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Certainly! Let's solve this problem step by step.
We are given a frequency distribution table of marks obtained by students, with one frequency, [tex]\( x \)[/tex], missing. Additionally, we are informed that the median mark is 24.
Here's the provided table for clarity:
| Marks obtained | [tex]\(0-10\)[/tex] | [tex]\(10-20\)[/tex] | [tex]\(20-30\)[/tex] | [tex]\(30-40\)[/tex] | [tex]\(40-50\)[/tex] |
|----------------|---|---|---|---|---|
| No. of students| 4 | 12 | [tex]\( x \)[/tex] | 9 | 5 |
### Step 1: Determine the total number of students
First, let's determine the total number of observations (students). Using the given frequencies and one missing frequency [tex]\( x \)[/tex]:
[tex]\[ \text{Total number of students (N)} = 4 + 12 + x + 9 + 5 \][/tex]
[tex]\[ \Rightarrow N = 30 + x \][/tex]
### Step 2: Identify the cumulative frequency
Next, we need to calculate the cumulative frequency (CF) up to each class interval. The cumulative frequency is the sum of all frequencies up to and including the current class interval.
[tex]\[ \begin{array}{c|c|c} \hline \text{Class Interval} & \text{Frequency (f)} & \text{Cumulative Frequency (CF)} \\ \hline 0-10 & 4 & 4 \\ 10-20 & 12 & 4 + 12 = 16 \\ 20-30 & x & 16 + x \\ 30-40 & 9 & 16 + x + 9 = 25 + x \\ 40-50 & 5 & 30 + x \\ \hline \end{array} \][/tex]
### Step 3: Determine the median class
The median class is the class interval wherein the cumulative frequency is greater than or equal to [tex]\( \frac{N}{2} \)[/tex].
Given that [tex]\( \text{Median} = 24 \)[/tex], this value lies in the [tex]\(20-30\)[/tex] interval.
From the cumulative frequency table:
[tex]\[ \left\{ \text{If N is even}, \quad \frac{N}{2} \quad \text{If N is odd}, \quad \left( \frac {N+1}{2} \right) \right\} \][/tex]
### Midway between Median class and it's interval
[tex]\[ \left( \frac {30+x+1}{2} - 1 \right) \][/tex]
### Step 4: Using the median formula
The formula for median in a continuous series is:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \cdot h \][/tex]
Where:
- [tex]\( L = 20 \)[/tex] (lower limit of median class)
- [tex]\( N = 30 + x \)[/tex] (total number of students)
- [tex]\( CF = 16 \)[/tex] (cumulative frequency of the class preceding the median class)
- [tex]\( f = x \)[/tex] (frequency of the median class)
- [tex]\( h = 10 \)[/tex] (class width)
Given that the median is [tex]\( 24 \)[/tex]:
[tex]\[ 24 = 20 + \left( \frac{\frac{30 + x}{2} - 16}{x} \right) \cdot 10 \][/tex]
### Step 5: Simplifying the equation
Simplify the right side step by step:
[tex]\[ 24 - 20 = \left( \frac{\frac{30 + x}{2} - 16}{x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{\frac{30 + x}{2} - 16}{x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{30 + x - 32}{2x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{x - 2}{2x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{10x - 20}{2x} \right) \][/tex]
[tex]\[ 4 = \left( \frac{10 - 20}{2x} \right) \][/tex]
### Step x:
[tex]\[ 4 = \frac{10(x - 2)}{2x} \][/tex]
[tex]\[ 4 = \frac{5(x - 2)}{x \right) }\][/tex]
[tex]\[ 4 = 5 - 10 / x) \][/tex]
[tex]\[ x = 10 \][/tex]
Hence, when the these interval lies upon (x-5) domain tends.asarray=[].
Thus, the value of [tex]\( x \)[/tex] is [tex]\(\boxed{10}\)[/tex].
We are given a frequency distribution table of marks obtained by students, with one frequency, [tex]\( x \)[/tex], missing. Additionally, we are informed that the median mark is 24.
Here's the provided table for clarity:
| Marks obtained | [tex]\(0-10\)[/tex] | [tex]\(10-20\)[/tex] | [tex]\(20-30\)[/tex] | [tex]\(30-40\)[/tex] | [tex]\(40-50\)[/tex] |
|----------------|---|---|---|---|---|
| No. of students| 4 | 12 | [tex]\( x \)[/tex] | 9 | 5 |
### Step 1: Determine the total number of students
First, let's determine the total number of observations (students). Using the given frequencies and one missing frequency [tex]\( x \)[/tex]:
[tex]\[ \text{Total number of students (N)} = 4 + 12 + x + 9 + 5 \][/tex]
[tex]\[ \Rightarrow N = 30 + x \][/tex]
### Step 2: Identify the cumulative frequency
Next, we need to calculate the cumulative frequency (CF) up to each class interval. The cumulative frequency is the sum of all frequencies up to and including the current class interval.
[tex]\[ \begin{array}{c|c|c} \hline \text{Class Interval} & \text{Frequency (f)} & \text{Cumulative Frequency (CF)} \\ \hline 0-10 & 4 & 4 \\ 10-20 & 12 & 4 + 12 = 16 \\ 20-30 & x & 16 + x \\ 30-40 & 9 & 16 + x + 9 = 25 + x \\ 40-50 & 5 & 30 + x \\ \hline \end{array} \][/tex]
### Step 3: Determine the median class
The median class is the class interval wherein the cumulative frequency is greater than or equal to [tex]\( \frac{N}{2} \)[/tex].
Given that [tex]\( \text{Median} = 24 \)[/tex], this value lies in the [tex]\(20-30\)[/tex] interval.
From the cumulative frequency table:
[tex]\[ \left\{ \text{If N is even}, \quad \frac{N}{2} \quad \text{If N is odd}, \quad \left( \frac {N+1}{2} \right) \right\} \][/tex]
### Midway between Median class and it's interval
[tex]\[ \left( \frac {30+x+1}{2} - 1 \right) \][/tex]
### Step 4: Using the median formula
The formula for median in a continuous series is:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \cdot h \][/tex]
Where:
- [tex]\( L = 20 \)[/tex] (lower limit of median class)
- [tex]\( N = 30 + x \)[/tex] (total number of students)
- [tex]\( CF = 16 \)[/tex] (cumulative frequency of the class preceding the median class)
- [tex]\( f = x \)[/tex] (frequency of the median class)
- [tex]\( h = 10 \)[/tex] (class width)
Given that the median is [tex]\( 24 \)[/tex]:
[tex]\[ 24 = 20 + \left( \frac{\frac{30 + x}{2} - 16}{x} \right) \cdot 10 \][/tex]
### Step 5: Simplifying the equation
Simplify the right side step by step:
[tex]\[ 24 - 20 = \left( \frac{\frac{30 + x}{2} - 16}{x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{\frac{30 + x}{2} - 16}{x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{30 + x - 32}{2x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{x - 2}{2x} \right) \cdot 10 \][/tex]
[tex]\[ 4 = \left( \frac{10x - 20}{2x} \right) \][/tex]
[tex]\[ 4 = \left( \frac{10 - 20}{2x} \right) \][/tex]
### Step x:
[tex]\[ 4 = \frac{10(x - 2)}{2x} \][/tex]
[tex]\[ 4 = \frac{5(x - 2)}{x \right) }\][/tex]
[tex]\[ 4 = 5 - 10 / x) \][/tex]
[tex]\[ x = 10 \][/tex]
Hence, when the these interval lies upon (x-5) domain tends.asarray=[].
Thus, the value of [tex]\( x \)[/tex] is [tex]\(\boxed{10}\)[/tex].
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