Question

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Numerically verify that the Central Limit Theorem is valid even when sampling from non-normal distributions. Suppose that a component has a probability of failure described by a Weibull distribution. Let X be the random variable that denotes time until failure; its probability density is:

for x ≥ 0, and zero elsewhere. In this problem, assume k = 1.5, γ = 100.
1- Simulate drawing a set of N = 20 sample values, repeated over M = 200 samples, from this component using the inverse sampling method.
2- Plot the histogram of all 4, 000 samples to verify that your samples represent the pdf above.
3- Compute the means associated with each of the M samples and plot a histogram of the sample means. Does this histogram appear to be normally distributed?

Answer 1

The CDF of the distribution is

Fx (x) = | fx (t) dt

1) The inverse function of the CDF is found as

$F_X\left ( x \right )=1-e^{-\left (x/\gamma \right ) ^k}\\ e^{-\left (x/\gamma \right ) ^k}=1-F_X\left ( x \right )\\ -\left (x/\gamma \right ) ^k=\ln \left [1-F_X\left ( x \right ) \right ]\\ x=\gamma \left \{ -\ln \left [1-F_X\left ( x \right ) \right ] \right \}^{1/k}$

Now by inverse transform method
1/k
are random sample from the given Weibull distribution when
U ~Unif (0, 1)
. The R code for generating 4,000 samples from Weibull distribution using inverse sampling is given below.

set.seed(47757)
N <- 4000
U <- runif(N)
k <- 1.5
gamma <- 100
X <- gamma*(-log(1-U))^(1/k)
plot(1:1)
dev.new()
hist(X, freq=FALSE, main= "Histogram of Weibull distribution", lwd=2, col = "sky blue", xlab="x", ylab = "Density")

2) The histogram is plotted below. (R Code as above).

Histogram of Weibull distribution it 髦3 0 100 200 300

3) The R code to generate M=200 samples of N=20 random numbers from Weibull distribution and plotting the histogram of samples means is given below.
set.seed(47757)
N <- 20
M <- 200
k <- 1.5
gamma <- 100
X_bar <- array(dim = M)
for (i in 1:M)
{
U <- runif(N)
X <- gamma*(-log(1-U))^(1/k)
X_bar[i] <- mean(X)
}

plot(1:1)
dev.new()
hist(X_bar, freq=FALSE, main= "Histogram of Sample Means of Weibull Distribution", lwd=2, col = "sky blue", xlab="x", ylab = "Density")

Histogram of Sample Means of Weibull Distribution 60 80 100 120

The histogram looks normally distribution now.