Question

The density of random variable X is f(x) = 5(Xº+1)(3-X) for 1<x<3 and 0 otherwise. Using the R integrate function: 68 a) Find the probability that X > 2.10 b) Find the probability that 1.5 < X < 2.5 c) Find the expected value of x d) Find the standard deviation of X e) In the following paste your R script for this problem

Answer 1

#Ans a) P(X>2.10)=0.4512593

  b) P(1.5<x<2.5)=0.6240809

c)  expected value of x=2

d) standard deviation of X =0.4936568

#e=R-script

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> f=function(x){5*(x^3+1)*(3-x)/68}
> #a=P(X>2.10)
> a=integrate(f,2.10,3);a
0.4512593 with absolute error < 5e-15
> #b=P(1.5<x<2.5)
> b=integrate(f,1.5,2.5);b
0.6240809 with absolute error < 6.9e-15
> #ux=expected value of x
> k=function(x){x*5*(x^3+1)*(3-x)/68}
> integrate(k,1,3)
2 with absolute error < 2.2e-14
> ux=2
> #ux2=expected value of x2
> d=function(x){x^2*5*(x^3+1)*(3-x)/68}
> ux2=integrate(d,1,3);ux2
4.243697 with absolute error < 4.7e-14
> ux2=4.243697
> var=ux2-ux^2;var
[1] 0.243697
> #sd=standard deviation
> sd=sqrt(var);sd
[1] 0.4936568