IDNLearn.com is designed to help you find the answers you need quickly and easily. Get the information you need from our community of experts who provide accurate and thorough answers to all your questions.
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.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your go-to source for dependable answers. Thank you for visiting, and we hope to assist you again.
