ \text{Median Position} = \frac{N}{2} = \frac{265}{2} = 132.5 \][/tex]
4. Identifying the Median Class:
- We look for the class where the cumulative frequency just exceeds the median position.
- Here, the median lies between 145 and 165, because [tex]\(132.5\)[/tex] (median position) falls in the cumulative frequency [tex]\(77\)[/tex] up until [tex]\(145\)[/tex] and [tex]\(133\)[/tex] up until [tex]\(165\)[/tex].
5. Class Interval Parameters:
[tex]\[ \text{Lower Class Boundary, } L = 145 \][/tex]
[tex]\[ \text{Frequency of Median Class, } f = 56 \][/tex]
[tex]\[ \text{Cumulative Frequency of the class before Median Class, } cf = 77 \][/tex]
[tex]\[ \text{Class Interval Width, } h = 165 - 145 = 20 \][/tex]
6. Calculating the Median:
- We use the formula for median in grouped data:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \][/tex]
Substituting the values:
[tex]\[ \text{Median} = 145 + \left( \frac{132.5 - 77}{56} \right) \times 20 \][/tex]
[tex]\[ \text{Median} = 145 + \left( \frac{55.5}{56} \right) \times 20 \][/tex]
[tex]\[ \text{Median} = 145 + 19.82142857142857 \][/tex]
[tex]\[ \text{Median} = 164.82142857142858 \][/tex]
Therefore, the median monthly consumption is approximately [tex]\(164.82\)[/tex].
