Find the first decile for the following data:
`{:("Size", 0.10,10-20,20-30,30-40,40-50,50-60,60-70,70-80),("Frequency "," "2," "18," "30, " "45, " "35, " "20, " "6, " "3):}`

Answer

To find the first decile (D1) for the given data, we will follow these steps:

### Step 1: Organize the Data
We have the following size intervals and their corresponding frequencies:

| Size Interval | Frequency |
|---------------|-----------|
| 0-10 | 2 |
| 10-20 | 18 |
| 20-30 | 30 |
| 30-40 | 45 |
| 40-50 | 35 |
| 50-60 | 20 |
| 60-70 | 6 |
| 70-80 | 3 |

### Step 2: Calculate Cumulative Frequency
Now, we will calculate the cumulative frequency for each interval.

| Size Interval | Frequency | Cumulative Frequency |
|---------------|-----------|----------------------|
| 0-10 | 2 | 2 |
| 10-20 | 18 | 20 |
| 20-30 | 30 | 50 |
| 30-40 | 45 | 95 |
| 40-50 | 35 | 130 |
| 50-60 | 20 | 150 |
| 60-70 | 6 | 156 |
| 70-80 | 3 | 159 |

### Step 3: Find Total Frequency (n)
The total frequency \( n \) is the last cumulative frequency, which is 159.

### Step 4: Calculate the Position of the First Decile (D1)
The formula for finding the position of the first decile is given by:

\[ D_1 = \frac{n + 1}{10} \]

Substituting the value of \( n \):

\[ D_1 = \frac{159 + 1}{10} = \frac{160}{10} = 16 \]

### Step 5: Identify the Class Interval for D1
Now, we need to find the cumulative frequency that is just greater than or equal to 16. From our cumulative frequency table, we see:

- The cumulative frequency for the interval 0-10 is 2.
- The cumulative frequency for the interval 10-20 is 20.

Since 16 lies between 2 and 20, the 16th term falls in the interval 10-20.

### Step 6: Determine the Values for the Decile Formula
For the interval 10-20:
- Lower boundary \( l = 10 \)
- Class width \( h = 20 - 10 = 10 \)
- Frequency \( f = 18 \) (frequency of the interval 10-20)
- Cumulative frequency just before this interval \( CF = 2 \)

### Step 7: Apply the Decile Formula
The formula for calculating the decile is:

\[ D_1 = l + \left( \frac{(D - CF)}{f} \right) \times h \]

Where:
- \( D = 16 \)
- \( CF = 2 \)
- \( f = 18 \)
- \( h = 10 \)

Substituting the values:

\[ D_1 = 10 + \left( \frac{(16 - 2)}{18} \right) \times 10 \]
\[ D_1 = 10 + \left( \frac{14}{18} \right) \times 10 \]
\[ D_1 = 10 + \left( \frac{7}{9} \right) \times 10 \]
\[ D_1 = 10 + \frac{70}{9} \]
\[ D_1 = 10 + 7.78 \approx 17.78 \]

### Final Answer
Thus, the first decile \( D_1 \) is approximately **17.78**.

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