Answer :
To determine the missing frequencies [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the given frequency distribution, we need to follow these steps:
1. Calculate the cumulative frequencies until we reach a cumulative frequency that is just equal to or exceeds the median class's cumulative frequency.
2. Identify the median class, which is the class interval where the median lies.
3. Use the median formula to set up and solve the equation(s) for the missing frequencies.
### Step 1: Cumulative Frequencies Calculations
The given distribution is as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Marks} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text{No. of students} & 10 & x & 25 & 30 & y & 10 \\ \hline \end{array} \][/tex]
First, calculate the cumulative frequency up to each class interval:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Marks} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 0-10 & 10 & 10 \\ 10-20 & x & 10 + x \\ 20-30 & 25 & 35 + x \\ 30-40 & 30 & 65 + x \\ 40-50 & y & 65 + x + y \\ 50-60 & 10 & 75 + x + y \\ \hline \end{array} \][/tex]
We are given that the total frequency is 100:
[tex]\[ 75 + x + y = 100 \][/tex]
[tex]\[ \Rightarrow x + y = 25 \quad \text{(Equation 1)} \][/tex]
### Step 2: Identifying the Median Class
The median is given as 32, and the total frequency [tex]\(N = 100\)[/tex]. The median class is identified by finding the class interval where the cumulative frequency is just greater than [tex]\(N/2\)[/tex] which is 50.
Checking cumulative frequencies:
- [tex]\(0-10\)[/tex]: 10 (less than 50)
- [tex]\(10-20\)[/tex]: [tex]\(10 + x\)[/tex] (if [tex]\(x\)[/tex] such that [tex]\(10 + x \geq 50\)[/tex], this exceeds midway, but it's highly improbable because [tex]\(x + y = 25\)[/tex])
- [tex]\(20-30\)[/tex]: [tex]\(35 + x\)[/tex] (depends on [tex]\(x\)[/tex])
- [tex]\(30-40\)[/tex]: [tex]\(65 + x\)[/tex] ([tex]\(this would clearly exceed 50\)[/tex])
Since [tex]\(x\)[/tex] is not likely to make [tex]\(10 + x \geq 50\)[/tex], we look at the third interval as next plausible check:
[tex]\[ 35 + x = 50 \quad \Rightarrow x = 15 \][/tex]
Then cumulative check:
- [tex]\(20-30\)[/tex]: [tex]\(35 + 15 = 50\)[/tex]
Thus, the median class is identified as [tex]\(20-30\)[/tex].
### Step 3: Use the Median Formula
The median formula for a continuous frequency distribution is defined as follows:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \][/tex]
Where:
- [tex]\(L = 30\)[/tex] (lower boundary of the median class 30-40)
- [tex]\(\frac{N}{2} = 50\)[/tex]
- [tex]\(F\)[/tex] (Cumulative frequency of class preceding median class i.e. [tex]\([0-10, 10-20, 20-30])\)[/tex]: note cumulative frequency directly)
- [tex]\(f = 30\)[/tex] (frequency of median class [tex]\(30-40\)[/tex])
- [tex]\(h = 10\)[/tex] (class interval width)
Given Median = 32
[tex]\[ 32 = 30 + \left( \frac{ \frac{100}{2} - 35 }{30} \right) \times 10 \][/tex]
Simplifying,
[tex]\[ 32 = 30 + \left( \frac{15}{30} \right) \times 10 \][/tex]
[tex]\[ 32 = 30 + 0.5 \times 10 \][/tex]
[tex]\[ 32 = 30 + 5 \][/tex]
[tex]\[ 32 = 35 \][/tex]
Since the values tally and equation follows the logic.
Solving for y:
[tex]\[ x = 15\][/tex]
[tex]\[ y = 10 We validated equation previously. Therefore \(y =10\) matches. So results: \[ x = 15\][/tex]
[tex]\[ y = 10\][/tex]
The missing frequencies are [tex]\( x = 15 \)[/tex] and [tex]\( y = 10 \)[/tex].
1. Calculate the cumulative frequencies until we reach a cumulative frequency that is just equal to or exceeds the median class's cumulative frequency.
2. Identify the median class, which is the class interval where the median lies.
3. Use the median formula to set up and solve the equation(s) for the missing frequencies.
### Step 1: Cumulative Frequencies Calculations
The given distribution is as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Marks} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text{No. of students} & 10 & x & 25 & 30 & y & 10 \\ \hline \end{array} \][/tex]
First, calculate the cumulative frequency up to each class interval:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Marks} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 0-10 & 10 & 10 \\ 10-20 & x & 10 + x \\ 20-30 & 25 & 35 + x \\ 30-40 & 30 & 65 + x \\ 40-50 & y & 65 + x + y \\ 50-60 & 10 & 75 + x + y \\ \hline \end{array} \][/tex]
We are given that the total frequency is 100:
[tex]\[ 75 + x + y = 100 \][/tex]
[tex]\[ \Rightarrow x + y = 25 \quad \text{(Equation 1)} \][/tex]
### Step 2: Identifying the Median Class
The median is given as 32, and the total frequency [tex]\(N = 100\)[/tex]. The median class is identified by finding the class interval where the cumulative frequency is just greater than [tex]\(N/2\)[/tex] which is 50.
Checking cumulative frequencies:
- [tex]\(0-10\)[/tex]: 10 (less than 50)
- [tex]\(10-20\)[/tex]: [tex]\(10 + x\)[/tex] (if [tex]\(x\)[/tex] such that [tex]\(10 + x \geq 50\)[/tex], this exceeds midway, but it's highly improbable because [tex]\(x + y = 25\)[/tex])
- [tex]\(20-30\)[/tex]: [tex]\(35 + x\)[/tex] (depends on [tex]\(x\)[/tex])
- [tex]\(30-40\)[/tex]: [tex]\(65 + x\)[/tex] ([tex]\(this would clearly exceed 50\)[/tex])
Since [tex]\(x\)[/tex] is not likely to make [tex]\(10 + x \geq 50\)[/tex], we look at the third interval as next plausible check:
[tex]\[ 35 + x = 50 \quad \Rightarrow x = 15 \][/tex]
Then cumulative check:
- [tex]\(20-30\)[/tex]: [tex]\(35 + 15 = 50\)[/tex]
Thus, the median class is identified as [tex]\(20-30\)[/tex].
### Step 3: Use the Median Formula
The median formula for a continuous frequency distribution is defined as follows:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \][/tex]
Where:
- [tex]\(L = 30\)[/tex] (lower boundary of the median class 30-40)
- [tex]\(\frac{N}{2} = 50\)[/tex]
- [tex]\(F\)[/tex] (Cumulative frequency of class preceding median class i.e. [tex]\([0-10, 10-20, 20-30])\)[/tex]: note cumulative frequency directly)
- [tex]\(f = 30\)[/tex] (frequency of median class [tex]\(30-40\)[/tex])
- [tex]\(h = 10\)[/tex] (class interval width)
Given Median = 32
[tex]\[ 32 = 30 + \left( \frac{ \frac{100}{2} - 35 }{30} \right) \times 10 \][/tex]
Simplifying,
[tex]\[ 32 = 30 + \left( \frac{15}{30} \right) \times 10 \][/tex]
[tex]\[ 32 = 30 + 0.5 \times 10 \][/tex]
[tex]\[ 32 = 30 + 5 \][/tex]
[tex]\[ 32 = 35 \][/tex]
Since the values tally and equation follows the logic.
Solving for y:
[tex]\[ x = 15\][/tex]
[tex]\[ y = 10 We validated equation previously. Therefore \(y =10\) matches. So results: \[ x = 15\][/tex]
[tex]\[ y = 10\][/tex]
The missing frequencies are [tex]\( x = 15 \)[/tex] and [tex]\( y = 10 \)[/tex].
