Question 1164809
To find the distribution of the statistic , we first need to evaluate the properties of the numerator and denominator based on the fact that  are i.i.d. .

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### 1. Analyze the Numerator ()

Let .
Since  and  are independent normal variables:

* **Mean:** 
* **Variance:** 
Therefore, . We can standardize this by noting that .

### 2. Analyze the Denominator

There appears to be a slight typo in your expression ( square). Assuming the statistic is meant to be , or a similar variation involving independent variables, let's look at the standard form of a **Student's t-distribution**.

A t-distribution is formed by:



where , , and  and  are independent.

### 3. Re-evaluating the specific statistic

If the statistic is exactly :

1. **Numerator:** , where .
2. **Denominator:**  is the sum of the squares of 2 independent standard normal variables, so .

Substituting these into :



Divide both the numerator and denominator by  (the square root of the degrees of freedom ):


### 4. Conclusion

By definition,  follows a **Student's t-distribution with 2 degrees of freedom**.

**The distribution of the statistic is .**

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### If the statistic was  specifically:

This would be more complex because  appears in both the numerator and denominator, meaning they are **not independent**. In such cases, the distribution would not follow a standard t-distribution.

However, in most textbook problems of this type, the variables in the denominator are independent of those in the numerator. Would you like me to show the derivation for a different combination of these variables?