Let X1,X2,…,Xn be i.i.d. random variables with mean μ and variance σ2 . Denote the sample mean by X¯¯¯¯n=∑ni=1Xin .
Assume that n is large enough that the central limit theorem (clt) holds. Find a random variable Z with approximate distribution N(0,1) , in terms of X¯¯¯¯n , n , μ and σ . (Note that μ and σ2 refers to the mean and variance of Xi , not X¯¯¯¯n .)
It is the sample mean minus the mean mu divided by the sample variance
(barX_n-mu)/(sigma/sqrt(n))
continuation of the above question:
Z∼N(0,1) for Z=
Correction: divided by sample standard deviation
To find a random variable Z with an approximate distribution N(0,1), we can make use of the Central Limit Theorem (CLT). According to the CLT, for a large enough sample size n, the distribution of the sample mean X̄n approaches a normal distribution with mean μ and standard deviation σ/√n.
Since we are interested in finding a random variable Z with a standard normal distribution N(0,1), we need to standardize X̄n.
The standardization formula is given by:
Z = (X̄n - μ) / (σ/√n)
Explanation of formula:
1. Subtract the mean μ from the sample mean X̄n. We do this to center the distribution at zero.
2. Divide the result by the standard deviation of the sample mean, which is σ/√n. This step scales the distribution.
Now, using the given information, we substitute X̄n, σ, μ, and n into the standardization formula:
Z = (X̄n - μ) / (σ/√n)
Therefore, the random variable Z with an approximate distribution N(0,1) can be found by standardizing X̄n using the formula Z = (X̄n - μ) / (σ/√n).