1 Suppose the function fx C sin z is a probability density

1. Suppose the function f(x) = C sin z is a probability density function for a random variable X on the interval [0, pi]. (a) Find the value of C. (b) Find P(X

Solution

P = C sinx

Then Pdx from x=0 to x=pi =1

= C sinx dx from x=0 to x=pi

= [C cosx] from x=pi to x=0

-2C = 1

C =-1/2 (ans)

b) P(X<=1/4) = Csinx dx from x=0 to pi/4

-1/2 sin x dx = [1/2 cosx ] x=0 to pi/4 = 0.146 (ans)

P( pi/4 < X <3pi/4) = -1/2 sin x dx from pi/4 to 3pi/4

1/2 cos(3pi/4) - 1/2 cos(pi/4) = -1/sqrt2 = 0.707 (ans)

c) Mean = Csinxdx from x=0 to pi/pi = 1/pi =0.318 (ans)

Median = 2/pi = 0.636 (ans)

2) Cx(10-x) dx from x=0 to 10 =1

C*(5x^2 - x^3/3) from x=0 to x=10 = 1

C(500/3) =1

C =3/500 = 0.006 (ans)

b)P(X>=2) = Cx(10-x) from x=2 to 10

C[5x^2 - x^3/3] = C(500/3 - 52/3] = 0.006*448/3 = 0.896 (ans)