Question

The random variable X has probability density function k(x25x-4) 1<x<4 otherwise -{ f(x) 1. Show thatk. (5pts) Find 2. Е (X), (5pts) 3. the mode of X, (5pts) 4. the cumulative distribution function F(X) for all x. (5pts) 5. Evaluate P(X < 2.5). (5pts) 6. Deduce the value of the median and comment on the shape of the distribution (10pts)

Answer 1

Thegiven PDF is

F(x) k(-x+5x- 4); 1x4

1) The condition for PDF is

4 f(x)dx 1 1 4 k(-x25x- 4)dx 1 k-x3/35x2/2-4x] 1 1 2 2 k OINN/o

2) The expected value is

4 E(X)= xf (x)dx 4 kx-x25x-4)dx E(X) E(X) k(-x4/4 5x3/3 2x211 2 45 E(X) 6 E(X) 2.5

3) The mode is such that

Xmode

f'(Xmode) = 0 k(-2xmode +5) 0 Xmode 2.5

4) The CDF is

Fx(x) f(t)dt 1 k(-t2 5t 4)dt Fx(x) = Fx(x) k(-t3/35t2/2 4t] x3 1 5(x2- 1) 4(x-1)) Fx(x) k 3 2 5x2 4Xt x3 2 11 ;1 x 4 6 Fx(x) 9 3 2

5) The probability,

P(X<2.5) Fx(2.5) 2.53 5(2.5)_4(2.5) 2 P(X2.5) 11 - 3 2 6 1 P(X 2.5)= 2 NaIN

6) The median is such that $F_X(x_m)=0.5$ .

From part (5), we see that

Xm 2.5

The shape of the distribution is an inverted parabola, symmetric about the lin .

X2.5

Answer 2

Thegiven PDF is

F(x) k(-x+5x- 4); 1x4

1) The condition for PDF is

4 f(x)dx 1 1 4 k(-x25x- 4)dx 1 k-x3/35x2/2-4x] 1 1 2 2 k OINN/o

2) The expected value is

4 E(X)= xf (x)dx 4 kx-x25x-4)dx E(X) E(X) k(-x4/4 5x3/3 2x211 2 45 E(X) 6 E(X) 2.5

3) The mode is such that

Xmode

f'(Xmode) = 0 k(-2xmode +5) 0 Xmode 2.5

4) The CDF is

Fx(x) f(t)dt 1 k(-t2 5t 4)dt Fx(x) = Fx(x) k(-t3/35t2/2 4t] x3 1 5(x2- 1) 4(x-1)) Fx(x) k 3 2 5x2 4Xt x3 2 11 ;1 x 4 6 Fx(x) 9 3 2

5) The probability,

P(X<2.5) Fx(2.5) 2.53 5(2.5)_4(2.5) 2 P(X2.5) 11 - 3 2 6 1 P(X 2.5)= 2 NaIN

6) The median is such that $F_X(x_m)=0.5$ .

From part (5), we see that

Xm 2.5

The shape of the distribution is an inverted parabola, symmetric about the lin .

X2.5