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Sagot :
Let's solve the given problem step by step.
We are given the following information:
1. Total number of families ([tex]\( n \)[/tex]) = 1000
2. Median expenditure = Rs. 650
3. Distribution of expenditures:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline Expenditure (Rs.) & 400-500 & 500-600 & 600-700 & 700-800 & 800-900 \\ \hline No. of families & 50 & - & 450 & - & 100 \\ \hline \end{tabular} \][/tex]
Here, we have missing frequencies in the second class (500-600) and the fourth class (700-800), which we will denote as [tex]\( f_2 \)[/tex] and [tex]\( f_4 \)[/tex].
### Step 1: Locate the Median Class
Since the median expenditure is given as Rs. 650, it falls in the class interval 600-700. We will use this information to find the missing frequencies.
### Step 2: Calculate Cumulative Frequencies
We need to calculate the cumulative frequencies up to the median class.
Given frequencies:
[tex]\[ \begin{align*} f_1 &= 50 \\ f_2 &= \text{unknown} \\ f_3 &= 450 \\ f_4 &= \text{unknown} \\ f_5 &= 100 \\ \end{align*} \][/tex]
The equations for cumulative frequencies are:
[tex]\[ \begin{align*} C_1 &= f_1 = 50 \\ C_2 &= f_1 + f_2 = 50 + f_2 \\ C_3 &= f_1 + f_2 + f_3 = 50 + f_2 + 450 = 500 + f_2 \\ C_4 &= f_1 + f_2 + f_3 + f_4 = 500 + f_2 + f_4 \\ C_5 &= f_1 + f_2 + f_3 + f_4 + f_5 = 500 + f_2 + f_4 + 100 = 600 + f_2 + f_4 \end{align*} \][/tex]
### Step 3: Use Median Formula to Find [tex]\( f_2 \)[/tex]
The formula for the median in a grouped frequency distribution is:
[tex]\[ \text{Median} = l + \left(\frac{N/2 - F}{f}\right) \cdot h \][/tex]
where:
- [tex]\( l \)[/tex] is the lower boundary of the median class (=600).
- [tex]\( N \)[/tex] is the total number of families (=1000).
- [tex]\( F \)[/tex] is the cumulative frequency of the class before the median class.
- [tex]\( f \)[/tex] is the frequency of the median class (=450).
- [tex]\( h \)[/tex] is the class interval width (=100).
Substitute the known values into the median formula:
[tex]\[ 650 = 600 + \left(\frac{1000/2 - F}{450}\right) \cdot 100 \][/tex]
[tex]\[ 650 = 600 + \left(\frac{500 - F}{450}\right) \cdot 100 \][/tex]
Solving for [tex]\( F \)[/tex]:
[tex]\[ 50 = \left(\frac{500 - F}{450}\right) \cdot 100 \][/tex]
[tex]\[ 50 \cdot 450 = 100 \cdot (500 - F) \][/tex]
[tex]\[ 22500 = 100 \cdot (500 - F) \][/tex]
[tex]\[ 22500 = 50000 - 100F \][/tex]
[tex]\[ 100F = 50000 - 22500 \][/tex]
[tex]\[ 100F = 27500 \][/tex]
[tex]\[ F = 275 \][/tex]
Since [tex]\( F = 50 + f_2 \)[/tex]:
[tex]\[ 275 = 50 + f_2 \][/tex]
[tex]\[ f_2 = 225 \][/tex]
### Step 4: Calculate [tex]\( f_4 \)[/tex]
We know the total frequency is 1000:
[tex]\[ 50 + 225 + 450 + f_4 + 100 = 1000 \][/tex]
[tex]\[ 825 + f_4 = 1000 \][/tex]
[tex]\[ f_4 = 175 \][/tex]
So, the missing frequencies are [tex]\( f_2 = 225 \)[/tex] and [tex]\( f_4 = 175 \)[/tex].
### Step 5: Calculate the Mean
To find the mean, we need to calculate the midpoints of the class intervals and then use the formula for the mean:
[tex]\[ \text{Mean} = \frac{\sum f_i x_i}{N} \][/tex]
where [tex]\( x_i \)[/tex] is the midpoint of each class.
Class midpoints [tex]\( (x_i) \)[/tex]:
[tex]\[ \begin{align*} 400-500 &\implies 450 \\ 500-600 &\implies 550 \\ 600-700 &\implies 650 \\ 700-800 &\implies 750 \\ 800-900 &\implies 850 \\ \end{align*} \][/tex]
The corresponding frequencies [tex]\( (f_i) \)[/tex]:
[tex]\[ \begin{align*} f_1 &= 50 \\ f_2 &= 225 \\ f_3 &= 450 \\ f_4 &= 175 \\ f_5 &= 100 \\ \end{align*} \][/tex]
Summing [tex]\( f_i x_i \)[/tex]:
[tex]\[ \begin{align*} 50 \cdot 450 &= 22500 \\ 225 \cdot 550 &= 123750 \\ 450 \cdot 650 &= 292500 \\ 175 \cdot 750 &= 131250 \\ 100 \cdot 850 &= 85000 \\ \end{align*} \][/tex]
Sum:
[tex]\[ 22500 + 123750 + 292500 + 131250 + 85000 = 655000 \][/tex]
Mean:
[tex]\[ \text{Mean} = \frac{655000}{1000} = 655 \][/tex]
Thus, the missing frequencies are [tex]\( f_2 = 225 \)[/tex] and [tex]\( f_4 = 175 \)[/tex], and the mean expenditure is Rs. 655.
We are given the following information:
1. Total number of families ([tex]\( n \)[/tex]) = 1000
2. Median expenditure = Rs. 650
3. Distribution of expenditures:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline Expenditure (Rs.) & 400-500 & 500-600 & 600-700 & 700-800 & 800-900 \\ \hline No. of families & 50 & - & 450 & - & 100 \\ \hline \end{tabular} \][/tex]
Here, we have missing frequencies in the second class (500-600) and the fourth class (700-800), which we will denote as [tex]\( f_2 \)[/tex] and [tex]\( f_4 \)[/tex].
### Step 1: Locate the Median Class
Since the median expenditure is given as Rs. 650, it falls in the class interval 600-700. We will use this information to find the missing frequencies.
### Step 2: Calculate Cumulative Frequencies
We need to calculate the cumulative frequencies up to the median class.
Given frequencies:
[tex]\[ \begin{align*} f_1 &= 50 \\ f_2 &= \text{unknown} \\ f_3 &= 450 \\ f_4 &= \text{unknown} \\ f_5 &= 100 \\ \end{align*} \][/tex]
The equations for cumulative frequencies are:
[tex]\[ \begin{align*} C_1 &= f_1 = 50 \\ C_2 &= f_1 + f_2 = 50 + f_2 \\ C_3 &= f_1 + f_2 + f_3 = 50 + f_2 + 450 = 500 + f_2 \\ C_4 &= f_1 + f_2 + f_3 + f_4 = 500 + f_2 + f_4 \\ C_5 &= f_1 + f_2 + f_3 + f_4 + f_5 = 500 + f_2 + f_4 + 100 = 600 + f_2 + f_4 \end{align*} \][/tex]
### Step 3: Use Median Formula to Find [tex]\( f_2 \)[/tex]
The formula for the median in a grouped frequency distribution is:
[tex]\[ \text{Median} = l + \left(\frac{N/2 - F}{f}\right) \cdot h \][/tex]
where:
- [tex]\( l \)[/tex] is the lower boundary of the median class (=600).
- [tex]\( N \)[/tex] is the total number of families (=1000).
- [tex]\( F \)[/tex] is the cumulative frequency of the class before the median class.
- [tex]\( f \)[/tex] is the frequency of the median class (=450).
- [tex]\( h \)[/tex] is the class interval width (=100).
Substitute the known values into the median formula:
[tex]\[ 650 = 600 + \left(\frac{1000/2 - F}{450}\right) \cdot 100 \][/tex]
[tex]\[ 650 = 600 + \left(\frac{500 - F}{450}\right) \cdot 100 \][/tex]
Solving for [tex]\( F \)[/tex]:
[tex]\[ 50 = \left(\frac{500 - F}{450}\right) \cdot 100 \][/tex]
[tex]\[ 50 \cdot 450 = 100 \cdot (500 - F) \][/tex]
[tex]\[ 22500 = 100 \cdot (500 - F) \][/tex]
[tex]\[ 22500 = 50000 - 100F \][/tex]
[tex]\[ 100F = 50000 - 22500 \][/tex]
[tex]\[ 100F = 27500 \][/tex]
[tex]\[ F = 275 \][/tex]
Since [tex]\( F = 50 + f_2 \)[/tex]:
[tex]\[ 275 = 50 + f_2 \][/tex]
[tex]\[ f_2 = 225 \][/tex]
### Step 4: Calculate [tex]\( f_4 \)[/tex]
We know the total frequency is 1000:
[tex]\[ 50 + 225 + 450 + f_4 + 100 = 1000 \][/tex]
[tex]\[ 825 + f_4 = 1000 \][/tex]
[tex]\[ f_4 = 175 \][/tex]
So, the missing frequencies are [tex]\( f_2 = 225 \)[/tex] and [tex]\( f_4 = 175 \)[/tex].
### Step 5: Calculate the Mean
To find the mean, we need to calculate the midpoints of the class intervals and then use the formula for the mean:
[tex]\[ \text{Mean} = \frac{\sum f_i x_i}{N} \][/tex]
where [tex]\( x_i \)[/tex] is the midpoint of each class.
Class midpoints [tex]\( (x_i) \)[/tex]:
[tex]\[ \begin{align*} 400-500 &\implies 450 \\ 500-600 &\implies 550 \\ 600-700 &\implies 650 \\ 700-800 &\implies 750 \\ 800-900 &\implies 850 \\ \end{align*} \][/tex]
The corresponding frequencies [tex]\( (f_i) \)[/tex]:
[tex]\[ \begin{align*} f_1 &= 50 \\ f_2 &= 225 \\ f_3 &= 450 \\ f_4 &= 175 \\ f_5 &= 100 \\ \end{align*} \][/tex]
Summing [tex]\( f_i x_i \)[/tex]:
[tex]\[ \begin{align*} 50 \cdot 450 &= 22500 \\ 225 \cdot 550 &= 123750 \\ 450 \cdot 650 &= 292500 \\ 175 \cdot 750 &= 131250 \\ 100 \cdot 850 &= 85000 \\ \end{align*} \][/tex]
Sum:
[tex]\[ 22500 + 123750 + 292500 + 131250 + 85000 = 655000 \][/tex]
Mean:
[tex]\[ \text{Mean} = \frac{655000}{1000} = 655 \][/tex]
Thus, the missing frequencies are [tex]\( f_2 = 225 \)[/tex] and [tex]\( f_4 = 175 \)[/tex], and the mean expenditure is Rs. 655.
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