The median from the following data is
`{:("Mid-value","Frequency"),(115,6),(125,25),(135,48),(145,72),(155,116),(165,60),(175,38),(185,22),(195,3):}`

Options

A. 1) `153.80`
B. 2) `153.91`
C. 3) `165.18`
D. 4) `165.93`

Answer

To find the median from the given data, we will follow these steps:

### Step 1: Organize the data
We have the mid-values and their corresponding frequencies. Let's summarize them in a table format:

| Mid-value | Frequency |
|-----------|-----------|
| 115 | 6 |
| 125 | 25 |
| 135 | 48 |
| 145 | 72 |
| 155 | 116 |
| 165 | 60 |
| 175 | 38 |
| 185 | 22 |
| 195 | 3 |

### Step 2: Calculate cumulative frequencies
Next, we will calculate the cumulative frequencies (CF):

| Mid-value | Frequency | Cumulative Frequency |
|-----------|-----------|----------------------|
| 115 | 6 | 6 |
| 125 | 25 | 31 |
| 135 | 48 | 79 |
| 145 | 72 | 151 |
| 155 | 116 | 267 |
| 165 | 60 | 327 |
| 175 | 38 | 365 |
| 185 | 22 | 387 |
| 195 | 3 | 390 |

### Step 3: Determine N and N/2
The total number of observations \( N \) is the last cumulative frequency, which is 390. Now, we calculate \( N/2 \):

\[ N/2 = 390 / 2 = 195 \]

### Step 4: Identify the median class
Now we need to find the median class, which is the class interval where the cumulative frequency is greater than or equal to \( N/2 \).

From the cumulative frequency table:
- The cumulative frequency just below 195 is 151 (for the mid-value 145).
- The cumulative frequency just above 195 is 267 (for the mid-value 155).

Thus, the median class is **155** to **165**.

### Step 5: Identify L, CF, F, and H
- **L** (lower limit of the median class) = 155
- **CF** (cumulative frequency of the class preceding the median class) = 151
- **F** (frequency of the median class) = 116
- **H** (class interval width) = 10 (since 165 - 155 = 10)

### Step 6: Apply the median formula
The formula for the median is:

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

Substituting the values:

\[ \text{Median} = 155 + \left( \frac{195 - 151}{116} \right) \times 10 \]

Calculating the values inside the parentheses:

\[ = 155 + \left( \frac{44}{116} \right) \times 10 \]

Calculating \( \frac{44}{116} \):

\[ = 0.3793 \quad (\text{approximately}) \]

Now, multiplying by 10:

\[ = 3.793 \quad (\text{approximately}) \]

Finally, adding to L:

\[ \text{Median} = 155 + 3.793 \approx 158.793 \]

### Final Answer
Thus, the median of the given data is approximately **158.79**.