Question

pter 3- ntro et on to statistics and 5. A continuous random variable X that can assume values between x-1 and x 3 has a density function given by fx)-1/2 a. Show that the area under the curve is equal to 1. b. Find P(2 <X<2.5). Find P(X S 1.6).

Answer 1

5.

a.

Area under the curve,

$\int_{1}^{3}f(x)~dx = \int_{1}^{3}(1/2)~dx = \left [ x/2 \right ]^3_1 = (3/2) - (1/2) = 1$

b.

P(2 < X < 2.5) =

f(z) dr (1/2) dr = Įr/2]2.52 = (2.5/2)-(2/2) = 0.25

c.

P(X $\le$ 1.6) =

r1.6 1.6

8.

P(1 < Y < 3 | X = 1) = P(X = 1, 1 < Y < 3 ) / P(X = 1)

$f(x) = \int_{2}^{4} (6 - x - y)/8 ~dy = \left [ (6y - xy - y^2/2)/8 \right ]^4_2$

(24 - 4r - 16/2)/8- (12 - 2r - 4/2)/8 (2 - x/2) - (5/4- r/4)

(3- x)/4

P(X = 1) = f(1) = (3 - 1)/4 = 1/2

Now,

P(X = x, 1 < Y < 3 ) =

$= \int_{1}^{3} (6 - x - y)/8 ~dy = \left [ (6y - xy - y^2/2)/8 \right ]^3_1$

P(X = 1, 1 < Y < 3 ) = 1 - 1/4 = 3/4

P(1 < Y < 3 | X = 1) = P(X = 1, 1 < Y < 3 ) / P(X = 1)

= (3/4) / (1/2)

= 3/2