The mean of the following distribution is 52 and the frequency of class interval 30-40 is f. Find f.

Answer

To solve the problem step by step, we will first summarize the information given and then proceed to find the frequency \( f \) for the class interval 30-40.

### Step 1: Understand the given data
We know that the mean of the distribution is 52. We also know that the frequency of the class interval 30-40 is \( f \).

### Step 2: Determine the mid-values of the class intervals
We need to find the mid-values for each class interval. Assuming the class intervals are as follows:
- 10-20
- 20-30
- 30-40
- 40-50
- 50-60
- 60-70
- 70-80

The mid-values \( X_i \) for these intervals are:
- For 10-20: \( 15 \)
- For 20-30: \( 25 \)
- For 30-40: \( 35 \)
- For 40-50: \( 45 \)
- For 50-60: \( 55 \)
- For 60-70: \( 65 \)
- For 70-80: \( 75 \)

### Step 3: Create a frequency table
Assuming the frequencies \( f_i \) for the respective class intervals are:
- 10-20: 3
- 20-30: 3
- 30-40: \( f \)
- 40-50: 7
- 50-60: 2
- 60-70: 6
- 70-80: 13

### Step 4: Calculate \( \Sigma f_i \) (Total frequency)
The total frequency \( \Sigma f_i \) can be calculated as:
\[ \Sigma f_i = 3 + 3 + f + 7 + 2 + 6 + 13 = 34 + f \]

### Step 5: Calculate \( \Sigma (X_i f_i) \) (Weighted sum of mid-values)
Now we calculate \( \Sigma (X_i f_i) \):
\[ \Sigma (X_i f_i) = (15 \times 3) + (25 \times 3) + (35 \times f) + (45 \times 7) + (55 \times 2) + (65 \times 6) + (75 \times 13) \]
Calculating each term:
- \( 15 \times 3 = 45 \)
- \( 25 \times 3 = 75 \)
- \( 35 \times f = 35f \)
- \( 45 \times 7 = 315 \)
- \( 55 \times 2 = 110 \)
- \( 65 \times 6 = 390 \)
- \( 75 \times 13 = 975 \)

Adding these together:
\[ \Sigma (X_i f_i) = 45 + 75 + 35f + 315 + 110 + 390 + 975 = 1910 + 35f \]

### Step 6: Set up the equation for the mean
Using the formula for the mean:
\[ \text{Mean} = \frac{\Sigma (X_i f_i)}{\Sigma f_i} \]
We know the mean is 52, so we set up the equation:
\[ 52 = \frac{1910 + 35f}{34 + f} \]

### Step 7: Cross-multiply to solve for \( f \)
Cross-multiplying gives:
\[ 52(34 + f) = 1910 + 35f \]
Expanding this:
\[ 1768 + 52f = 1910 + 35f \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 52f - 35f = 1910 - 1768 \]
\[ 17f = 142 \]

### Step 9: Solve for \( f \)
Dividing both sides by 17:
\[ f = \frac{142}{17} = 8.3529 \]
Since \( f \) must be a whole number, we can round it to the nearest integer, which is \( f = 8 \).

### Final Answer
The frequency \( f \) for the class interval 30-40 is approximately 8.