The following data gives the distribution of height of students :
`{:(underset(("in cm"))"Height",160,150,152,161,156,154,155),(underset("students")"Numbers of ",12,8,4,4,3,3,7):}`
The median of the distribution is

Options

A. 154
B. 155
C. 160
D. 161

Explanation

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

### Step 1: Organize the Data
We will first organize the data into a frequency table with heights and their corresponding frequencies.

| Height (cm) | Frequency |
|-------------|-----------|
| 150 | 8 |
| 152 | 4 |
| 154 | 3 |
| 155 | 7 |
| 156 | 3 |
| 160 | 12 |
| 161 | 4 |

### Step 2: Calculate the Total Frequency
Next, we will calculate the total frequency (N) by summing all the frequencies.

\[ N = 8 + 4 + 3 + 7 + 3 + 12 + 4 = 41 \]

### Step 3: Calculate the Cumulative Frequency
Now, we will calculate the cumulative frequency for each height.

| Height (cm) | Frequency | Cumulative Frequency |
|-------------|-----------|----------------------|
| 150 | 8 | 8 |
| 152 | 4 | 12 |
| 154 | 3 | 15 |
| 155 | 7 | 22 |
| 156 | 3 | 25 |
| 160 | 12 | 37 |
| 161 | 4 | 41 |

### Step 4: Determine the Median Position
Since the total frequency \( N = 41 \) is odd, the median position can be found using the formula:

\[ \text{Median Position} = \frac{N + 1}{2} = \frac{41 + 1}{2} = 21 \]

### Step 5: Locate the Median
Now, we need to find the cumulative frequency that is greater than or equal to 21. From the cumulative frequency table, we see that:

- The cumulative frequency just before 21 is 15 (for height 154 cm).
- The cumulative frequency that is greater than or equal to 21 is 22 (for height 155 cm).

### Step 6: Conclusion
Thus, the median height of the students is:

\[ \text{Median} = 155 \text{ cm} \]