section 3.6 recall that y –1/2 e –y dy = 0 (multivariable calculus is required to prove this!)...
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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
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