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Recall that

(Multivariable Calculus is required to prove this!)

(1/2) =

y–1/2e–y dy =

0

Perform the following change of variables in the integral:

w = 2yy = dy =

< w < < y <

w dw

w2 / 2

0

0

– w2 / 2

2 e dw =

0

– w2 / 2

2 e

————— dw = 1

From this, we see that

0

– w2 / 2

– w2 / 2

2 e

————— dw = 2

e

——— dw = 1

2

–

–

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

x = a + bww = dw =

< x < < w <

We shall come back to this derivation later.

Right now skip to the following:

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.

– z2 / 2

e

f(z) = ——— for –< z <

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

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

1. The random variable Z is N(0, 1). Find each of the following:

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

P(– 1 < Z < 2) =

(2)– (– 1) =

(2)– (1 – (1)) =

0.8185

P(– 2 < Z < – 1) =

(– 1)– (– 2) =

(1 – (1))– (1 – (2)) =

0.1359

P(Z < 6) =

(6) =

practically 1

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) = 0.123

1 – (– c) = 0.123

(– c) = 0.877

– c = 1.16

c = –1.16

a constant c such that P(Z > c) = 0.25

P(Z > c) = 0.25

1 – (c) = 0.25

c 0.67

a constant c such that P(Z > c) = 0.90

P(Z > c) = 0.90

1 – (c) = 0.90

(– c) = 0.90

– c = 1.28

c = –1.28

z0.10

P(Z > z) = 1 – (z) =

z0.10 = 1.282

z0.90

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

P(Z > –z) = 1 – z1– = –z

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

– w2 / 2

– w2 / 2

2 e

————— dw = 2

e

——— dw = 1

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

–

–

(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.

e

———— dx = 1

b2

–

The moment generating function of X is M(t) = E(etX) =

(x – a)2

– ———

2b2

(x – a)2 –2b2tx

– ——————

2b2

etx e

———— dx =

b2

e

———— dx =

b2

–

–

Let us consider the exponent

(x – a)2 –2b2tx

– ——————

2b2

exp{ }

—————————— dx

b2

(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

– ———————————— . Therefore, M(t) =

2b2

(x – a)2 –2b2tx

– ——————

2b2

exp{ }

—————————— dx =

b2

–

[x –(a+b2t)]2

– ——————

2b2

b2t2

exp{at + ——}

2

b2t2

at + ——

2

exp{ }

—————————— dx =

b2

e

–

b2t2

at + ——

2

M(t) =

e for –< t <

b2t2

at + ——

2

M(t) =

(a + b2t) e

b2t2

at + ——

2

b2t2

at + ——

2

M(t) =

(a + b2t)2e +

b2 e

a

E(X2) = M(0) =

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

(x –)2

– ———

22

e

f(x) = ———— for –< x <

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.

– z2 / 2

e

f(z) = ——— for –< z <

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

The random variable X is N(10, 9). Use Theorem 3.6-1 to find each of the following:

6 – 10 X– 10 12 – 10

P( ——— < ——— < ———— ) =

3 3 3

P(6 < X < 12) =

P(– 1.33 < Z < 0.67) =

(0.67)– (– 1.33) =

(0.67)– (1 – (1.33)) =

0.7486 – (1 – 0.9082) = 0.6568

X– 10 25 – 10

P( ——— > ———— ) =

3 3

P(X > 25) =

P(Z > 5) =

1 – (5) =

practically 0

a constant c such that P(|X – 10| < c) = 0.95

X– 10 c

P( ——— < — ) = 0.95

3 3

P(|X – 10| < c) = 0.95

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

The random variable X is N(–7, 100). Find each of the following:

X+ 7 0 + 7

P( ——— > —— ) =

10 10

P(X > 0) =

P(Z > 0.7) =

1 – (0.7) =

0.2420

a constant c such that P(X > c) = 0.98

X+ 7 c +7

P( —— > —— ) = 0.98

10 10

P(X > c) = 0.98

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

X2+ 14X + 49

——————

100

the distribution for the random variable Q =

X2+ 14X + 49

—————— =

100

From Theorem 3.6-2, we know that Q =

2

X + 7

——

10

must have a distribution.

2(1)

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