1. Section 3.6 Recall that y –1/2 e –y 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 <  0  2 edw=  – w 2 / 2 w 2 / 2 0  0  w dw From this, we see that  0  2 e ————— dw = 1   – w 2 / 2

2.  –   2 e ————— dw = 2   – w 2 / 2  –  e ——— dw= 1  2  – w 2 / 2 Let –  < a <  and 0 < b < , and perform the following change of variables in the integral: x = a + bww =dw = < x < < w < A 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. e f(z) =——— for –  < z <   2  – z 2 / 2 We shall come back to this derivation later. Right now skip to the following:

3. 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 We shall discuss these theorems later. Right now go to Class Exercise #1:

4. P(Z < 1.25) =  (1.25) = 0.8944 P(Z > 0.75) = 1 –  (0.75) = 0.2266 P(Z < – 1.25) =  (– 1.25) =1 –  (1.25) = 0.1056 P(Z > – 0.75) = 1 –  (– 0.75) =1 – (1 –  (0.75)) =  (0.75) = 0.7734 1. 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)) = 0.1359 P(Z < 6) =  (6) = practically 1 P(– 2 < Z < – 1) =

5. a constant c such that P(Z < c) = 0.591 P(Z < c) = 0.591  (c) = 0.591 c = 0.23 a constant c such that P(Z < c) = 0.123 P(Z < c) = 0.123  (c) = 0.1231 –  (– c) = 0.123  (– c) = 0.877 – c = 1.16c = – 1.16 a constant c such that P(Z > c) = 0.25 a constant c such that P(Z > c) = 0.90 P(Z > c) = 0.25 1 –  (c) = 0.25c  0.67 P(Z > c) = 0.90 1 –  (c) = 0.90  (– c) = 0.90 – c = 1.28c = – 1.28

6. 1.-continued P(Z > z  ) =   1 –  (z  ) =  z 0.10 = 1.282 z 0.90 = – z 0.10 = – 1.282 a constant c such that P(|Z| < c) = 0.99 P(– c < Z < c) = 0.99P(Z < c) – P(Z < – c) = 0.99  (c) –  (– c) = 0.99  (c) – (1 –  (c)) = 0.99  (c) = 0.995 c = z 0.005 = 2.576 z 0.10 z 0.90 P(Z > z  ) =   1 –  (z  ) =  (–z  ) =   1 –  (–z  ) = 1 –   P(Z > –z  ) = 1 –   z 1–  = –z 

7.  –   2 e ————— dw = 2   – w 2 / 2  –  e ——— dw= 1  2  – w 2 / 2 Let –  < a <  and 0 < b < , and perform the following change of variables in the integral: x = a + bww =dw = < x < < w < (x – a) / b  –   (1/b) dx  –  e ———— dx = 1 b  2  (x – a) 2 – ——— 2b 2 The function of x being integrated can be the p.d.f. for a random variable X which has all real numbers as its space.

8. The moment generating function of X is M(t) = E(e tX ) =  –  e tx e ———— dx = b  2  (x – a) 2 – ——— 2b 2  –  e ———— dx = b  2  (x – a) 2 – 2b 2 tx – —————— 2b 2  –  exp{} —————————— dx b  2  (x – a) 2 – 2b 2 tx – —————— 2b 2 Let us consider the exponent (x – a) 2 – 2b 2 tx – ——————. 2b 2 (x – a) 2 – 2b 2 tx – —————— = 2b 2 x 2 – 2ax + a 2 – 2b 2 tx – ————————— = 2b 2 x 2 – 2(a + b 2 t)x + (a + b 2 t) 2 – 2ab 2 t – b 4 t 2 – —————————————————= 2b 2

9. [x – (a + b 2 t)] 2 – 2ab 2 t – b 4 t 2 – ————————————.Therefore, M(t) = 2b 2  –  exp{} —————————— dx = b  2  (x – a) 2 – 2b 2 tx – —————— 2b 2  –  exp{} —————————— dx = b  2  [x – (a+b 2 t)] 2 – —————— 2b 2 b 2 t 2 exp{at + ——} 2 b 2 t 2 at + —— 2 e b 2 t 2 at + —— 2 efor –  < t <  M (t) = M  (t) = b 2 t 2 at + —— 2 (a + b 2 t) e b 2 t 2 at + —— 2 (a + b 2 t) 2 e+ b 2 t 2 at + —— 2 M(t) = b 2 e

10. E(X) = M (0) =E(X 2 ) = M  (0) = aa 2 + b 2 Var(X) =a 2 + b 2 – a 2 = b 2 Since X has mean  = and variance  2 =, we can write the p.d.f of X as ab2b2 e f(x) =———— for –  < x <   2  (x –  ) 2 – ——— 2  2 A 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. e f(z) =——— for –  < z <   2  – z 2 / 2

11. 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

12. 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 – 10 P( ——— < ——— < ———— ) = 3 3 3 P(– 1.33 < Z < 0.67) =  (0.67) –  (– 1.33) =  (0.67) – (1 –  (1.33)) = 0.7486 – (1 – 0.9082) = 0.6568 P(X > 25) = X – 10 25 – 10 P( ——— > ———— ) = 3 3 P(Z > 5) = 1 –  (5) = practically 0

13. P(|X – 10| < c) = 0.95 X – 10 c P( ——— < — ) = 0.95 3 3 P(|Z| < c/3) = 0.95  (c/3) –  (– c/3) = 0.95  (c/3) – (1 –  (c/3)) = 0.95  (c/3) = 0.975 c/3 = z 0.025 = 1.960c = 5.880 2.-continued a constant c such that P(|X – 10| < c) = 0.95

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

15. the distribution for the random variable Q = X 2 + 14X + 49——— 100 From Theorem 3.6-2, we know that Q = X 2 + 14X + 49 —————— = 100 X + 7 —— 10 2 must have a distribution.  2 (1) 3.-continued