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Sagot :
Let's tackle each part of the question step by step:
(क) What does [tex]\( L \)[/tex] represent in the formula [tex]\( M_d = L + \frac{i}{f} \left( \frac{N}{2} - cf \right) \)[/tex] of a continuous series?
In the formula for finding the median of a continuous series, [tex]\( L \)[/tex] represents the lower limit of the median class. The median class is the class interval in which the median lies. We'll identify the median class during our calculation of the median.
---
(ख) Find the mean [tex]\((\bar{x})\)[/tex].
To find the mean, follow these steps:
1. Calculate the midpoint of each class interval. The midpoint [tex]\( x_i \)[/tex] of a class interval [tex]\([a, b]\)[/tex] is given by [tex]\( x_i = \frac{a + b}{2} \)[/tex].
2. Multiply each midpoint by the frequency of the corresponding class interval.
3. Sum these products to get the total sum of values [tex]\( \sum f_i x_i \)[/tex].
4. Divide the total sum by the total number of observations [tex]\( N \)[/tex] to get the mean.
Given the data:
| Expenditure (in thousands) | No. of families |
|----------------------------|------------------|
| 10-20 | 3 |
| 20-30 | 2 |
| 30-40 | 5 |
| 40-50 | 6 |
| 50-60 | 4 |
Midpoints for each class interval:
- Midpoint of 10-20: [tex]\( \frac{10+20}{2} = 15 \)[/tex]
- Midpoint of 20-30: [tex]\( \frac{20+30}{2} = 25 \)[/tex]
- Midpoint of 30-40: [tex]\( \frac{30+40}{2} = 35 \)[/tex]
- Midpoint of 40-50: [tex]\( \frac{40+50}{2} = 45 \)[/tex]
- Midpoint of 50-60: [tex]\( \frac{50+60}{2} = 55 \)[/tex]
Summations:
[tex]\[ \begin{aligned} \sum f_i x_i &= (3 \times 15) + (2 \times 25) + (5 \times 35) + (6 \times 45) + (4 \times 55) \\ &= 45 + 50 + 175 + 270 + 220 \\ &= 760 \end{aligned} \][/tex]
Total number of families [tex]\( N \)[/tex]:
[tex]\[ N = 3 + 2 + 5 + 6 + 4 = 20 \][/tex]
Mean [tex]\( \bar{x} \)[/tex]:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{N} = \frac{760}{20} = 38 \][/tex]
So, the mean [tex]\( \bar{x} \)[/tex] is [tex]\( 38 \)[/tex] thousand rupees.
---
(ग) By what percentage is the mean [tex]\((\bar{x})\)[/tex] less than the median [tex]\((M_d)\)[/tex]? Find it by calculation.
To find the percentage difference between the mean and the median, we need to calculate the median first.
Steps to find the median [tex]\( M_d \)[/tex]:
1. Determine the total number of families [tex]\( N \)[/tex]. [tex]\( N = 20 \)[/tex].
2. Find [tex]\( \frac{N}{2} \)[/tex]. [tex]\( \frac{N}{2} = 10 \)[/tex].
3. Identify the median class. The median class is the class whose cumulative frequency just exceeds [tex]\( \frac{N}{2} \)[/tex].
Cumulative frequencies:
- Cumulative frequency up to 10-20: 3
- Cumulative frequency up to 20-30: 3 + 2 = 5
- Cumulative frequency up to 30-40: 5 + 5 = 10 (This is where [tex]\( \frac{N}{2} = 10 \)[/tex], so this is the median class)
- Cumulative frequency up to 40-50: 10 + 6 = 16
- Cumulative frequency up to 50-60: 16 + 4 = 20
The median class is 30-40. Now apply the median formula [tex]\( M_d = L + \frac{i}{f} \left( \frac{N}{2} - cf \right) \)[/tex], where:
- [tex]\( L = 30 \)[/tex]: the lower limit of the median class
- [tex]\( i = 10 \)[/tex]: the class interval
- [tex]\( f = 5 \)[/tex]: the frequency of the median class
- [tex]\( cf = 5 \)[/tex]: the cumulative frequency of the class before the median class
[tex]\[ M_d = 30 + \frac{10}{5} \left( 10 - 5 \right) = 30 + 2 \times 5 = 30 + 10 = 40 \][/tex]
The median [tex]\( M_d \)[/tex] is [tex]\( 40 \)[/tex] thousand rupees.
Percentage difference:
[tex]\[ \text{Percentage difference} = \left( \frac{M_d - \bar{x}}{M_d} \right) \times 100 = \left( \frac{40 - 38}{40} \right) \times 100 = \frac{2}{40} \times 100 = 5\% \][/tex]
So, the mean is 5% less than the median.
---
(घ) How many maximum no. of families are there who expense less than median expenditure?
To find the number of families who spend less than the median expenditure:
- Cumulative frequency just before the median class (30-40) gives us this information.
- The cumulative frequency up to but not including 30-40 is 5.
So, the maximum number of families who expense less than the median expenditure is 5.
---
Summary of the results:
1. [tex]\( L \)[/tex] represents the lower limit of the median class.
2. The mean expenditure [tex]\( \bar{x} = 38 \)[/tex] thousand rupees.
3. The mean is 5% less than the median.
4. The maximum number of families who expense less than the median expenditure is 5.
(क) What does [tex]\( L \)[/tex] represent in the formula [tex]\( M_d = L + \frac{i}{f} \left( \frac{N}{2} - cf \right) \)[/tex] of a continuous series?
In the formula for finding the median of a continuous series, [tex]\( L \)[/tex] represents the lower limit of the median class. The median class is the class interval in which the median lies. We'll identify the median class during our calculation of the median.
---
(ख) Find the mean [tex]\((\bar{x})\)[/tex].
To find the mean, follow these steps:
1. Calculate the midpoint of each class interval. The midpoint [tex]\( x_i \)[/tex] of a class interval [tex]\([a, b]\)[/tex] is given by [tex]\( x_i = \frac{a + b}{2} \)[/tex].
2. Multiply each midpoint by the frequency of the corresponding class interval.
3. Sum these products to get the total sum of values [tex]\( \sum f_i x_i \)[/tex].
4. Divide the total sum by the total number of observations [tex]\( N \)[/tex] to get the mean.
Given the data:
| Expenditure (in thousands) | No. of families |
|----------------------------|------------------|
| 10-20 | 3 |
| 20-30 | 2 |
| 30-40 | 5 |
| 40-50 | 6 |
| 50-60 | 4 |
Midpoints for each class interval:
- Midpoint of 10-20: [tex]\( \frac{10+20}{2} = 15 \)[/tex]
- Midpoint of 20-30: [tex]\( \frac{20+30}{2} = 25 \)[/tex]
- Midpoint of 30-40: [tex]\( \frac{30+40}{2} = 35 \)[/tex]
- Midpoint of 40-50: [tex]\( \frac{40+50}{2} = 45 \)[/tex]
- Midpoint of 50-60: [tex]\( \frac{50+60}{2} = 55 \)[/tex]
Summations:
[tex]\[ \begin{aligned} \sum f_i x_i &= (3 \times 15) + (2 \times 25) + (5 \times 35) + (6 \times 45) + (4 \times 55) \\ &= 45 + 50 + 175 + 270 + 220 \\ &= 760 \end{aligned} \][/tex]
Total number of families [tex]\( N \)[/tex]:
[tex]\[ N = 3 + 2 + 5 + 6 + 4 = 20 \][/tex]
Mean [tex]\( \bar{x} \)[/tex]:
[tex]\[ \bar{x} = \frac{\sum f_i x_i}{N} = \frac{760}{20} = 38 \][/tex]
So, the mean [tex]\( \bar{x} \)[/tex] is [tex]\( 38 \)[/tex] thousand rupees.
---
(ग) By what percentage is the mean [tex]\((\bar{x})\)[/tex] less than the median [tex]\((M_d)\)[/tex]? Find it by calculation.
To find the percentage difference between the mean and the median, we need to calculate the median first.
Steps to find the median [tex]\( M_d \)[/tex]:
1. Determine the total number of families [tex]\( N \)[/tex]. [tex]\( N = 20 \)[/tex].
2. Find [tex]\( \frac{N}{2} \)[/tex]. [tex]\( \frac{N}{2} = 10 \)[/tex].
3. Identify the median class. The median class is the class whose cumulative frequency just exceeds [tex]\( \frac{N}{2} \)[/tex].
Cumulative frequencies:
- Cumulative frequency up to 10-20: 3
- Cumulative frequency up to 20-30: 3 + 2 = 5
- Cumulative frequency up to 30-40: 5 + 5 = 10 (This is where [tex]\( \frac{N}{2} = 10 \)[/tex], so this is the median class)
- Cumulative frequency up to 40-50: 10 + 6 = 16
- Cumulative frequency up to 50-60: 16 + 4 = 20
The median class is 30-40. Now apply the median formula [tex]\( M_d = L + \frac{i}{f} \left( \frac{N}{2} - cf \right) \)[/tex], where:
- [tex]\( L = 30 \)[/tex]: the lower limit of the median class
- [tex]\( i = 10 \)[/tex]: the class interval
- [tex]\( f = 5 \)[/tex]: the frequency of the median class
- [tex]\( cf = 5 \)[/tex]: the cumulative frequency of the class before the median class
[tex]\[ M_d = 30 + \frac{10}{5} \left( 10 - 5 \right) = 30 + 2 \times 5 = 30 + 10 = 40 \][/tex]
The median [tex]\( M_d \)[/tex] is [tex]\( 40 \)[/tex] thousand rupees.
Percentage difference:
[tex]\[ \text{Percentage difference} = \left( \frac{M_d - \bar{x}}{M_d} \right) \times 100 = \left( \frac{40 - 38}{40} \right) \times 100 = \frac{2}{40} \times 100 = 5\% \][/tex]
So, the mean is 5% less than the median.
---
(घ) How many maximum no. of families are there who expense less than median expenditure?
To find the number of families who spend less than the median expenditure:
- Cumulative frequency just before the median class (30-40) gives us this information.
- The cumulative frequency up to but not including 30-40 is 5.
So, the maximum number of families who expense less than the median expenditure is 5.
---
Summary of the results:
1. [tex]\( L \)[/tex] represents the lower limit of the median class.
2. The mean expenditure [tex]\( \bar{x} = 38 \)[/tex] thousand rupees.
3. The mean is 5% less than the median.
4. The maximum number of families who expense less than the median expenditure is 5.
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