Among the list `{1, 2, 3, 4, 7, 7, 10, 10, 11, 14, 19, 19, 23, 24, 25, 26}`. what is the ratio of the largest item in Quartile 2 to the average value in Quartile 4?

Answer

To solve the problem, we need to find the ratio of the largest item in Quartile 2 to the average value in Quartile 4 from the given list of numbers.

### Step-by-Step Solution:

1. **Identify the Quartiles**:
The given list is:
`{1, 2, 3, 4, 7, 7, 10, 10, 11, 14, 19, 19, 23, 24, 25, 26}`.
There are 16 items in total. Since we have 16 items, we can divide them into 4 quartiles, each containing 4 items.

- **Quartile 1 (Q1)**: 1, 2, 3, 4
- **Quartile 2 (Q2)**: 7, 7, 10, 10
- **Quartile 3 (Q3)**: 11, 14, 19, 19
- **Quartile 4 (Q4)**: 23, 24, 25, 26

2. **Find the Largest Item in Quartile 2**:
From Quartile 2, the items are: 7, 7, 10, 10.
The largest item is **10**.

3. **Calculate the Average Value in Quartile 4**:
From Quartile 4, the items are: 23, 24, 25, 26.
To find the average, we sum these values and divide by the number of items (4):
\[ \text{Sum} = 23 + 24 + 25 + 26 = 98 \]
\[ \text{Average} = \frac{98}{4} = 24.5 \]

4. **Calculate the Ratio**:
Now we need to find the ratio of the largest item in Quartile 2 (which is 10) to the average value in Quartile 4 (which is 24.5):
\[ \text{Ratio} = \frac{10}{24.5} \]

5. **Simplify the Ratio**:
To simplify \( \frac{10}{24.5} \), we can express 24.5 as a fraction:
\[ 24.5 = \frac{245}{10} \]
Thus, the ratio becomes:
\[ \frac{10}{\frac{245}{10}} = \frac{10 \times 10}{245} = \frac{100}{245} \]
Now, we can simplify \( \frac{100}{245} \) by dividing both the numerator and denominator by 5:
\[ \frac{100 \div 5}{245 \div 5} = \frac{20}{49} \]

### Final Answer:
The required ratio is \( \frac{20}{49} \).