Section 3.6

Recall thaty1/2 ey dy = 0(Multivariable Calculus is required to prove this!)(1/2) =Perform the following change of variables in the integral:w = 2y y =dy = < w < < y

2 e dw = 2 w2 / 2 e dw= 12 w2 / 2Let < a < and 0 < b < , and perform the following change of variables in the integral:x = a + bww =dw = < x < < w

We let (z) = P(Z z), the distribution function of Z. Since f(z) is a symmetric function, it is easy to see that ( z) = 1 (z). Tables Va and Vb in Appendix B of the text display a graph of f(z) and values of (z) and ( z).Important Theorems in the Text:If X is N(,2), then Z = (X ) / is N(0,1). Theorem 3.6-1

If X is N(,2), then V = [(X ) / ]2 is 2(1). Theorem 3.6-2We shall discuss these theorems later. Right now go to Class Exercise #1:

P(Z < 1.25) =(1.25) =0.8944P(Z > 0.75) =1 (0.75) =0.2266P(Z < 1.25) =( 1.25) =1 (1.25) =0.1056P(Z > 0.75) =1 ( 0.75) =1 (1 (0.75)) = (0.75) =0.77341. The random variable Z is N(0, 1). Find each of the following:P( 1 < Z < 2) =(2) ( 1) =(2) (1 (1)) =0.8185( 1) ( 2) =(1 (1)) (1 (2)) =0.1359P(Z < 6) =(6) =practically 1P( 2 < Z < 1) =

a constant c such that P(Z < c) = 0.591P(Z < c) = 0.591(c) = 0.591c = 0.23a constant c such that P(Z < c) = 0.123P(Z < c) = 0.123(c) = 0.1231 ( c) = 0.123( c) = 0.877 c = 1.16c = 1.16a constant c such that P(Z > c) = 0.25a constant c such that P(Z > c) = 0.90P(Z > c) = 0.251 (c) = 0.25c 0.67P(Z > c) = 0.901 (c) = 0.90( c) = 0.90 c = 1.28c = 1.28

1.-continuedP(Z > z) = 1 (z) = z0.10 = 1.282z0.90 = z0.10 = 1.282a constant c such that P(|Z| < c) = 0.99P( c < Z < c) = 0.99P(Z < c) P(Z < c) = 0.99(c) ( c) = 0.99(c) (1 (c)) = 0.99(c) = 0.995c = z0.005 = 2.576z0.10z0.90P(Z > z) = 1 (z) = (z) = 1 (z) = 1 P(Z > z) = 1 z1 = z

2 e dw = 2 w2 / 2 e dw= 12 w2 / 2Let < a < and 0 < b < , and perform the following change of variables in the integral:x = a + bww =dw = < x < < w

The moment generating function of X is M(t) = E(etX) = etx e dx = b2 (x a)2 2b2 e dx = b2 (x a)2 2b2tx 2b2 exp{} dx b2 (x a)2 2b2tx 2b2Let us consider the exponent (x a)2 2b2tx . 2b2 (x a)2 2b2tx = 2b2 x2 2ax + a2 2b2tx = 2b2 x2 2(a + b2t)x + (a + b2t)2 2ab2t b4t2 =2b2

[x (a + b2t)]2 2ab2t b4t2 .Therefore, M(t) =2b2 exp{} dx = b2 (x a)2 2b2tx 2b2 exp{} dx = b2 [x (a+b2t)]2 2b2 b2t2exp{at + } 2 b2t2at + 2e b2t2at + 2efor < t < M (t) =M (t) = b2t2at + 2(a + b2t) e b2t2at + 2(a + b2t)2 e+ b2t2at + 2M(t) =b2 e

E(X) = M (0) =E(X2) = M (0) =aa2 + b2Var(X) =a2 + b2 a2 = b2Since X has mean = and variance 2 = , we can write the p.d.f of X asab2ef(x) = for < x < 2 (x )2 22A random variable having this p.d.f. is said to have a normal distribution with mean and variance 2, that is, a N(,2) distribution.A random variable Z having a N(0,1) distribution, called a standard normal distribution, has p.d.f.ef(z) = for < z < 2 z2 / 2

We let (z) = P(Z z), the distribution function of Z. Since f(z) is a symmetric function, it is easy to see that ( z) = 1 (z). Tables Va and Vb in Appendix B of the text display a graph of f(z) and values of (z) and ( z).Important Theorems in the Text:If X is N(,2), then Z = (X ) / is N(0,1). Theorem 3.6-1

If X is N(,2), then V = [(X ) / ]2 is 2(1). Theorem 3.6-2

2.The random variable X is N(10, 9). Use Theorem 3.6-1 to find each of the following:P(6 < X < 12) = 6 10 X 10 12 10P( < < ) = 3 3 3P( 1.33 < Z < 0.67) =(0.67) ( 1.33) =(0.67) (1 (1.33)) =0.7486 (1 0.9082) = 0.6568P(X > 25) = X 10 25 10P( > ) = 3 3P(Z > 5) =1 (5) =practically 0

P(|X 10| < c) = 0.95 X 10 cP( < ) = 0.95 3 3P(|Z| < c/3) = 0.95(c/3) ( c/3) = 0.95(c/3) (1 (c/3)) = 0.95(c/3) = 0.975c/3 = z0.025 = 1.960c = 5.8802.-continueda constant c such that P(|X 10| < c) = 0.95

3.The random variable X is N(7, 100). Find each of the following:P(X > 0) = X + 7 0 + 7P( > ) = 10 10P(Z > 0.7) =1 (0.7) =0.2420a constant c such that P(X > c) = 0.98P(X > c) = 0.98 X + 7 c +7P( > ) = 0.98 10 10P(Z > (c+7) / 10) = 0.981 ((c+7) /10) = 0.98((c+7) /10) = 0.02(c+7) /10 = z0.98 = z0.02 = 2.054c = 27.54

the distribution for the random variable Q = X2 + 14X + 49 100From Theorem 3.6-2, we know that Q = X2 + 14X + 49 = 100X + 7 102must have a distribution.2(1)3.-continued