The length of 50 ears of barley (to the nearest millimeter) gave the following frequency distribution .
`{:("Length (mm)",15-19,20-24,25-29,30-34,35-39),("Frequency",1,8,10,16,15):}`
Find the median.

Answer

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

### Step 1: Organize the Data
We have the following frequency distribution:

| Length (mm) | Frequency |
|--------------|-----------|
| 15 - 19 | 1 |
| 20 - 24 | 8 |
| 25 - 29 | 10 |
| 30 - 34 | 16 |
| 35 - 39 | 15 |

### Step 2: Convert to Inclusive Classes
To convert the classes into inclusive classes, we adjust the boundaries:

| Length (mm) | Frequency |
|--------------|-----------|
| 14.5 - 19.5 | 1 |
| 19.5 - 24.5 | 8 |
| 24.5 - 29.5 | 10 |
| 29.5 - 34.5 | 16 |
| 34.5 - 39.5 | 15 |

### Step 3: Calculate Cumulative Frequency
Now, we calculate the cumulative frequency:

| Length (mm) | Frequency | Cumulative Frequency |
|--------------|-----------|----------------------|
| 14.5 - 19.5 | 1 | 1 |
| 19.5 - 24.5 | 8 | 1 + 8 = 9 |
| 24.5 - 29.5 | 10 | 9 + 10 = 19 |
| 29.5 - 34.5 | 16 | 19 + 16 = 35 |
| 34.5 - 39.5 | 15 | 35 + 15 = 50 |

### Step 4: Find Median Class
The total number of observations (n) is 50. To find the median, we need to find the class that contains the 25th observation:

- Since the cumulative frequency just exceeds 25 at the interval 29.5 - 34.5 (cumulative frequency = 35), this is our median class.

### Step 5: Identify Values for Median Formula
- **Median Class (L)**: 29.5
- **n**: 50
- **Cumulative Frequency of Preceding Class (cf)**: 19 (for the class 24.5 - 29.5)
- **Frequency of Median Class (f)**: 16
- **Class Width (h)**: 5 (from 29.5 to 34.5)

### Step 6: Apply the Median Formula
The formula for the median in a grouped frequency distribution is:

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

Substituting the values:

\[ \text{Median} = 29.5 + \left( \frac{50/2 - 19}{16} \right) \times 5 \]

Calculating:

\[ \text{Median} = 29.5 + \left( \frac{25 - 19}{16} \right) \times 5 \]

\[ \text{Median} = 29.5 + \left( \frac{6}{16} \right) \times 5 \]

\[ \text{Median} = 29.5 + \left( 0.375 \right) \times 5 \]

\[ \text{Median} = 29.5 + 1.875 \]

\[ \text{Median} = 31.375 \]

### Final Answer
The median of the given frequency distribution is **31.375 mm**.

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