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Sagot :
Let's solve the given problems step-by-step.
### Given Data:
[tex]\[ \begin{array}{|l|c|c|c|c|c|c|} \hline \text{Jw;~\|x~\} Rs. ma (Wages in Rs.)} & 500-600 & 600-700 & 700-800 & 800-900 & 900-1000 & 1000-1100 \\ \hline \text{Kamdar-ekho samkhyak (Number of workers)} & 3 & 5 & 6 & 2 & 3 & 1 \\ \hline \end{array} \][/tex]
### Part (a): Modal Class
The modal class is the class interval with the highest frequency of workers.
- The frequencies are: 3, 5, 6, 2, 3, 1
- The maximum frequency is 6 which corresponds to the class interval 700-800.
Thus, the modal class is 700-800.
### Part (b): Median Class
The median class is the class interval that contains the median of the data.
1. Calculate the total number of workers:
[tex]\[ \text{Total workers} = 3 + 5 + 6 + 2 + 3 + 1 = 20 \][/tex]
2. The position of the median is given by:
[tex]\[ \text{Median position} = \frac{N + 1}{2} = \frac{20 + 1}{2} = 10.5 \][/tex]
3. To find the median class, we need to calculate the cumulative frequency and determine the class interval where the 10.5th worker lies.
[tex]\[ \begin{array}{|l|c|} \hline \text{Class Interval} & \text{Cumulative Frequency} \\ \hline 500-600 & 3 \\ 600-700 & 3 + 5 = 8 \\ 700-800 & 8 + 6 = 14 \\ 800-900 & 14 + 2 = 16 \\ 900-1000 & 16 + 3 = 19 \\ 1000-1100 & 19 + 1 = 20 \\ \hline \end{array} \][/tex]
Since the 10.5th worker falls into the cumulative frequency that reaches 14, which is the third interval:
Thus, the median class is 700-800.
### Part (c): Average Daily Income
To calculate the average daily income, we first need to find the midpoint of each class interval, then use these midpoints to calculate the weighted average.
1. Calculate the midpoints ([tex]\(x_i\)[/tex]):
[tex]\[ x_1 = \frac{500 + 600}{2} = 550, \quad x_2 = \frac{600 + 700}{2} = 650, \quad x_3 = \frac{700 + 800}{2} = 750, \][/tex]
[tex]\[ x_4 = \frac{800 + 900}{2} = 850, \quad x_5 = \frac{900 + 1000}{2} = 950, \quad x_6 = \frac{1000 + 1100}{2} = 1050 \][/tex]
2. The number of workers ([tex]\(f_i\)[/tex]) are given as: 3, 5, 6, 2, 3, 1
3. Calculate the total sum of the wages:
[tex]\[ \text{Total wages} = (3 \times 550) + (5 \times 650) + (6 \times 750) + (2 \times 850) + (3 \times 950) + (1 \times 1050) \][/tex]
[tex]\[ = 1650 + 3250 + 4500 + 1700 + 2850 + 1050 = 15000 \][/tex]
4. Thus, the average daily income is:
[tex]\[ \text{Average income} = \frac{\text{Total wages}}{\text{Total workers}} = \frac{15000}{20} = 750 \][/tex]
Therefore, the average daily income of a worker is Rs. 750.
### Part (d): Median and Modal Class
The median and modal class do not always lie in the same class.
- The modal class is the interval with the highest frequency (most common value).
- The median class is the interval in which the middle value of the dataset lies (50th percentile).
In this specific case, both the median and modal class are 700-800. However, this is not always true as they are determined by different criteria. The modal class relates to frequency, while the median class relates to position within a cumulative frequency distribution. So, while they can coincide, it is not a necessity.
Thus, they do not always lie in the same class.
### Given Data:
[tex]\[ \begin{array}{|l|c|c|c|c|c|c|} \hline \text{Jw;~\|x~\} Rs. ma (Wages in Rs.)} & 500-600 & 600-700 & 700-800 & 800-900 & 900-1000 & 1000-1100 \\ \hline \text{Kamdar-ekho samkhyak (Number of workers)} & 3 & 5 & 6 & 2 & 3 & 1 \\ \hline \end{array} \][/tex]
### Part (a): Modal Class
The modal class is the class interval with the highest frequency of workers.
- The frequencies are: 3, 5, 6, 2, 3, 1
- The maximum frequency is 6 which corresponds to the class interval 700-800.
Thus, the modal class is 700-800.
### Part (b): Median Class
The median class is the class interval that contains the median of the data.
1. Calculate the total number of workers:
[tex]\[ \text{Total workers} = 3 + 5 + 6 + 2 + 3 + 1 = 20 \][/tex]
2. The position of the median is given by:
[tex]\[ \text{Median position} = \frac{N + 1}{2} = \frac{20 + 1}{2} = 10.5 \][/tex]
3. To find the median class, we need to calculate the cumulative frequency and determine the class interval where the 10.5th worker lies.
[tex]\[ \begin{array}{|l|c|} \hline \text{Class Interval} & \text{Cumulative Frequency} \\ \hline 500-600 & 3 \\ 600-700 & 3 + 5 = 8 \\ 700-800 & 8 + 6 = 14 \\ 800-900 & 14 + 2 = 16 \\ 900-1000 & 16 + 3 = 19 \\ 1000-1100 & 19 + 1 = 20 \\ \hline \end{array} \][/tex]
Since the 10.5th worker falls into the cumulative frequency that reaches 14, which is the third interval:
Thus, the median class is 700-800.
### Part (c): Average Daily Income
To calculate the average daily income, we first need to find the midpoint of each class interval, then use these midpoints to calculate the weighted average.
1. Calculate the midpoints ([tex]\(x_i\)[/tex]):
[tex]\[ x_1 = \frac{500 + 600}{2} = 550, \quad x_2 = \frac{600 + 700}{2} = 650, \quad x_3 = \frac{700 + 800}{2} = 750, \][/tex]
[tex]\[ x_4 = \frac{800 + 900}{2} = 850, \quad x_5 = \frac{900 + 1000}{2} = 950, \quad x_6 = \frac{1000 + 1100}{2} = 1050 \][/tex]
2. The number of workers ([tex]\(f_i\)[/tex]) are given as: 3, 5, 6, 2, 3, 1
3. Calculate the total sum of the wages:
[tex]\[ \text{Total wages} = (3 \times 550) + (5 \times 650) + (6 \times 750) + (2 \times 850) + (3 \times 950) + (1 \times 1050) \][/tex]
[tex]\[ = 1650 + 3250 + 4500 + 1700 + 2850 + 1050 = 15000 \][/tex]
4. Thus, the average daily income is:
[tex]\[ \text{Average income} = \frac{\text{Total wages}}{\text{Total workers}} = \frac{15000}{20} = 750 \][/tex]
Therefore, the average daily income of a worker is Rs. 750.
### Part (d): Median and Modal Class
The median and modal class do not always lie in the same class.
- The modal class is the interval with the highest frequency (most common value).
- The median class is the interval in which the middle value of the dataset lies (50th percentile).
In this specific case, both the median and modal class are 700-800. However, this is not always true as they are determined by different criteria. The modal class relates to frequency, while the median class relates to position within a cumulative frequency distribution. So, while they can coincide, it is not a necessity.
Thus, they do not always lie in the same class.
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