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 e dw = – w2 / 2

w2 / 2

0 0

w dw

From this, we see that

0

2 e————— dw = 1

– w2 / 2

–

2 e————— dw = 2

– w2 / 2

–

e——— dw = 12

– w2 / 2

Let – < a < and 0 < b < , and perform the following change of variables in the integral:

x = a + bw w = 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.

ef(z) = ——— for – < z <

2

– z2 / 2

We shall come back to this derivation later.

Right now skip to the following:

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:

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) =

a constant c such that P(Z < c) = 0.591P(Z < c) = 0.591 (c) = 0.591 c = 0.23

a constant c such that P(Z < c) = 0.123P(Z < c) = 0.123 (c) = 0.123 1 – (– c) = 0.123

(– c) = 0.877 – c = 1.16 c = – 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) = 0.25 c 0.67

P(Z > c) = 0.90 1 – (c) = 0.90 (– c) = 0.90

– c = 1.28 c = – 1.28

1.-continued

P(Z > z) = 1 – (z) = z0.10 = 1.282

z0.90 = – z0.10 = – 1.282

a constant c such that P(|Z| < c) = 0.99

P(– c < Z < c) = 0.99 P(Z < c) – P(Z < – c) = 0.99

(c) – (– c) = 0.99 (c) – (1 – (c)) = 0.99

(c) = 0.995 c = z0.005 = 2.576

z0.10

z0.90

P(Z > z) = 1 – (z) = (–z) = 1 – (–z) = 1 –

P(Z > –z) = 1 – z1– = –z

–

2 e————— dw = 2

– w2 / 2

–

e——— dw = 12

– w2 / 2

Let – < a < and 0 < b < , and perform the following change of variables in the integral:

x = a + bw w = dw =

< x < < w <

(x – a) / b

– –

(1/b) dx

–

e———— dx = 1 b2

(x – a)2

– ——— 2b2

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.

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– —————— 2b2

Let 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

b2t2

exp{at + ——} 2

b2t2

at + —— 2e

b2t2

at + —— 2e for – < t <

M (t) =

M (t) =

b2t2

at + —— 2(a + b2t) e

b2t2

at + —— 2(a + b2t)2 e +

b2t2

at + —— 2

M(t) =

b2 e

E(X) = M (0) = E(X2) = M (0) =a a2 + b2

Var(X) = a2 + b2 – a2 = b2

Since X has mean = and variance 2 = , we can write the p.d.f of X as

a b2

ef(x) = ———— for – < x <

2

(x – )2

– ——— 22

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.

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 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 – 10P( ——— > ———— ) = 3 3

P(Z > 5) =

1 – (5) = practically 0

P(|X – 10| < c) = 0.95 X – 10 cP( ——— < — ) = 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 = z0.025 = 1.960 c = 5.880

2.-continued

a 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 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 +7P( —— > —— ) = 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 = z0.98 = – z0.02 = – 2.054

c = – 27.54

the distribution for the random variable Q = X2 + 14X + 49—————— 100

From Theorem 3.6-2, we know that Q = X2 + 14X + 49—————— = 100

X + 7—— 10

2

must have a distribution.2(1)

3.-continued