Answer :
Sure, let's solve this step-by-step.
### 1. Given Data
We are provided with the mean of the marks obtained, \(\bar{X} = 32.5\). The marks obtained and the number of students in each range are as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Marks obtained} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Number of students} & 5 & 10 & k & 35 & 15 \\ \hline \end{array} \][/tex]
### 2. Interval midpoints
First, we calculate the midpoints of each interval:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Midpoint} & 5 & 15 & 25 & 35 & 45 \\ \hline \end{array} \][/tex]
### 3. Total number of students
The total number of students can be expressed as:
[tex]\[ \text{Total number of students} = 5 + 10 + k + 35 + 15 = k + 65 \][/tex]
### 4. Total marks
The total marks can be calculated by multiplying the number of students in each interval by the interval midpoint and summing:
[tex]\[ \text{Total marks} = 5 \cdot 5 + 10 \cdot 15 + k \cdot 25 + 35 \cdot 35 + 15 \cdot 45 \][/tex]
[tex]\[ = 25 + 150 + 25k + 1225 + 675 = 25k + 2075 \][/tex]
### 5. Mean formula
The mean of the marks obtained is given as:
[tex]\[ \bar{X} = \frac{\text{Total marks}}{\text{Total number of students}} \][/tex]
Substituting the known values:
[tex]\[ 32.5 = \frac{25k + 2075}{k + 65} \][/tex]
### 6. Solve for \(k\)
Rewrite the equation:
[tex]\[ 32.5 (k + 65) = 25k + 2075 \][/tex]
Distribute on the left side:
[tex]\[ 32.5k + 2112.5 = 25k + 2075 \][/tex]
Isolate the \(k\) terms to one side:
[tex]\[ 32.5k - 25k = 2075 - 2112.5 \][/tex]
[tex]\[ 7.5k = -37.5 \][/tex]
Solve for \(k\):
[tex]\[ k = \frac{-37.5}{7.5} = -5 \][/tex]
So, the value of \(k\) is:
[tex]\[ k = -5 \][/tex]
### 1. Given Data
We are provided with the mean of the marks obtained, \(\bar{X} = 32.5\). The marks obtained and the number of students in each range are as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Marks obtained} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Number of students} & 5 & 10 & k & 35 & 15 \\ \hline \end{array} \][/tex]
### 2. Interval midpoints
First, we calculate the midpoints of each interval:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{Interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Midpoint} & 5 & 15 & 25 & 35 & 45 \\ \hline \end{array} \][/tex]
### 3. Total number of students
The total number of students can be expressed as:
[tex]\[ \text{Total number of students} = 5 + 10 + k + 35 + 15 = k + 65 \][/tex]
### 4. Total marks
The total marks can be calculated by multiplying the number of students in each interval by the interval midpoint and summing:
[tex]\[ \text{Total marks} = 5 \cdot 5 + 10 \cdot 15 + k \cdot 25 + 35 \cdot 35 + 15 \cdot 45 \][/tex]
[tex]\[ = 25 + 150 + 25k + 1225 + 675 = 25k + 2075 \][/tex]
### 5. Mean formula
The mean of the marks obtained is given as:
[tex]\[ \bar{X} = \frac{\text{Total marks}}{\text{Total number of students}} \][/tex]
Substituting the known values:
[tex]\[ 32.5 = \frac{25k + 2075}{k + 65} \][/tex]
### 6. Solve for \(k\)
Rewrite the equation:
[tex]\[ 32.5 (k + 65) = 25k + 2075 \][/tex]
Distribute on the left side:
[tex]\[ 32.5k + 2112.5 = 25k + 2075 \][/tex]
Isolate the \(k\) terms to one side:
[tex]\[ 32.5k - 25k = 2075 - 2112.5 \][/tex]
[tex]\[ 7.5k = -37.5 \][/tex]
Solve for \(k\):
[tex]\[ k = \frac{-37.5}{7.5} = -5 \][/tex]
So, the value of \(k\) is:
[tex]\[ k = -5 \][/tex]
