The median of following frequency distribution is
`{:(,"Class",,fi),(,60-70,,5),(,70-80,,15),(,80-90,,20),(,90-100,,30),(,100-110,,20),(,110-120,,8):}`

Options

A. 92
B. 92.5
C. 93
D. 93.5

Explanation

To find the median of the given frequency distribution, we will follow these steps:

### Step 1: Create a Cumulative Frequency Table
We will first create a cumulative frequency table from the given frequency distribution.

| Class | Frequency (fi) | Cumulative Frequency (CF) |
|----------|----------------|---------------------------|
| 60-70 | 5 | 5 |
| 70-80 | 15 | 5 + 15 = 20 |
| 80-90 | 20 | 20 + 20 = 40 |
| 90-100 | 30 | 40 + 30 = 70 |
| 100-110 | 20 | 70 + 20 = 90 |
| 110-120 | 8 | 90 + 8 = 98 |

### Step 2: Find the Total Frequency (N)
Now, we will find the total frequency \( N \) by adding all the frequencies.

\[ N = 5 + 15 + 20 + 30 + 20 + 8 = 98 \]

### Step 3: Determine the Median Class
To find the median, we need to find \( \frac{N}{2} \):

\[ \frac{N}{2} = \frac{98}{2} = 49 \]

Now, we look for the cumulative frequency that is greater than or equal to 49. From the cumulative frequency table, we see that:

- The cumulative frequency just before 49 is 40 (for the class 80-90).
- The cumulative frequency that is greater than 49 is 70 (for the class 90-100).

Thus, the median class is **90-100**.

### Step 4: Apply the Median Formula
We will use the median formula:

\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \]

Where:
- \( L \) = lower boundary of the median class = 90
- \( N \) = total frequency = 98
- \( F \) = cumulative frequency of the class preceding the median class = 40 (for class 80-90)
- \( f \) = frequency of the median class = 30 (for class 90-100)
- \( h \) = width of the median class = 10 (from 90 to 100)

Substituting these values into the formula:

\[ \text{Median} = 90 + \left( \frac{49 - 40}{30} \right) \times 10 \]

Calculating further:

\[ \text{Median} = 90 + \left( \frac{9}{30} \right) \times 10 \]

\[ \text{Median} = 90 + 3 = 93 \]

### Final Answer
The median of the given frequency distribution is **93**.

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