Question

Calculation to be done using R

Suppose X has a standard normal distribution. The pnorm(), norm(), and qnorm() functions should be useful in the following. Remember, if X has a standard normal distribution then E(X) = 0 and Var(x)=1. a) What is the expected value of b) What is the variance of Xt2 ? c) What is the expected value of x6? d) Using R, calculate the probability X > .83) e) Using R, calculate the probability that -0.2 < X < 1.1 f) Using R, calculate the probability that 0 g) If the probability that X > t = .29 then what is t?

Answer 1

a)we have $E(aX&plus;b)=aE(X)&plus;b$ .Here E(X)=0

hence $E(\frac{X&plus;1}{3})=\frac{1}{3}(E(X)&plus;1)=\frac{1}{3}(0&plus;1)=\frac{1}{3}$

b)we have $V(aX&plus;b)=a^{2}V(X)$ .Here V(X)=1

Hence $V(\frac{X&plus;1}{3})=\frac{1}{3}^{2}V(X&plus;1)=\frac{1}{9}V(X)=\frac{1}{9}$

c)we have $f(X)=\frac{1}{\sqrt{2\Pi }}e^{\frac{-X^{2}}{2}} ,-\infty <X<\infty$

$E(X^{6})=\int_{-\infty }^{\infty }X^{6}\frac{1}{\sqrt{2\Pi }}e^{\frac{-X^{2}}{2}}dX= \frac{1}{\sqrt{2\Pi }}\int_{-\infty }^{\infty }X^{6}e^{\frac{-X^{2}}{2}}dX$

=$\frac{1}{\sqrt{2\Pi }}2\int_{0}^{\infty }X^{6}e^{\frac{-X^{2}}{2}}dX$(as the function is even)

=$\frac{2}{\sqrt{2\Pi }}\frac{1}2{}\frac{Gamma(\frac{6&plus;1}{2})}{\frac{1}{2}^{\frac{6&plus;1}{2}}}$since $\int_{0}^{\infty }X^{n}e^{-ax^{m}}dx=\frac{1}{m}\frac{Gamma(\frac{n&plus;1}{m})}{a^{\frac{n&plus;1}{m}}}$

and Gamma(n)=(n-1)Gamma(n-1) and Gamma(1/2)=$\sqrt{\Pi }$

=$\frac{1}{\sqrt{2\Pi }}\frac{gamma(7/2)}{(1/2)^{7/2}}=\frac{1}{\sqrt{2\Pi }}\frac{5/2*3/2*1/2*\sqrt{\Pi }}{\frac{1}{8\sqrt{2}}}=15$

d)give the command >pnorm(0.83,mean=0,s.d=1)

it gives the probability that X<0.83

(1- above value) gives the required answer