Compute the missing frequencies in the following distribution , given that is `sum f_(i)` = 100 and the median is 32
`{:("Marks",0-10,10-20,20-30,30-40,40-50,50-60),("Number of students "," "9," "?, " "26, " "30, " " ?, " "10):}`

Answer

To solve the problem of computing the missing frequencies in the given distribution, we will follow these steps:

### Step 1: Define the missing frequencies
Let the frequency for the class interval 10-20 be \( x \) and for the class interval 40-50 be \( y \). The frequencies for the other intervals are given as follows:
- 0-10: 9
- 10-20: \( x \)
- 20-30: 26
- 30-40: 30
- 40-50: \( y \)
- 50-60: 10

### Step 2: Set up the equation for the sum of frequencies
We know that the total sum of frequencies is 100. Therefore, we can write the equation:
\[ 9 + x + 26 + 30 + y + 10 = 100 \]
Simplifying this gives:
\[ 75 + x + y = 100 \]
Thus, we can express this as:
\[ x + y = 25 \quad \text{(Equation 1)} \]

### Step 3: Use the median to set up another equation
The median is given as 32. To find the median class, we need to determine the cumulative frequency (cf) just before the median. The cumulative frequencies are:
- 0-10: 9
- 10-20: \( 9 + x \)
- 20-30: \( 35 + x \)
- 30-40: \( 65 + x \)
- 40-50: \( 65 + x + y \)
- 50-60: \( 75 + x + y \)

Since the median is 32, it falls in the class interval 30-40. The frequency for this class is 30, and the cumulative frequency just before it is \( 35 + x \).

Using the median formula:
\[ \text{Median} = l + \left( \frac{n}{2} - cf \right) \cdot \frac{h}{f} \]
Where:
- \( l = 30 \) (lower boundary of the median class)
- \( n = 100 \) (total frequency)
- \( cf = 35 + x \) (cumulative frequency before the median class)
- \( f = 30 \) (frequency of the median class)
- \( h = 10 \) (class width)

Substituting the values into the formula:
\[ 32 = 30 + \left( \frac{100}{2} - (35 + x) \right) \cdot \frac{10}{30} \]
This simplifies to:
\[ 32 = 30 + \left( 50 - (35 + x) \right) \cdot \frac{10}{30} \]
\[ 32 = 30 + \left( 15 - x \right) \cdot \frac{10}{30} \]
\[ 32 = 30 + \frac{10(15 - x)}{30} \]
\[ 32 = 30 + \frac{5(15 - x)}{15} \]
\[ 32 = 30 + \frac{5(15 - x)}{15} \]
Multiplying through by 15 to eliminate the fraction:
\[ 2 \cdot 15 = 30 + 5(15 - x) \]
\[ 30 = 30 + 75 - 5x \]
\[ 0 = 75 - 5x \]
\[ 5x = 75 \]
\[ x = 15 \]

### Step 4: Substitute \( x \) back into Equation 1
Now substituting \( x = 15 \) into Equation 1:
\[ 15 + y = 25 \]
\[ y = 10 \]

### Final Result
The missing frequencies are:
- Frequency for the class interval 10-20 (\( x \)) = 15
- Frequency for the class interval 40-50 (\( y \)) = 10

### Summary
The completed frequency distribution is:
- 0-10: 9
- 10-20: 15
- 20-30: 26
- 30-40: 30
- 40-50: 10
- 50-60: 10