Let X₁, X₂, ... be i.i.d. random variables having a χ²-distribution with 5 degrees of freedom.
Let a ∈ \(\mathbb{R}\) be constant. Then the limiting distribution of \(a\left(\frac{X_1+\cdots+X_n-5 n}{\sqrt{n}}\right)\) is 

Given:-

X1, X2, ... are i.i.d. random variables having a χ2-distribution with 5 degrees of freedom.

Concept Used:-

The limiting distribution of the given expression can be found using the central limit theorem.

The central limit theorem states that the sum of many independent and identically distributed random variables, properly normalized, converges in distribution to a normal distribution.

Explanation:-

Here, we have n i.i.d. random variables with a χ²-distribution with 5 degrees of freedom.

The mean of each χ²-distributed variable is 5 and the variance is,

2 × 5 = 10

Therefore, the mean of the sum of n such variables is n5, and the variance is,

⇒ variance = (n × 10)

We can normalize the expression by subtracting the mean and dividing by the standard deviation. That is,

\(⇒a[\dfrac{(X_1 + X_ 2 + ... + X_ n - 5n)}{\sqrt{n×10}}] \)

\(=(\dfrac{a}{\sqrt{10}}) [\dfrac{(X_ 1 + X_ 2 + ... + X_ n - 5n)}{n}] \sqrt{n} \)

The term in the brackets on the right-hand side is the sum of n i.i.d. random variables with a mean of 0 and a variance of 1/2.

Therefore, by the CLT, this term converges in distribution to a standard normal distribution as n goes to infinity.

The overall expression converges in distribution to a normal distribution with mean zero and variance a²/10.

So, the limiting distribution of \(a\left(\frac{X_1+\cdots+X_n-5 n}{\sqrt{n}}\right)\) is ​​the standard normal distribution for an appropriate value of a.

Hence, the correct option is 3.