Question

Problem #3. X is a random variable with an exponential distribution with rate 1 = 3 Thus the pdf of X is f(x) = le-ix for 0 < x where = 3. a) Using the f(x) above and the R integrate function calculate the expected value of X. b) Using the dexp function and the R integrate command calculate the expected value of X. c) Using the pexp function find the probability that .4 SX 5.7 d) Calculate the probability that X > .5. Use the pexp function e) Calculate the probability that X is at least 1 more than its expected value. Use the pexp function f) Copy your R script for the above into the text box here.

Answer 1

R RGUI (32-bit IR Console R File Edit View Misc Packages Windows Help ## part (a) integrand <function (x) x 3*exp (-3*x) k integrate (integrand, lower 0, upper Inf) o.3333333 with absolute error < 8.le-08 part (b) integrand <- function (x) xdexp (x, rate 3, log FALSE) intearate (integrand, lower 0, upper Inf) 0.3333333 with absolute error< 8.le-08 *part (c) pexp (.7, rate 3, log FALSE)-pexp (-4, rate 3, log FALSE) [ 0.1787378 part (d) 1-pexp (.5, rate-3, log FALSE) 0.2231302 part (5) d-o.1.0.33 1-pexp (d, rate 3, log FALSE) 0.9057427 0651 O Type here to search ENG 10-11-2019

## part(a)
integrand <- function(x) {x*3*exp(-3*x)}
integrate(integrand, lower = 0, upper = Inf)
## part(b)
integrand <- function(x) {x*dexp(x, rate = 3, log = FALSE)}
integrate(integrand, lower = 0, upper = Inf)
## part(c)
pexp(.7, rate = 3, log = FALSE)-pexp(.4, rate = 3, log = FALSE)
## part(d)
1-pexp(.5, rate = 3, log = FALSE)
## part(5)
d=0.1*0.33
1-pexp(d, rate = 3, log = FALSE)