1. Section 3.3 If the space of a random variable X consists of discrete points, then X is said to be a random variable of the discrete type. If the space of a random variable X is a set S consisting of an interval (possibly unbounded) of real numbers or a union of such intervals, then X is said to be a random variable of the continuous type. The probability density function (p.d.f.) of a continuous type random variable X is an integrable function f(x) satisfying the following conditions: (1) (2) (3) f(x) > 0 for x  S (Note: for convenience, we may allow S to contain certain discrete points where f(x) = 0.) f(x) dx = 1 S P(A) = P(X  A) = f(x) dx A i.e, P(a < X < b) =f(x) dx a b

2. Suppose X is a continuous-type random variable with outcome space S and p.d.f f(x). The mean of X is The variance of X is The standard deviation of X is x f(x) dx =E(X) = . (x –  ) 2 f(x) dx =E[(X –  ) 2 ] =  2 = Var(X). E[X 2 – 2X  +  2 ] = E(X 2 ) – 2  E(X) + E(  2 ) = E(X 2 ) – 2  2 +  2 = E(X 2 ) –  2 =  =  Var(X). Recall that for any random variable X, the cumulative distribution function (c.d.f.) of X is defined to be F(x) = P(X  x). If X is a continuous-type random variable, we may write F(x) = S S –– f(t) dt x The Fundamental Theorem of Calculus impliesF / (x) = f(x).

3. For a continuous-type random variable X with outcome space S and p.d.f f(x), we can define the following: The m.g.f. of X (if it exists) is and (just as with discrete-type random variables) The (100p)th percentile of the distribution of X, where 0 < p < 1, is defined to be a number  p such that The median of the distribution of X is defined to be The first, second, and third quartiles of the distribution of X are defined to be M(t) = E(e tX ) = e tx f(x) dx S M (n) (0) = E(X n ).  0.5.  0.25,  0.5, and  0.75 respectively. F(  p ) = p.

4. 1. (a) (b) A random variable Y has p.d.f. f(y) =if 1 < y < 3. Find each of the following: P(2 < Y < 3) 3 – y —— 2 2 3 f(y) dy = 2 3 (3 – y) ——— dy= 2 3y y 2 — – —= 2 4 y = 2 3 1 — 4 P(1.5 < Y < 2.5) 1.5 2.5 f(y) dy = 1.5 2.5 (3 – y) ——— dy= 2 3y y 2 — – —= 2 4 y = 1.5 2.5 1 — 2

5. (c) (d) P(Y < 2) E(Y) –– 2 f(y) dy = 1 2 (3 – y) ——— dy= 2 3y y 2 — – —= 2 4 y = 1 2 3 — 4 ––  y f(y) dy = 1 3 (3 – y) y ——— dy= 2 1 3 (3y – y 2 ) ———— dy = 2 3y 2 y 3 — – —= 4 6 y = 1 3 5 — 3

6. (e)Var(Y) E(Y 2 ) = ––  y 2 f(y) dy = 1 3 (3 – y) y 2 ——— dy= 2 1 3 (3y 2 – y 3 ) ———— dy = 2 3y 3 y 4 — – —= 6 8 y = 1 3 3 Var(Y) = 5 3 – — = 3 2 2 — 9

7. (f) M(t) =E(e tY ) = ––  e ty f(y) dy = 1 3 (3 – y) e ty ——— dy= 2 1 3 (3e ty – ye ty ) ————— dy = 2 y = 1 3 3y y 2 — – —= 2 4 if t = 0 3te ty – (yte ty – e ty ) ——————— = 2t 2 y = 1 3 e 3t – (2t + 1)e t —————— 2t 2 the m.g.f. (moment generating function) of Y 1 if t  0

8. F(y) = P(Y  y) = –– y f(t) dt = 0if y < 1 1if 3  y 1 y (3 – t) ——— dt = 2 3t t 2 — – —= 2 4 t = 1 y 3y y 2 5 — – — – — if 1  y < 3 2 4 4 (g)the c.d.f. (cumulative distribution function) of Y

9. F(  0.25 ) = P(Y   0.25 ) = 0.25 3  2 5 — – — – — = 0.25 2 4 4 6  –  2 – 5 = 1 6  –  2 – 6 = 0  2 – 6  + 6 = 0 From the quadratic formula, we find  = 3 –  3  1.268 or  = 3 +  3  4.732 Considering the space of Y, we see that  0.25 = 3 –  3  1.268 (h)the quartiles of the distribution of Y

10. F(  0.50 ) = P(Y   0.50 ) = 0.5 3  2 5 — – — – — = 0.5 2 4 4 6  –  2 – 5 = 2 6  –  2 – 7 = 0  2 – 6  + 7 = 0 From the quadratic formula, we find  = 3 –  2  1.586 or  = 3 +  2  4.414 Considering the space of Y, we see that  0.50  1.586

11. F(  0.75 ) = P(Y   0.75 ) = 0.75 3  2 5 — – — – — = 0.75 2 4 4 6  –  2 – 5 = 3 6  –  2 – 8 = 0  2 – 6  + 8 = 0 From the quadratic formula, we find  = 2 or  = 4. Considering the space of Y, we see that  0.75 = 2

12. 2. (a) A random variable X has p.d.f. f(x) = Find each of the following: 1/6 if –2 < x < 1 1/2 if 1  x < 2 P(–0.5 < X < 0.5) – 0.5 0.5 f(x) dx = – 0.5 0.5 1 — dx= 6 x — = 6 x = –0.5 0.5 1 — 6

13.  0.75 f(x) dx = 0.75 1 1 — dx+ 6 1 2 1 — dx= 2 x — + 6 x = 0.75 1 x — = 2 x = 1 2 1 — + 24 1 — = 2 13 — 24 (b)P(X > 0.75) 2 0.75 f(x) dx =

14. P(0.75 < X < 1.25) 0.75 1.25 f(x) dx = 0.75 1 1 — dx+ 6 1 1.25 1 — dx= 2 1 — 6 (c)

15. ––  x f(x) dx = – 2 1 x — dx+ 6 1 2 x — dx= 2 x 2 — + 12 x = – 2 1 x 2 — = 4 x = 1 2 –3 — + 12 3 — = 4 1 — 2 (d)E(X)

16. E(X 2 ) = ––  x 2 f(x) dx = – 2 1 x 2 — dx+ 6 1 2 x 2 — dx= 2 x 3 — + 18 x = – 2 1 x 3 — = 6 x = 1 2 1 — + 2 7 — = 6 5 — 3 Var(X) = 5 1 — – — = 3 2 2 17 — 12 (e)Var(X)

17. M(t) = E(e tX ) = ––  e tx f(x) dx = – 2 1 e tx — dx+ 6 1 2 e tx — dx= 2 x — + 6 x = – 2 1 x — = 2 x = 1 2 1 2 e tx —+—= 6t2t x = – 2 1 3e 2t – 2e t – e –2t —————— 6t (f)the m.g.f. (moment generating function) of X if t = 0 if t  0

18. F(x) = P(X  x) = –– x f(t) dt = 0if x < – 2 1if 2  x – 2 x 1 — dt = 6 t — = 6 t = – 2 x x + 2 —— if – 2  x < 1 6 – 2 1 1 — dt + 6 1 x 1 — dt = 2 1 t — +— = 2 t = 1 x x — if 1  x < 2 2 (g)the c.d.f. (cumulative distribution function) of X

19. P(X   0.25 ) = F(  0.25 ) = 0.25  + 2 ——– = 0.25 6  + 2 = 3/2  0.25 = –1/2 Obviously,  0.50 = P(X   0.75 ) = F(  0.75 ) = 0.75  — = 0.75 2  0.75 = 3/2 (h)the quartiles of the distribution of X 1