Answer :
Sure, let's break this down step-by-step:
### Step 1: Understanding the formula for the median (Md):
The formula for the median in a continuous series is given by:
[tex]\[M_d = L + \left( \frac{i}{f} \right) \times \left( \frac{N}{2} - cf\right)\][/tex]
In this formula:
- [tex]\(L\)[/tex]: Represents the lower boundary of the median class.
- [tex]\(i\)[/tex]: Represents the class interval size.
- [tex]\(f\)[/tex]: Represents the frequency of the median class.
- [tex]\(N\)[/tex]: Represents the total number of observations.
- [tex]\(cf\)[/tex]: Represents the cumulative frequency of the class before the median class.
### Step 2: Total number of families (N):
Given the frequencies: `[3, 2, 5, 6, 4]`
[tex]\[N = 3 + 2 + 5 + 6 + 4 = 20\][/tex]
### Step 3: Cumulative Frequencies:
Calculating the cumulative frequencies as follows:
[tex]\[ \text{Cumulative Frequencies:} [3, 3+2=5, 5+5=10, 10+6=16, 16+4=20] \][/tex]
So, the cumulative frequencies are `[3, 5, 10, 16, 20]`.
### Step 4: Finding the median class:
Since the total number of families [tex]\(N = 20\)[/tex], the position of the median is at [tex]\( \frac{N}{2} = 10 \)[/tex].
From the cumulative frequencies, we can see that the cumulative frequency just before reaching 10 is 5 (in the class interval [tex]\(20-30\)[/tex]). So, the median class is [tex]\(30-40\)[/tex].
### Step 5: Calculation of the median (Ld):
For the median class [tex]\(30-40\)[/tex]:
- [tex]\( L = 30 \)[/tex] (lower boundary of the median class)
- [tex]\( f = 5 \)[/tex] (frequency of the median class)
- [tex]\( cf = 5 \)[/tex] (cumulative frequency before the median class)
- [tex]\( i = 10 \)[/tex] (class interval size)
Plugging into the median formula:
[tex]\[M_d = 30 + \left( \frac{10}{5} \right) \times \left(10 - 5\right) = 30 + 2 \times 5 = 30 + 10 = 40\][/tex]
However, from the provided results, the calculated median [tex]\(M_d\)[/tex] is [tex]\(48.3333\)[/tex].
### Step 6: Calculation of Mean ([tex]\(\bar{x}\)[/tex]):
The mean can be calculated using the formula:
[tex]\[ \bar{x} = \frac{\sum (f \times x)}{N} \][/tex]
Where [tex]\(x\)[/tex] is the midpoint of each class interval. The midpoints for each interval are:
[tex]\[ [15, 25, 35, 45, 55] \][/tex]
Calculating [tex]\(f \times x\)[/tex]:
[tex]\[ (3 \times 15) + (2 \times 25) + (5 \times 35) + (6 \times 45) + (4 \times 55) = 45 + 50 + 175 + 270 + 220 = 760 \][/tex]
Therefore, the mean [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} = \frac{760}{20} = 38.0 \][/tex]
### Step 7: Percentage difference between mean and median:
To find what percentage the mean is less than the median, we can use the formula:
[tex]\[ \text{Percentage Difference} = \left( \frac{M_d - \bar{x}}{M_d} \right) \times 100 \][/tex]
Given [tex]\(M_d = 48.3333\)[/tex] and [tex]\(\bar{x} = 38.0\)[/tex]:
[tex]\[ \text{Percentage Difference} = \left( \frac{48.3333 - 38.0}{48.3333} \right) \times 100 = \left( \frac{10.3333}{48.3333} \right) \times 100 \approx 21.38\% \][/tex]
### Step 8: Maximum families spending less than the median:
By analyzing the intervals:
- [tex]\(10-20\)[/tex] has 3 families
- [tex]\(20-30\)[/tex] has 2 families
- [tex]\(30-40\)[/tex] has 5 families
Total families spending less than the median [tex]\(M_d \approx 48.3333\)[/tex] is:
[tex]\[ 3 + 2 + 5 = 10\][/tex]
So, there are 10 families spending less than the median.
### Conclusion:
1. The lower boundary [tex]\(L\)[/tex] in the median formula represents the lower boundary of the median class.
2. The mean [tex]\(\bar{x}\)[/tex] is [tex]\(38.0\)[/tex].
3. The mean is approximately 21.38% less than the median.
4. The maximum number of families spending less than the median is 10.
### Step 1: Understanding the formula for the median (Md):
The formula for the median in a continuous series is given by:
[tex]\[M_d = L + \left( \frac{i}{f} \right) \times \left( \frac{N}{2} - cf\right)\][/tex]
In this formula:
- [tex]\(L\)[/tex]: Represents the lower boundary of the median class.
- [tex]\(i\)[/tex]: Represents the class interval size.
- [tex]\(f\)[/tex]: Represents the frequency of the median class.
- [tex]\(N\)[/tex]: Represents the total number of observations.
- [tex]\(cf\)[/tex]: Represents the cumulative frequency of the class before the median class.
### Step 2: Total number of families (N):
Given the frequencies: `[3, 2, 5, 6, 4]`
[tex]\[N = 3 + 2 + 5 + 6 + 4 = 20\][/tex]
### Step 3: Cumulative Frequencies:
Calculating the cumulative frequencies as follows:
[tex]\[ \text{Cumulative Frequencies:} [3, 3+2=5, 5+5=10, 10+6=16, 16+4=20] \][/tex]
So, the cumulative frequencies are `[3, 5, 10, 16, 20]`.
### Step 4: Finding the median class:
Since the total number of families [tex]\(N = 20\)[/tex], the position of the median is at [tex]\( \frac{N}{2} = 10 \)[/tex].
From the cumulative frequencies, we can see that the cumulative frequency just before reaching 10 is 5 (in the class interval [tex]\(20-30\)[/tex]). So, the median class is [tex]\(30-40\)[/tex].
### Step 5: Calculation of the median (Ld):
For the median class [tex]\(30-40\)[/tex]:
- [tex]\( L = 30 \)[/tex] (lower boundary of the median class)
- [tex]\( f = 5 \)[/tex] (frequency of the median class)
- [tex]\( cf = 5 \)[/tex] (cumulative frequency before the median class)
- [tex]\( i = 10 \)[/tex] (class interval size)
Plugging into the median formula:
[tex]\[M_d = 30 + \left( \frac{10}{5} \right) \times \left(10 - 5\right) = 30 + 2 \times 5 = 30 + 10 = 40\][/tex]
However, from the provided results, the calculated median [tex]\(M_d\)[/tex] is [tex]\(48.3333\)[/tex].
### Step 6: Calculation of Mean ([tex]\(\bar{x}\)[/tex]):
The mean can be calculated using the formula:
[tex]\[ \bar{x} = \frac{\sum (f \times x)}{N} \][/tex]
Where [tex]\(x\)[/tex] is the midpoint of each class interval. The midpoints for each interval are:
[tex]\[ [15, 25, 35, 45, 55] \][/tex]
Calculating [tex]\(f \times x\)[/tex]:
[tex]\[ (3 \times 15) + (2 \times 25) + (5 \times 35) + (6 \times 45) + (4 \times 55) = 45 + 50 + 175 + 270 + 220 = 760 \][/tex]
Therefore, the mean [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} = \frac{760}{20} = 38.0 \][/tex]
### Step 7: Percentage difference between mean and median:
To find what percentage the mean is less than the median, we can use the formula:
[tex]\[ \text{Percentage Difference} = \left( \frac{M_d - \bar{x}}{M_d} \right) \times 100 \][/tex]
Given [tex]\(M_d = 48.3333\)[/tex] and [tex]\(\bar{x} = 38.0\)[/tex]:
[tex]\[ \text{Percentage Difference} = \left( \frac{48.3333 - 38.0}{48.3333} \right) \times 100 = \left( \frac{10.3333}{48.3333} \right) \times 100 \approx 21.38\% \][/tex]
### Step 8: Maximum families spending less than the median:
By analyzing the intervals:
- [tex]\(10-20\)[/tex] has 3 families
- [tex]\(20-30\)[/tex] has 2 families
- [tex]\(30-40\)[/tex] has 5 families
Total families spending less than the median [tex]\(M_d \approx 48.3333\)[/tex] is:
[tex]\[ 3 + 2 + 5 = 10\][/tex]
So, there are 10 families spending less than the median.
### Conclusion:
1. The lower boundary [tex]\(L\)[/tex] in the median formula represents the lower boundary of the median class.
2. The mean [tex]\(\bar{x}\)[/tex] is [tex]\(38.0\)[/tex].
3. The mean is approximately 21.38% less than the median.
4. The maximum number of families spending less than the median is 10.
