0 otherwise. Page 100 Exercise 9: Suppose that a random variable X has a discrete distribution with the following p.m.f.: { c. 2 x. 0 otherwise.

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1 Stat 42 Solutions for Homework Set 4 Page Exercise 5: Suppose that a box contains seven red balls and three blue balls. If five balls are selected at random, without replacement, determine the p.m.f. of the number of red balls that will be obtained. Solution: For x = 2, 3, 4, 5, the probability of obtaining exactly x red balls is ( ) ( ) 7 3 x 5 x ( ) for x = 2, 3, 4, 5 f(x) = 5 Page Exercise 9: Suppose that a random variable X has a discrete distribution with the following p.m.f.: { c f(x) = 2 x for x =,, 2,..., Find the value of the constant c. Solution: We need x= f X(x) =, which means that c = / x=. The last sum is known from Calculus 2 x to equal /( /2) = 2, so c = /2. Page 7 Exercise 6: Let X be a random variable for which the p.d.f. is: { x for x 4, f(x) = 8 After the value of X has been observed, let Y be the integer closest to X. Find the p.m.f. of the random variable Y. { Solution: Let X have the pdf given by: f(x) = 8 x, x 4. Letting Y = integer closest to X, we otherwise have: P r(y = ) = P r(x < /2) = 8 xdx = /2 6 x2 = 64 = /64, P r(y = ) = P r(/2 X < ) = P r(y = 2) = P r( X < ) = P r(y = 3) = P r( X < ) = P r(y = 4) = P r(x ) = 4 /2 8 xdx = 8 xdx = 8 xdx = 8 xdx = 6 x2 6 x2 /2 6 x2 6 x2 4 = = /8, = = /4, = = 3/8, = = 5/64. Page 7 Exercise : Suppose that the p.d.f. of a random variable X is as follows: { c for < x <, f(x) = ( x) /2 a. Find the value of the constant c and sketch the p.d.f. b. Find the value of P r(x /2).

2 Solution: a. We must have: f(x)dx = c ( x) dx = 2c( /2 x) /2 = 2c. Therefore c = /2. f(x) x The R-codes used to generate the above graph is: Figure : Figure for 3.3. x <- seq(.,.99,.) y <- (/2)*(-x)^(-/2) plot(x,y,type="l",xlab="x",ylab=expression(f(x)),col="blue",xlim=c(-.5,.5), ylim=c(-.2,5),bty="n",xaxt="n",yaxt="n") points(,,pch=6,col="blue") arrows(,,.3,,code=2,length=.,col="blue") points(,,pch=6,col="blue") arrows(,,-.3,,code=2,length=.,col="blue") axis(,pos=c(,),at=seq(-.5,.5,.4),lty="dotted") axis(2,at=seq(.,5,),lab=c("",seq(,5,)),lty="dotted") abline(v=,lty="dashed") b. f X (x)dx = 2( x) dx /2 = ( x) /2 /2 = ( /2) /2. 2

3 Page 6 Exercise 4: Suppose that the c.d.f. F of a random variable X is as sketched in Figure 3.9. Find each of the probabilities given. a. P r(x = ) =. since there is a jump of. at x =. b. P r(x < ) =. since P (X < ) = F ( ) =.. c. P r(x ) =.2 since F () = P (X ) =.2. d. P r(x = ) = since there is no discrete jump at x =. e. P r( < X 3) = F (3) F () =.8.2 =.6. f. P r( < X < 3) = F (3 ) F () =.6.2 =.4. g. P r( X 3) = F (3) F ( ) =.8. =.7. h. P r( < X 2) = F (2) F () =.3.3 =. i. P r( X 2) = F (2) F ( ) =.3.3 =. j. P r(x > 5) = P (X 5) = F (5) = =. k. P r(x 5) = P (X < 5) = F (5 ) = =. l. P r(3 X 4) = F (4) F (3 ) =.8.6 =.2. Page 6 Exercise 6: Suppose that the c.d.f. of a random variable X is as follows: F (x) = Find and sketch the pdf of X. Solution: By definition, for X a continuous random variable: f X (x) = d dx F X(x) = d dx ex 3 = e x 3, x 3. The pdf of X is as follows: { e x 3 for x 3 for x > 3 } f(x) x R-codes used to generate the above graphs is: Figure 2: Figure for

4 x <- seq(-3,3,.) y <- exp(x-3)*(x <= 3) + *(x > 3) plot(x,y,type="l",xlab="x",ylab="",xlim=c(-3,4),ylim=c(,.2),bty="n",xaxt="n",yaxt="n",col="blue") axis(,pos=c(,),at=seq(-3,4,),lty="dotted") axis(2,pos=c(,),at=seq(.,.2,.2),lab=c("",seq(.2,.2,.2)),lty="dotted") text(-.6,.7,expression(f(x))) points(3,,pch=6,col="blue") arrows(3,,3.5,,code=2,length=.,col="blue") points(3,,pch=,col="blue") Page 6 Exercise 8: Suppose that a point in the xy-plane is chosen at random from the interior of a circle for which the equation is x 2 + y 2 = ; and suppose that the probability that the point will belong to each region inside the circle is proportional to the area of that region. Let Z denote a random variable representing the distance from the center of the circle to the point. Find and sketch the c.d.f. of Z. Solution: P r(z z) is the probability that Z lies within a circle of radius z centered at the origin. This probability is The c.d.f. is: P r(z z) = Area of circle of radius z Area of circle of radius = πz2 π 2 = z2, for z F(z) z Figure 3: Figure for The R-codes used to generate the above graph is: x <- seq(-.5,.5,.) y <- *(x<)+ x^2*(x <= & x >=) + *(x>) plot(x,y,type="l",xlab="z",ylab="",col="blue",xlim=c(-.,.),ylim=c(-.,.), bty="n",xaxt="n",yaxt="n") axis(,pos=c(,),at=seq(-.2,.2,.2),lab=c("","",seq(.2,.2,.2)),lty="dashed") axis(2,pos=c(,),at=seq(-.2,.2,.2),lab=c("","",seq(.2,.2,.2)),lty="dashed") text(-.,.7,expression(f(z))) 4

5 Page 6 Exercise 2: For the c.d.f. in exercise 6, find the quantile function. Solution: We set F (x) = p and solve for x. e (x 3) = p x 3 = log (p) x = 3 log (p) Thus, the quantile function of X is F (p) = 3 log (p). 5