Joint Probability Distributions (Chapter 5)

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¡ :¯œ0}º

Joint Probability Distributions (Chapter 5)

Min Wang

Industrial Engineering & Management Chaoyang University of Technology

Min Wang

¡ :¯œ0}º Joint Probability Distributions (Chapter 5)

:¯œ0ƒb u‰b¸óÉ[b ‰bí(4 ¯

I. :¯œ0ƒb

(Joint Probability Mass/Density Function)

:¯œ0}º

v, BbEí1ÝÉuø_‰bíœ0}º, 7 uÖ_‰bÈíÉ[£wœ0}º, Ä¤Bb.Û«n Ö_‰bÈí:¯œ0ƒb

[p.154 (p.142)] In general, if X and Y are two random variables, the probability distribution that deﬁnes their smultaneous behavior is called a joint probability distribution.

Min Wang

1

Example

IÕø_t£íäsŸ, I

S = {(x, y )|1 ≤ x ≤ 6, 1 ≤ y ≤ 6}

w2 x H[øŸ|Ûíõb, y H[ùŸ|Ûíõb, à (3, 5) H[øŸ|ÛíõbÑ 3, ùŸ|ÛíõbÑ5, †uª?

6× 6 = 36 .°í!‹

2

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Example (Cont’)

cq©ø_ª?!‹êÞíª?4uøší, †LSø _!‹ (LS x £ y M) êÞíœ0Ñ ₃₆¹ ,

f (x = 3, y = 5) = 1 36,

Ä¤Bb˚ P(X = x, Y = y) = f (x, y) Ñ‰b X ¸ Y í:¯œ0”¾ƒb (joint probability mass

function)

Min Wang

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:¯œ0”¾ƒb-s_‰b (joint probability mass function), p.155

The joint probability mass function of the discrete random variables X and Y , denoted as fXY(x, y), satisﬁes

(1) fXY(x, y) ≥ 0 (2)

x

y fXY(x, y) = 1

(3) f_XY(x, y) = P(X = x, Y = y)

4

:¯œ0òƒb (joint probability density function), p.167

A joint probability density function for the

continuous random variables X and Y , denoted as f_XY(x, y), satisﬁes the following properties:

(1) fXY(x, y) ≥ 0 for all x, y (2) ^∞

−∞

∞

−∞ fXY(x, y)dxdy = 1

(3) For any region R of two-dimensional space P([X , Y ] ∈ R) =

R

fXY(x, y)dxdy

Min Wang

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Example (Cont’)

˛ø fXY(x, y ) = 1/6, ∀x, y = 1, 2, . . . , 6 7

fX(x) = P{X = x} ˚Ñ X íiÒœ0”¾ƒb (marginal probability mass function), à:

fX(3) = P{X = 3} = f (3, 1) + f (3, 2)

+ f (3, 3) + f (3, 4) + f (3, 5) + f (3, 6)

= 1 36 + 1

36 +· · · + 1

36

u 6 _

= 1 6

6ÿuÉ5? X = x víœ0

6

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iÒœ0”¾ƒb (marginal probability mass function), p.156 If X and Y are discrete random variables with joint probability mass function fXY(x, y ), then the marginal probability mass functions of X and Y are

fX(x) = P(X = x) =

y

fXY(x, y ) and fY(y ) = P(Y = y ) =

x

fXY(x, y )

where the ﬁrst sum is over all points in the range of (X , Y ) for which X = x and the second sum is over all points in the range of (X , Y ) for which Y = y .

Min Wang

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Exercise 5-1, p.163

Show that the following function satisﬁes the properties of a joint probability mass function.

x y fXY(x, y)

1 1 1/4

1.5 2 1/8

1.5 3 1/4

2.5 4 1/4

3 5 1/8

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Solution

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:¯œ0}ºí‚M¸‰æb

If the marginal probability distribution of X has the probability mass function fX(x), the

E (X ) = μX =

x

xfX(x) =

x

y

fXY(x, y)

=fX(x)

=

x

y

xfXY(x, y)

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:¯œ0}ºí‚M¸‰æb (Cont’)

and

V (X ) = σ²_X =

x

(x − μX)²fX(x)

=

x

(x − μX)²

y

fXY(x, y)

=

x

y

(x − μ_X)²f_XY(x, y)

= E (X²)− μ²_X

Min Wang

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Exercise 5-1, p.163

:¯œ0”¾ƒbÑ:

x y fXY(x, y )

1 1 1/4

1.5 2 1/8

1.5 3 1/4

2.5 4 1/4

3 5 1/8

Determine the following proba- bilities:

(a) P(X < 2.5, Y < 3)

(b) P(X < 2.5)

(c) P(Y < 3)

(d) P(X > 1.8, Y > 4.7)

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Exercise 5-1, p.163

X ¸ Y í:¯œ0”¾ƒbÑ:

x y fXY(x, y)

1 1 1/4

1.5 2 1/8

1.5 3 1/4

2.5 4 1/4

3 5 1/8

Determine E (X ), E (Y ), V (X ) and V (Y )

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Solution

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Solution (Cont’)

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‘Kœ0”¾ƒb (Conditional Probability Mass Function), p.157

Given discrete random variables X and Y with joint probability mass function fXY(x, y) the conditional probability mass function of Y given X = x is

fY |x(y) = fXY(x, y)

fX(x) for fX(x) > 0

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‘Kœ0”¾ƒbíÔ4, p.157

Because a conditional probability mass function fY |x(y) is a probability mass function, the following properties are satisﬁed:

(1) f_{Y |x}(y) ≥ 0 (2)

y f_{Y |x}(y) = 1

(3) P(Y = y|X = x) = f_{Y |x}(y)

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‘K‚M (Conditional Mean) ¸‘K‰æb (Condition Variance), p.157

The conditional mean of Y given X = x, denoted as E (Y |x) or μY |x, is

E (Y |x) =

y

yfY |x(y )

and the conditional variance of Y given X = x, denoted as V (Y |x) or σ_{Y |x}² , is

V (Y |x) =

y

(y − μY |x)²fY |x(y ) =

y

y²fY |x(y ) − μ²_{Y |x}

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Exercise 5-1, p.163

Continuation of Exercise 5-1. Determine

(g) The conditional probability distribution of Y given that X = 1.5.

(h) The conditional probability distribution of X given that Y = 2.

(i) E (Y |X = 1.5)

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Solution

(g) The conditional probability distribution of Y given that X = 1.5.

:¯œ0”¾ƒbÑ:

x y fXY(x, y )

1 1 1/4

1.5 2 1/8

1.5 3 1/4

2.5 4 1/4

3 5 1/8

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Solution (Cont’)

(h) The conditional

probability distribution of X given that Y = 2.

:¯œ0”¾ƒbÑ:

x y fXY(x, y )

1 1 1/4

1.5 2 1/8

1.5 3 1/4

2.5 4 1/4

3 5 1/8

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Solution

(i) E (Y |X = 1.5)

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Ö (Independent), p.159

For discrete random variable X and Y , if any one of the following properties is true, the others are also true, and X and Y are independent.

(1) fXY(x, y) = fX(x)fY(y) for all x and y

(2) f_{Y |x}(y) = fY(y) for all x and y with fX(x) > 0 (3) f_X|y(x) = fX(x) for all x and y with fY(y) > 0 (4) P(X ∈ A, Y ∈ B) = P(X ∈ A)P(Y ∈ B) for

any sets A and B in the range of X and Y , respectively.

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Exercise 5-1, p.163

Continuation of Exercise 5-1. Determine (j) Are X and Y independent?

(M1)

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Solution (Cont’)

(M2) :¯œ0”¾ƒbÑ:

x y fXY(x, y)

1 1 1/4

1.5 2 1/8

1.5 3 1/4

2.5 4 1/4

3 5 1/8

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:¯œ0”¾ƒb-Ö_‰b (joint probability mass function), p.159

The joint probability mass function of X1, X², . . . , Xp is

f_X1,X²,...,Xp(x1, x2, . . . , xp)

= P(X1 = x1, X2 = x2, . . . , Xp = x_p) for all points (x1, x2, . . . , xp) in the range of X1, X2, . . . , Xp.

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iÒœ0}º-Ö_‰b, p.160

If X1, X2, X3, . . . , Xp are discrete random variables with joint probability mass function

fX1,X2,...,Xp(x1, x2, . . . , xp), the marginal probability mass function of any Xi is

fXi(xi) = P(Xi = xi) =

fX1,X2,...,Xp(x1, x2, . . . , xp) where the sum is over the points in the range of (X1, X2, . . . , Xp) for which Xi = xi

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œ0}ºí‚M¸‰æb-Ö_‰b, p.160

E (Xi) =

xifX1,X2,...,Xp(x1, x2, . . . , xp) and

V (Xi) =

(xi − μXi)²fX1,X2,...,Xp(x1, x², . . . , xp) where the sum is over all points in the range of

X1, X2, . . . , Xp. ₂₈

‘Kœ0}º (Conditional Probability Distributions), p.151

Conditional probability distribution can be

developed for multiple discrete random variables by an an extension of the ideas used for two discrete random variables. For example, the conditional joint probability mass function of X1, X2, X3, given X4, X5

is

f_X1,X²,X³|X⁴,X⁵(x1, x2, x3) = fX1,X2,X3,X4,X5(x1, x2, x3, x4, x5) fX4,X5(x4, x⁵)

for fX4,X5(x4, x5)> 0.

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Ö (Indpendent), p.153

Discrete random variables X1, X2, . . . , Xp are independent if and only if

fX1,X2,...,Xp(x1, x2, . . . , xp) = fX1(x1)fX2(x2)· · · fXp(xp) for all x1, x2, . . . , xp.

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Exercise 5-8(5-17), p.164

Suppose the random variables X , Y , and Z have the following joint probability distribution

x y z f (x, y , z)

1 1 1 0.05

1 1 2 0.10

1 2 1 0.15

1 2 2 0.20

2 1 1 0.20

2 1 2 0.15

2 2 1 0.10

2 2 2 0.05

Determine the following:

(a) P(X = 2)

(b) P(X = 1, Y = 2)

(c) P(Z < 1.5)

(d) P(X = 1 or Z = 2)

(e) E (X )

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Exercise 5-8, p.164

x y z f (x, y , z)

1 1 1 0.05

1 1 2 0.10

1 2 1 0.15

1 2 2 0.20

2 1 1 0.20

2 1 2 0.15

2 2 1 0.10

2 2 2 0.05

Determine the following:

(f) P(X = 1|Y = 1)

(g) P(X = 1, Y = 1|Z = 1)

(h) P(X = 1|Y = 1, Z = 2)

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:¯œ0òƒb (joint probability density function), p.167

A joint probability density function for the

continuous random variables X and Y , denoted as fXY(x, y), satisﬁes the following properties:

(1) fXY(x, y) ≥ 0 for all x, y (2) _∞

−∞

_∞

−∞fXY(x, y)dxdy = 1

(3) For all region R of two-dimensional space P((X , Y ) ∈ R) =

R f_XY(x, y) dxdy

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iÒœ0òƒb (marginal probability density function), p.169 If the joint probability density function of continuous random variable X and Y is fXY(x, y ), themarginal probability density functions of X and Y are

fX(x) =

yfXY(x, y )dy and fY(y ) =

xfXY(x, y )dx

where the ﬁrst integral is over all points in the range of (X , Y ) for which X = x and the second integral is over all points in

the range of (X , Y ) for which Y = y . 34

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‘Kœ0òƒb (conditional probability density function), p.170

Given continuous random variables X and Y with joint probability density function fXY(x, y) the conditional probability density function of Y given X = x is

f_{Y |x}(y) = f_XY(x, y)

fX(x) for f_X(x) > 0

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‘Kœ0òƒbíÔ4, p.157

Because a conditional probability density function f_{Y |x}(y) is a probability density function, the

following properties are satisﬁed:

(1) f_{Y |x}(y) ≥ 0 (2)

f_{Y |x}(y)dy = 1

(3) P(Y ∈ B|X = x) =

B f_{Y |x}(y)dy for any set B in the range of Y .

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‘KÌb£‰æb (conditional mean and variance), p.171

The conditional mean of Y given X = x, denoted as E (Y |x) or μ_{Y |x} is

E (Y |x) =

yfY |x(y)dy

and the conditional variance of Y given X = x, denoted as V (Y |x) or σ_{Y |x}² , is

V (Y |x) =

(y − μ_{Y |x})²f_{Y |x}(y)dy

=

y²fY |x(y)dy − μ²_{Y |x}

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Ö (independent), p.172

For continuous random variables X and Y , if any one of the following properties is true, the others are also true, and X and Y are said to be

independent.

(1) f_XY(x, y) = f_X(x)f_Y(y) for all x and y

(2) f_{Y |x}(y) = f_Y(y) for all x and y with f_X(x) > 0 (3) f_X|y(x) = f_X(x) for all x and y with f_Y(y) > 0 (4) P(X ∈ A, Y ∈ B) = P(X ∈ A)P(Y ∈ B) for

any sets A and B in the range of X and Y ,

respectively. ³⁸

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Exercise 5-17, p. 177

Determine the value of c such that the function f (x, y ) = cxy for 0< x < 3 and 0 < y < 3 satisﬁes the properties of a joint probability density function. Determine

(a) P(X < 2, Y < 3) (b) P(X < 2.5) (c) P(1 < Y < 2.5)

(d) P(X > 1.8, 1 < Y < 2.5) (e) E (X )

(f) P(X < 0, Y < 4)

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Exercise 5-17 (Cont.), p. 177

(g) Marginal probability distribution of the random variable X

(h) Conditional probability of Y given that X = 1.5 (i) E (Y |X = 1.5)

(j) P(Y < 2|X = 1.5)

(k) Conditional probability distribution of X given that Y = 2

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Solution

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Solution (Cont’)

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Solution (Cont’)

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Solution (Cont’)

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II. u‰b¸óÉ[b

(Covariance and Correlation)

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ƒbí‚M (Expected Value of a Function), p.179

E [h(X , Y )] =

h(x, y )fXY(x, y ) X , Y discrete

h(x, y )fXY(x, y ) dxdy X , Y continuous

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Example 5-24, p.179

For the joint probability distribution of the two random variables in Fig. 5-12, calculated E [(X − μX)(Y − μY)].

1 1

2 3

3 y

2

x 0.1

0.2

0.2 0.2 0.3

(Sol.)

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u‰b (covariance), p.179

The covariance between the random variables X and Y , denoted as cov(X , Y ) or σ_XY, is

σXY = E [(X − μ_X)(Y − μ_Y)] = E (XY ) − μ_XμY

Moreover,

σXY = cov(X , Y ) = cov(Y , X ) = σYX

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u‰bíÔ4, p.181

u‰b (Covariance) ªJéý‰b5Èí(4É[ (linear relationship), J‰bÈíÉ[1Ý(4, †u‰b (covariance) Ì¶éý‰bÈíÉ[, àÇ 5-13(a) Fý, ç X ¸ Y 5Èí covariance Ñ£Mv (positive), X ¸ Y 5ÈÑ£óÉ, 6ÿu‰

b Y M}ÓO X MíÓ‹7Ó‹, óúË, ?}ÓO X MíÁý 7Áý 7Ç 5-13(c) F×ÛíÑ X ¸ Y 5ÈÑŠóÉ, w covariance ÑŠM (negative), 6ÿuç X MÓ‹v, Y üóú ËÁý, 7ç X MÁýv, Y MüóúËÓ‹ 7Ç (b) ¸ (d) 2 X ¸ Y 1ÌéOíò(óÉ4, ] X ¸ Y ÑÉ (zero) óÉ

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Figure 5-13 Joint probability distribution and the sign of the covariance betwen X and Y , p.181

x y

x

y

(a) Positive covariance (b) Zero covariance

(c) Negative covariance (d) Zero covariance

All points are of equal probability

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óÉ[b (correlation), p.181

The correlation between random variables X and Y , denoted as ρXY, is

ρXY = cov(X , Y )

V (X )V (Y ) = σXY

σXσY

For any two random variables X and Y

−1 ≤ ρXY ≤ +1

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Example 5-26, p.182

For the discrete random variables X and Y with the joint distribution shown in Fig. 5-14, determine σXY and ρXY.

1 1

2 3

3 y

2

x 0.1

0.1

0.1 0.1

0.4

0 0.2 0

(Sol.)

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Solution (Cont’)

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III. ‰bí(4 ¯

(Linear Combinations of Random Variables)

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‰bí(4 ¯ (linear combination of random variables)

I X1 ¸ X2 Ñs_Óœ‰b, c1 ¸ c2 Ñs_b, † Y = c1X1 + c2X2

˚Ñ X1 ¸ X2 í(4 ¯ (linear combination) J Y = c1X1+ c2X2 Ñ X1 ¸ X2 í(4 ¯ (linear combination), † Y ?Ñø_Óœ‰b, ‚MÑ

E (Y ) = E (c1X1+ c2X2) = c1E (X1) + c2E (X2)

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Y í‰æbÑ

σ_Y² = V (Y ) = E [(Y − μY)²]

= E {[(c1X1+ c2X2)− (c1μX1+ c2μX2)]²}

= E {[(c1X1− c1μX1) + (c2X2− c2μX2)]²}

= E {c1²(X¹− μX1)²+ c2²(X²− μX2)² + 2c1c2(X1− μX1)(X2− μX2)}

= c1²E [(X1− μX1)²] + c2²E [(X2− μX2)²] + 2c1c2E [(X1− μX1)(X2− μX2)]

= c1²V (X ) + c2²V (Y ) + 2c1c2cov(X1, X2) J X¹ ¸ X² Ñ independent v, † cov(X¹, X²) = 0, FJ

V (c1X1+ c2X2) = c1²V (X ) + c2²V (X ) ⁵⁴

(4 ¯ (Linear Combination), p.188

Given random variables X1, X2, . . . , Xp and constants c1, c2, . . . , cp,

Y = c1X1+ c2X2 +· · · + cpXp

is a linear combination of X1, X2, . . . , Xp. The mean(Ìb) of linear combination

E (Y ) = c1E (X1) + c2E (X2) +· · · + cpE (Xp)

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(4 ¯í‰æb (variance), p.189

If X1, X2, . . . , Xp are random variables, and Y = c1X1+ c2X2+· · · + cpXp, then in general

V (Y ) = c1²V (X1) + c2²V (X2) +· · · + c_p²V (Xp) + 2

i<j

cicjcov (Xi, Xj)

If X1, X2, . . . , Xp are independent,

V (Y ) = c1²V (X¹)c2²V (X²) + C_p²V (Xp)

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Exercise 5-54, p.191

X and Y are independent, normal random variables with E (X ) = 0, V (X ) = 4, E (Y ) = 10, and V (Y ) = 9. Determine E (2X + 3Y ) and V (2X + 3Y ).

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G‰b (Normal Random Variable), p.122

A random variable X with probability density function f (x) = 1

√2πσe⁻^(x−μ)2^2σ2 − ∞ < x < ∞ is a normal random variable with parameters μ, where

−∞ < μ < ∞, and σ > 0. Also,

E (X ) = μ and V (X ) = σ²

and the notation N (μ, σ²) is used to denote the distribution.

The mean and variance of X are shown to equal μ and σ²,

respectively. ⁵⁸

™ÄGÓœ‰b (standard normal random variable), p.123

A normal random variable with μ = 0 and σ² = 1

is call a standard normal random variable and is denoted as Z .

The cumulative distribution function of a standard normal random variable is denoted as

Φ(z) = P(Z ≤ z)

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G‰b™Ä“ (Standarding), p.125

If X is a normal random variable with E (X ) = μ and V (X ) = σ², the random variable

Z = X − μ σ

is a normal random variable with E (Z ) = 0 and V (Z ) = 1. That is, Z is a standard normal random variable.

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G‰bíœ0, p.126

Suppose X is a normal random variable with mean μ and variance σ². Then,

P(X ≤ x) = P

X − μ

σ ≤ x − μ σ

= P(Z ≤ z) where Z is a standard normal random variable, and z = ^x−μ_σ is the z-value obtained by stnadardizing X .

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G‰bíÚœ0, p.126

654 APPENDIX A STATISTICAL TABLES AND CHARTS

Table II Cumulative Standard Normal Distribution (continued)

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.500000 0.503989 0.507978 0.511967 0.515953 0.519939 0.532922 0.527903 0.531881 0.535856 0.1 0.539828 0.543795 0.547758 0.551717 0.555760 0.559618 0.563559 0.567495 0.571424 0.575345 0.2 0.579260 0.583166 0.587064 0.590954 0.594835 0.598706 0.602568 0.606420 0.610261 0.614092 0.3 0.617911 0.621719 0.625516 0.629300 0.633072 0.636831 0.640576 0.644309 0.648027 0.651732 0.4 0.655422 0.659097 0.662757 0.666402 0.670031 0.673645 0.677242 0.680822 0.684386 0.687933 0.5 0.691462 0.694974 0.698468 0.701944 0.705401 0.708840 0.712260 0.715661 0.719043 0.722405 0.6 0.725747 0.729069 0.732371 0.735653 0.738914 0.742154 0.745373 0.748571 0.751748 0.754903 0.7 0.758036 0.761148 0.764238 0.767305 0.770350 0.773373 0.776373 0.779350 0.782305 0.785236 0.8 0.788145 0.791030 0.793892 0.796731 0.799546 0.802338 0.805106 0.807850 0.810570 0.813267 0.9 0.815940 0.818589 0.821214 0.823815 0.826391 0.828944 0.831472 0.833977 0.836457 0.838913 1.0 0.841345 0.843752 0.846136 0.848495 0.850830 0.853141 0.855428 0.857690 0.859929 0.862143 1.1 0.864334 0.866500 0.868643 0.870762 0.872857 0.874928 0.876976 0.878999 0.881000 0.882977 1.2 0.884930 0.886860 0.888767 0.890651 0.892512 0.894350 0.896165 0.897958 0.899727 0.901475 1.3 0.903199 0.904902 0.906582 0.908241 0.909877 0.911492 0.913085 0.914657 0.916207 0.917736 1.4 0.919243 0.920730 0.922196 0.923641 0.925066 0.926471 0.927855 0.929219 0.930563 0.931888 1.5 0.933193 0.934478 0.935744 0.936992 0.938220 0.939429 0.940620 0.941792 0.942947 0.944083 1.6 0.945201 0.946301 0.947384 0.948449 0.949497 0.950529 0.951543 0.952540 0.953521 0.954486 1.7 0.955435 0.956367 0.957284 0.958185 0.959071 0.959941 0.960796 0.961636 0.962462 0.963273 1.8 0.964070 0.964852 0.965621 0.966375 0.967116 0.967843 0.968557 0.969258 0.969946 0.970621 1.9 0.971283 0.971933 0.972571 0.973197 0.973810 0.974412 0.975002 0.975581 0.976148 0.976705 2.0 0.977250 0.977784 0.978308 0.978822 0.979325 0.979818 0.980301 0.980774 0.981237 0.981691

z 0

Φ (z)

⌽1z2 ⫽ P1Z ⱕ z2 ⫽冮_⫺⬁^z22␲¹ e^⫺1²^u

2

du

62

Assume X is normal distributed with a mean of 10 and a standard deviation of 2. Determine P(X < 13), P(X > 9), and P(6 < X < 14).

(Sol.)

Min Wang

63

Ìbí‚M¸‰æb, p.190

If ¯X = ^(X¹^+X²^+···+X_p ^p⁾ with E (Xi) = μ for i = 1, 2, . . . , p

E ( ¯X ) = μ

if X1, X², . . . , Xp are also independent with V (Xi) = σ² for i = 1, 2, . . . , p,

V ( ¯X ) = σ²

p 64

(18)

G‰b(4 ¯íÔ4, p.190

If X1, X2, . . . , Xp are independent, normal random variables with E (Xi) = μi and V (Xi) = σ_i², for i = 1, 2, . . . , p

Y = c1X1 + c2X2 +· · · + cpXp

is a normal random variable with

E (Y ) = c1μ¹ + c2μ² +· · · + cpμp

and

V (Y )c1²σ1²+ c2²σ2²+· · · + c_p²σ_p²

Min Wang

65

X and Y are independent, normal random variables with E (X ) = 0, V (X ) = 4, E (Y ) = 10, and V (Y ) = 9. Determine P(2X + 3Y < 30) and P(2X + 3Y < 40).

(Sol.)

66