Question

I need a function in R or just code in R that will solve this. (Thanks in advance)

1. The probability density function for Exponentiated Weibull distribution is and its corresponding cdf is given by F(x) 1-exp(x) This distribution is a popular distribution in analyzing lifetime data. a) b) Find the mean and variance when-10, α 2 and β-4 Compute the probability when the random variable X is between two standard deviation of the sample mean, where λ-10, α-2 and β-4

Answer 1

a) The R code for finding the mean and variance of the given distribution is given below.

lamda <- 10
alpha <- 2
beta <- 4

integrand1 <- function (x)
{
return(x*alpha*beta*lamda^beta*x^(beta-1)*(1-exp(-(lamda*x)^beta))^(alpha-1)*exp(-(lamda*x)^beta))
}
integrand2 <- function (x)
{
return(x^2*alpha*beta*lamda^beta*x^(beta-1)*(1-exp(-(lamda*x)^beta))^(alpha-1)*exp(-(lamda*x)^beta))
}

meanX <- integrate(integrand1, lower = 0, upper = Inf)$value
meanX
varX <- integrate(integrand2, lower = 0, upper = Inf)$value- meanX ^2
stdX <- sqrt(varX)
stdX
The computed mean is ${\color{Blue} \mu =0.1051}$ and standard deviation is  ${\color{Blue} \sigma =0.0205}$

b) The required probabilty

${\color{Blue} P\left ( \mu -2\sigma <X<\mu &plus;2\sigma \right )=0.9549}$
R code below.

lamda <- 10
alpha <- 2
beta <- 4
integrand3 <- function (x)
{
return(alpha*beta*lamda^beta*x^(beta-1)*(1-exp(-(lamda*x)^beta))^(alpha-1)*exp(-(lamda*x)^beta))
}
pr <- integrate(integrand3, lower = meanX - 2*stdX, upper = meanX + 2*stdX)$value
pr