# Lecture 7: Continuous Distributions and Probability Plots · PDF file 2019. 5....

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Lecture 7: Continuous Distributions and Probability Plots

MSU-STT-351-Sum-19A

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 1 / 43

Some Standard Continuous Distributions

In this lecture, we discuss some important continuous distributions and also a graphical method to check of thedata is form a continuous distribution. Gamma Function The gamma function Γ(α), for α > 0, is defined by

Γ(α) =

∫ ∞ 0

xα−1e−xdx.

Gamma Distribution G(α, β) A continuous random variable X is said to have a gamma distribution, denoted by G(α, β), if the pdf of X is

f(x |α, β) = { 1

βαΓ(α) xα−1e−x/β, x ≥ 0

0, otherwise.

where α > 0 and β > 0. (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 2 / 43

Continuous Distributions and Probability Plots

Here, E(X) = αβ and V(X) = αβ2. The gamma distribution G(α, 1) is called standard gamma distribution.

The cdf F(x) does not admit closed form. However, for the the standard gamma distribution, F(x) can be found using the Table A .4 on p. A -8.

Note also that if X ∼ G(α, β), then Xβ ∼ G(α, 1). Using this result, the probabilities of the gamma distribution can be calculated, using the Table A .4.

The density functions of some gamma distributions are given below.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 3 / 43

Continuous Distributions and Probability Plots

0 5 10 15 0

0.1

0.2

0.3

α = 5 β = 1 α = 4 β = 0.9 α = 3 β = 0.8 α = 6 β = 1.3

Figure: Gamma densities

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 4 / 43

Continuous Distributions and Probability Plots

Example 1 (Eg. 4.24) Suppose the survival time X in weeks of a randomly selected male mouse exposed of 240 rads of gamma radiation has a gamma distribution with α = 8 and β = 15. What is the probability that a mouse survives between 60 and 120 weeks?

Note here that X ∼ G(8, 15). Let Y = X/15 so that Y ∼ G(8, 1).

Hence,

P(60 < X < 120) = P(4 < Y < 8) = 0.547 − 0.051 = 0.496,

using Table A -4.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 5 / 43

Continuous Distributions and Probability Plots

Example 2 (Ex 66): Suppose the time spent by a randomly selected student who uses a terminal connected to a local time-sharing computer facility has a gamma distribution with mean 20 min and variance 80 min2. (a) What are the values of α? and β? (b) What is the probability that a student uses the terminal for at most 24 min? (c) What is the probability that a student spends between 20 and 40 min using the terminal?

Solution: Let F(x |α, β) denote the cdf of X ∼ G(α, β). (a) µ = 20, σ2 = 80⇒ αβ = 20, αβ2 = 80⇒ β = 8020 = 4, α = 5. (b) P(X ≤ 24) = F

( 24 4 |5, 1

) = F(6|5, 1) = 0.715 (Using table A.4)

(c) P(20 ≤ X ≤ 40) = F(10|5, 1) − F(5|5, 1) = 0.411.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 6 / 43

Continuous Distributions and Probability Plots

Exponential Distribution The gamma distribution with α = 1 and β = 1/λ is also called an exponential distribution with parameter λ > 0. It has the pdf

f(x |λ) = { λe−λx , x ≥ 0 0, otherwise.

If X is an exponential random variable with parameter λ, then

E(X) = 1 λ

and V(X) = 1 λ2 .

The cdf is F(x) = ∫ x

0 λe −λtdt = 1 − e−λx , x > 0. Also, the reliability

function of X is R(t) = P(X > t) = e−λt , t > 0.

The Chi-squared Distribution The distribution G(γ/2, 2) is called χ2 distribution with γ degrees of freedom. That is, χ2ν ≡ G(ν/2, 2). (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 7 / 43

Continuous Distributions and Probability Plots

0 1 2 3 4

0.5

1

1.5

2 λ = 0.5 λ = 1 λ = 2

Figure: Exponential densities

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 8 / 43

Continuous Distributions and Probability Plots

Example 3 ( Ex. 71) (a) The event {X2 ≤ y} is equivalent to what event involving X itself? (b) If X ∼ N(0, 1), show that, using Part (a), X2 ∼ χ21. Solution. Let fX (t) be the pdf and FX (t) be the cdf of X . Let Y = X2. Then (a) {X2 ≤ y} = {Y ≤ y} = {−√y ≤ X ≤ √y}, since y > 0. (b) Note FY (y) = P(X2 ≤ y) = FX (

√ y) − FX (−

√ y). Differentiating wrt y,

fY (y) = fX ( √

y) ( 1 2 √

y

) − fX (−

√ y)

( − 1

2 √

y

) =

1 √

2π e−y/2

(1 2

y−1/2 )

+ 1 √

2π e−y/2

(1 2

y−1/2 )

= 1 √

2π e−y/2y−1/2 =

1

21/2Γ( 12 ) e−y/2y−1/2,

which is G( 12 , 2) = χ 2 1. Note we have used the fact fX (x) =

1√ 2π

e−x 2/2.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 9 / 43

Continuous Distributions and Probability Plots

Example 4 (Ex 69 b, c): A system consists of five identical components connected in series as shown:

1 2 3 4 5

The entire system fails if one component fails. Suppose each component has a lifetime that is exponentially distributed with λ = .01 and that components fail independently.

Let Xi denote the life-time of the i-th component. Then X1, . . . ,X5 are independent rvs. Also, (Xi > t) denotes the event survives till time t . Note X = min{X1,X2, . . . ,Xn} denotes the life-time of the system.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 10 / 43

Continuous Distributions and Probability Plots

(a) Compute P(X ≥ t), F(t) = P(X ≤ t) and the pdf f(t) of X. What type of distribution does X have? (b) Suppose there are n components, each having exponential lifetime with parameter λ. What type of distribution does X have?

Solution: Note first that X = min{X1,X2, . . . ,Xn} denotes the life-time of the system. Now (a) P(X ≥ t) = P(X1 ≥ t)P(X2 ≥ t) . . .P(X5 ≥ t) = (e−λt )5 = e−0.05t . So,

Fx(t) = P(X ≤ t) = 1 − e−0.05t ; fx(t) = 0.05e−0.05t for t ≥ 0.

Thus, X also has an exponential distribution with parameter λ = 0.05.

(b) By the same reasoning, P(X ≤ t) = 1 − e−nλt , so X has an exponential distribution with parameter nλ.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 11 / 43

Continuous Distributions and Probability Plots

Weibull Distribution: W(α, β)

This is an extension of exponential distribution. Definition: A continuous variable X follows W(α, β) if its density is

f(x;α, β) =

αβα e−( x β )

α

xα−1, x > 0 0, otherwise

(i) Different values of α and β give positively and negatively skewed distributions. (ii)

µ = E(X) = β

α Γ(1/α);σ2 = V(X) =

2β2

α Γ (2 α

) .

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 12 / 43

Continuous Distributions and Probability Plots

0 2 4 6 8 10 12 14 16 0

0.1

0.2

0.3

0.4 α = 2 β = 2 α = 2 β = 4 α = 1.5 β = 5 α = 2.5 β = 6

Figure: Weibull densities

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 13 / 43

Continuous Distributions and Probability Plots

(iii)

P(X ≤ t) = ∫ t

0 f(x |α, β)dx

= α

βα

∫ t 0

e−( x β )

α

xα−1dx

=

∫ ( tβ )α 0

e−ydy,

using the transformation y = ( xβ ) α so that dy = αβα x

α−1dx. Thus, we get

P(X ≤ t) = [ − e−y

]( tβ )α 0

= 1 − e−( t β )

α

.

Therefore, P(X > t) = e(− t β )

α

, t > 0.

(iv) If X ∼ W(α, β), then Y = kX ∼ W(α, kβ).

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 14 / 43

Continuous Distributions and Probability Plots

Example 5. Suppose X ∼ W(2, 10). Then

(i) P(X < 10) = 1 − e − (

10 10

)2 = 1 − e−1 � 0.632.

(ii) Suppose c is P(X < c) = 0.95. Then

1 − e − (

c 10

)2 = 0.95

Solving for c, we get c � 17.3 which is the 95-th percentile.

(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 15 / 43

Continuous Distributions and Probability Plots

Example 6(Ex 76): Let X follow Weibull distribution with parameters α = 2 and β = 3. (a) What is median lifetime of such tubes? (b) If X has a Weibull distribution with the cdf form expression, obtain a general expression for the (100p)th percentile of the distribution.

Solution The median m satisfies (a) F(m) = 1 − e−(m/3)2 ⇒ e−m2/9 = 0.5⇒ m2 = −9 ln(.5) = 6.2383⇒ m = 2.50

(b) p = F(xp) = 1− e(xp/β)

α ⇒ (xp/β)α = − ln(1− p)⇒ xp = β[− ln(1− p)]1/α.