Marks obtaind by 40 students in a short assessment is given below, where a and b are two missing data. If the mean of the distribution is 7.2 find a and b

Answer

To find the values of \( a \) and \( b \) given that the mean of the marks obtained by 40 students is 7.2, we can follow these steps:

### Step 1: Set Up the Data
We have the following marks and their corresponding frequencies:
- Marks \( x \): 6, 7, 8, 9
- Frequencies \( f \): 6, \( a \), 16, 13, \( b \)

### Step 2: Calculate \( xf \) for Each Mark
We calculate \( xf \) (the product of marks and their frequencies):
- For 6: \( 6 \times 6 = 36 \)
- For \( a \): \( 6 \times a = 6a \)
- For 7: \( 7 \times 16 = 112 \)
- For 8: \( 8 \times 13 = 104 \)
- For 9: \( 9 \times b = 9b \)

The total for \( xf \) becomes:
\[ \text{Total } xf = 36 + 6a + 112 + 104 + 9b = 252 + 6a + 9b \]

### Step 3: Calculate the Total Frequency
The total frequency \( \sum f \) is:
\[ \sum f = 6 + a + 16 + 13 + b = 35 + a + b \]

### Step 4: Set Up the Mean Equation
The mean is given by the formula:
\[ \text{Mean} = \frac{\sum (xf)}{\sum f} \]
Substituting the known values:
\[ 7.2 = \frac{252 + 6a + 9b}{40} \]

### Step 5: Multiply Both Sides by 40
To eliminate the fraction, multiply both sides by 40:
\[ 7.2 \times 40 = 252 + 6a + 9b \]
Calculating \( 7.2 \times 40 \):
\[ 288 = 252 + 6a + 9b \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ 6a + 9b = 288 - 252 \]
\[ 6a + 9b = 36 \]

### Step 7: Simplify the Equation
Dividing the entire equation by 3:
\[ 2a + 3b = 12 \quad \text{(Equation 1)} \]

### Step 8: Set Up the Second Equation
From the total frequency equation:
\[ 35 + a + b = 40 \]
Rearranging gives:
\[ a + b = 5 \quad \text{(Equation 2)} \]

### Step 9: Solve the System of Equations
Now we have two equations:
1. \( 2a + 3b = 12 \)
2. \( a + b = 5 \)

From Equation 2, we can express \( a \) in terms of \( b \):
\[ a = 5 - b \]

### Step 10: Substitute into Equation 1
Substituting \( a \) into Equation 1:
\[ 2(5 - b) + 3b = 12 \]
Expanding gives:
\[ 10 - 2b + 3b = 12 \]
Combining like terms:
\[ 10 + b = 12 \]
Solving for \( b \):
\[ b = 12 - 10 = 2 \]

### Step 11: Find \( a \)
Substituting \( b = 2 \) back into Equation 2:
\[ a + 2 = 5 \]
Thus,
\[ a = 5 - 2 = 3 \]

### Final Values
The values of \( a \) and \( b \) are:
\[ a = 3, \quad b = 2 \]