Let 𝑋1,𝑋2,…,𝑋500∼i.i.d Normal(0,1)X 1​ ,X 2​ ,…,X 500​ ∼i.i.d Normal(0,1). Evaluate 𝑃(𝑋12+𝑋22+…+𝑋5002>570)P(X 12​ +X 22​ +…+X 5002​ >570) using Central Limit theorem. Enter the answer correct to 2 decimal places.Hint:(𝑋12+𝑋22+…+𝑋5002)∼ Gamma(250,0.5).Hint:(X 12​ +X 22​ +…+X 5002​ )∼ Gamma(250,0.5).

The Central Limit Theorem (CLT) states that the sum of a large number of independent and identically distributed (i.i.d.) random variables will be approximately normally distributed. However, in this case, we are dealing with the sum of the squares of i.i.d. normal random variables, which follows a chi-square distribution, not a normal distribution.

The hint provided in the question tells us that the sum of the squares of these variables follows a Gamma distribution with parameters 250 and 0.5. The Gamma distribution is a two-parameter family of continuous probability distributions, which includes the exponential and chi-square distributions as special cases.

The problem asks us to find the probability that the sum of the squares of these variables is greater than 570. This is equivalent to finding the cumulative distribution function (CDF) of the Gamma distribution at 570, and subtracting this value from 1.

Unfortunately, the Central Limit Theorem cannot be directly applied here, because it applies to the sum of i.i.d. variables, not the sum of their squares. However, we can use the properties of the Gamma distribution to solve this problem.

The CDF of a Gamma distribution with parameters k and θ is given by:

F(x; k, θ) = 1 - Γ(k, x / θ) / Γ(k)

where Γ(k, x / θ) is the incomplete gamma function, and Γ(k) is the gamma function.

So, we need to calculate:

P(X1^2 + X2^2 + ... + X500^2 > 570) = 1 - F(570; 250, 0.5)

This calculation involves special mathematical functions and cannot be done by hand, but it can be done using statistical software or a scientific calculator that includes these functions. The result should be rounded to two decimal places.