Answer :
### Step-by-Step Solution:
#### (क) [tex]$L$[/tex] ले के जनाउँछ ?
The symbol [tex]$L$[/tex] in the formula for finding the median [tex]$M_d = L + \frac{i}{f} \left( \frac{N}{2} - c \right)$[/tex] represents the lower boundary of the median class.
#### (ख) मध्यक [tex]$\bar{x}$[/tex] पत्ता लगाउनुहोस्।
To find the mean [tex]$\bar{x}$[/tex] of the given distribution, we use the formula for the mean of a grouped data:
[tex]\[ \bar{x} = \frac{\sum{(f_i \cdot x_i)}}{\sum{f_i}} \][/tex]
where [tex]\( f_i \)[/tex] is the frequency of the [tex]\(i\)[/tex]-th class and [tex]\( x_i \)[/tex] is the midpoint of the [tex]\(i\)[/tex]-th class interval. Let's calculate the midpoints and use them for finding the mean.
1. Class Intervals: [tex]\([10, 20)\)[/tex], [tex]\([20, 30)\)[/tex], [tex]\([30, 40)\)[/tex], [tex]\([40, 50)\)[/tex], [tex]\([50, 60)\)[/tex]
2. Number of Families: [tex]\(3, 2, 5, 6, 4\)[/tex]
3. Midpoints: Midpoint of each class is given by [tex]\(\frac{\text{lower limit} + \text{upper limit}}{2}\)[/tex]
[tex]\[ \begin{align*} \text{Midpoint of } [10, 20) & = \frac{10 + 20}{2} = 15 \\ \text{Midpoint of } [20, 30) & = \frac{20 + 30}{2} = 25 \\ \text{Midpoint of } [30, 40) & = \frac{30 + 40}{2} = 35 \\ \text{Midpoint of } [40, 50) & = \frac{40 + 50}{2} = 45 \\ \text{Midpoint of } [50, 60) & = \frac{50 + 60}{2} = 55 \\ \end{align*} \][/tex]
Next, we calculate the sum of the product of frequency and midpoint, and the total number of families:
[tex]\[ \begin{align*} \sum{(f \cdot x)} & = 3 \cdot 15 + 2 \cdot 25 + 5 \cdot 35 + 6 \cdot 45 + 4 \cdot 55 \\ & = 45 + 50 + 175 + 270 + 220 \\ & = 760 \end{align*} \][/tex]
The total number of families [tex]\(\sum{f}\)[/tex] is:
[tex]\[ 3 + 2 + 5 + 6 + 4 = 20 \][/tex]
So, the mean expenditure [tex]\(\bar{x}\)[/tex] is:
[tex]\[ \bar{x} = \frac{760}{20} = 38.0 \][/tex]
#### (ग) मध्यक [tex]$(\bar{x})$[/tex] मध्यिका [tex]$\left( M _{ d }\right)$[/tex] भन्दा कति प्रतिशतले थोरै छ? गणना गर्नुहोस्।
To find the median [tex]\(M_d\)[/tex], we follow the formula:
[tex]\[ M_d = L + \frac{i}{f} \left(\frac{N}{2} - c \right) \][/tex]
1. Total number of families [tex]\(N = 20\)[/tex].
2. We find [tex]\( \frac{N}{2} = 10\)[/tex].
Cumulative frequency calculation:
[tex]\[ \begin{align*} 3 & \\ 3 + 2 & = 5 \\ 5 + 5 & = 10 \\ 10 + 6 & = 16 \\ 16 + 4 & = 20 \end{align*} \][/tex]
From the cumulative frequency, the median class is the class interval where the cumulative frequency first reaches or exceeds [tex]\( \frac{N}{2} = 10 \)[/tex].
Here, the median class is [tex]\([30, 40)\)[/tex] because the cumulative frequency is 10 at this class interval.
- [tex]\(L = 30\)[/tex]
- [tex]\(i = 10\)[/tex] (class width)
- [tex]\(f_m = 5\)[/tex] (frequency of the median class)
- [tex]\(c = 5\)[/tex] (cumulative frequency of the class preceding the median class)
Now calculate [tex]\(M_d\)[/tex]:
[tex]\[ M_d = 30 + \frac{10}{5} \left(10 - 5 \right) = 30 + 2 \cdot 5 = 30 + 10 = 40 \][/tex]
Percentage difference between mean and median:
[tex]\[ \text{Percentage Difference} = \left( \frac{\bar{x} - M_d}{M_d} \right) \times 100 = \left(\frac{38.0 - 40.0}{40.0}\right) \times 100 = -5.0\% \][/tex]
#### (घ) मध्यिका खर्चं भन्दा कम खर्च गर्ने अधिकतम परिवार कति जना रहेछछ्?
Families with expenditure less than the median expenditure of 40 are from the intervals [tex]\([10, 20)\)[/tex], [tex]\([20, 30)\)[/tex], and part of [tex]\([30, 40)\)[/tex].
Totalling:
[tex]\[ 3 + 2 + (5 \cdot \frac{5}{10}) = 3 + 2 + 2.5 = 7.5 \approx 10 \text{ (nearest integer)} \][/tex]
So, there's a total of 10 families with expenditures less than the median expenditure.
### Final Answers:
1. [tex]\(L\)[/tex] represents the lower boundary of the median class.
2. Mean expenditure [tex]$\bar{x} = 38.0$[/tex]
3. Mean expenditure is [tex]\(5.0\%\)[/tex] less than the median.
4. There are 10 families with expenditure less than the median.
#### (क) [tex]$L$[/tex] ले के जनाउँछ ?
The symbol [tex]$L$[/tex] in the formula for finding the median [tex]$M_d = L + \frac{i}{f} \left( \frac{N}{2} - c \right)$[/tex] represents the lower boundary of the median class.
#### (ख) मध्यक [tex]$\bar{x}$[/tex] पत्ता लगाउनुहोस्।
To find the mean [tex]$\bar{x}$[/tex] of the given distribution, we use the formula for the mean of a grouped data:
[tex]\[ \bar{x} = \frac{\sum{(f_i \cdot x_i)}}{\sum{f_i}} \][/tex]
where [tex]\( f_i \)[/tex] is the frequency of the [tex]\(i\)[/tex]-th class and [tex]\( x_i \)[/tex] is the midpoint of the [tex]\(i\)[/tex]-th class interval. Let's calculate the midpoints and use them for finding the mean.
1. Class Intervals: [tex]\([10, 20)\)[/tex], [tex]\([20, 30)\)[/tex], [tex]\([30, 40)\)[/tex], [tex]\([40, 50)\)[/tex], [tex]\([50, 60)\)[/tex]
2. Number of Families: [tex]\(3, 2, 5, 6, 4\)[/tex]
3. Midpoints: Midpoint of each class is given by [tex]\(\frac{\text{lower limit} + \text{upper limit}}{2}\)[/tex]
[tex]\[ \begin{align*} \text{Midpoint of } [10, 20) & = \frac{10 + 20}{2} = 15 \\ \text{Midpoint of } [20, 30) & = \frac{20 + 30}{2} = 25 \\ \text{Midpoint of } [30, 40) & = \frac{30 + 40}{2} = 35 \\ \text{Midpoint of } [40, 50) & = \frac{40 + 50}{2} = 45 \\ \text{Midpoint of } [50, 60) & = \frac{50 + 60}{2} = 55 \\ \end{align*} \][/tex]
Next, we calculate the sum of the product of frequency and midpoint, and the total number of families:
[tex]\[ \begin{align*} \sum{(f \cdot x)} & = 3 \cdot 15 + 2 \cdot 25 + 5 \cdot 35 + 6 \cdot 45 + 4 \cdot 55 \\ & = 45 + 50 + 175 + 270 + 220 \\ & = 760 \end{align*} \][/tex]
The total number of families [tex]\(\sum{f}\)[/tex] is:
[tex]\[ 3 + 2 + 5 + 6 + 4 = 20 \][/tex]
So, the mean expenditure [tex]\(\bar{x}\)[/tex] is:
[tex]\[ \bar{x} = \frac{760}{20} = 38.0 \][/tex]
#### (ग) मध्यक [tex]$(\bar{x})$[/tex] मध्यिका [tex]$\left( M _{ d }\right)$[/tex] भन्दा कति प्रतिशतले थोरै छ? गणना गर्नुहोस्।
To find the median [tex]\(M_d\)[/tex], we follow the formula:
[tex]\[ M_d = L + \frac{i}{f} \left(\frac{N}{2} - c \right) \][/tex]
1. Total number of families [tex]\(N = 20\)[/tex].
2. We find [tex]\( \frac{N}{2} = 10\)[/tex].
Cumulative frequency calculation:
[tex]\[ \begin{align*} 3 & \\ 3 + 2 & = 5 \\ 5 + 5 & = 10 \\ 10 + 6 & = 16 \\ 16 + 4 & = 20 \end{align*} \][/tex]
From the cumulative frequency, the median class is the class interval where the cumulative frequency first reaches or exceeds [tex]\( \frac{N}{2} = 10 \)[/tex].
Here, the median class is [tex]\([30, 40)\)[/tex] because the cumulative frequency is 10 at this class interval.
- [tex]\(L = 30\)[/tex]
- [tex]\(i = 10\)[/tex] (class width)
- [tex]\(f_m = 5\)[/tex] (frequency of the median class)
- [tex]\(c = 5\)[/tex] (cumulative frequency of the class preceding the median class)
Now calculate [tex]\(M_d\)[/tex]:
[tex]\[ M_d = 30 + \frac{10}{5} \left(10 - 5 \right) = 30 + 2 \cdot 5 = 30 + 10 = 40 \][/tex]
Percentage difference between mean and median:
[tex]\[ \text{Percentage Difference} = \left( \frac{\bar{x} - M_d}{M_d} \right) \times 100 = \left(\frac{38.0 - 40.0}{40.0}\right) \times 100 = -5.0\% \][/tex]
#### (घ) मध्यिका खर्चं भन्दा कम खर्च गर्ने अधिकतम परिवार कति जना रहेछछ्?
Families with expenditure less than the median expenditure of 40 are from the intervals [tex]\([10, 20)\)[/tex], [tex]\([20, 30)\)[/tex], and part of [tex]\([30, 40)\)[/tex].
Totalling:
[tex]\[ 3 + 2 + (5 \cdot \frac{5}{10}) = 3 + 2 + 2.5 = 7.5 \approx 10 \text{ (nearest integer)} \][/tex]
So, there's a total of 10 families with expenditures less than the median expenditure.
### Final Answers:
1. [tex]\(L\)[/tex] represents the lower boundary of the median class.
2. Mean expenditure [tex]$\bar{x} = 38.0$[/tex]
3. Mean expenditure is [tex]\(5.0\%\)[/tex] less than the median.
4. There are 10 families with expenditure less than the median.
