For the following frequency distribution, determine the mode and the median:
`{:("Class",100-120,120-140,140-160,160-180,180-200),("Frequency",12,14,8,6,10):}`

Answer

To solve the problem, we need to determine the mode and the median for the given frequency distribution.

### Given Data:
- **Class Intervals**: 100-120, 120-140, 140-160, 160-180, 180-200
- **Frequencies**: 12, 14, 8, 6, 10

### Step 1: Calculate Cumulative Frequency
We first calculate the cumulative frequency for the given frequency distribution.

- For the class 100-120: Cumulative Frequency = 12
- For the class 120-140: Cumulative Frequency = 12 + 14 = 26
- For the class 140-160: Cumulative Frequency = 26 + 8 = 34
- For the class 160-180: Cumulative Frequency = 34 + 6 = 40
- For the class 180-200: Cumulative Frequency = 40 + 10 = 50

**Cumulative Frequency Table:**
```
Class Frequency Cumulative Frequency
100-120 12 12
120-140 14 26
140-160 8 34
160-180 6 40
180-200 10 50
```

### Step 2: Determine the Mode
The mode is the value that appears most frequently in a data set. In this case, we look for the class with the highest frequency.

- The highest frequency is 14, which corresponds to the class interval 120-140.

**Mode Formula:**
\[ \text{Mode} = L + \frac{f_1 - f_0}{(2f_1 - f_0 - f_2)} \times h \]
Where:
- \( L \) = lower limit of the modal class = 120
- \( f_1 \) = frequency of the modal class = 14
- \( f_0 \) = frequency of the class before the modal class = 12
- \( f_2 \) = frequency of the class after the modal class = 8
- \( h \) = width of the class interval = 20

**Calculating the Mode:**
\[ \text{Mode} = 120 + \frac{14 - 12}{(2 \times 14 - 12 - 8)} \times 20 \]
\[ = 120 + \frac{2}{(28 - 20)} \times 20 \]
\[ = 120 + \frac{2}{8} \times 20 \]
\[ = 120 + \frac{1}{4} \times 20 \]
\[ = 120 + 5 \]
\[ = 125 \]

### Step 3: Determine the Median
To find the median, we need to locate the median class. The median class is the class where the cumulative frequency is greater than or equal to \( \frac{n}{2} \), where \( n \) is the total frequency.

- Total frequency \( n = 50 \)
- \( \frac{n}{2} = 25 \)

The cumulative frequency just greater than 25 is 26, which corresponds to the class interval 120-140.

**Median Formula:**
\[ \text{Median} = L + \frac{\frac{n}{2} - CF}{f} \times h \]
Where:
- \( L \) = lower limit of the median class = 120
- \( CF \) = cumulative frequency of the class before the median class = 12
- \( f \) = frequency of the median class = 14
- \( h \) = width of the class interval = 20

**Calculating the Median:**
\[ \text{Median} = 120 + \frac{25 - 12}{14} \times 20 \]
\[ = 120 + \frac{13}{14} \times 20 \]
\[ = 120 + \frac{260}{14} \]
\[ = 120 + 18.57 \]
\[ \approx 138.57 \]

### Final Results:
- **Mode**: 125
- **Median**: 138.57