Download - Lecture 7: Continuous Distributions and Probability Plots · 2019. 5. 29. · Continuous Distributions and Probability Plots Here, E(X) = and V(X) = 2:The gamma distribution G( ;1)

Transcript
  • Lecture 7: Continuous Distributions and ProbabilityPlots

    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 andalso a graphical method to check of thedata is form a continuousdistribution.Gamma FunctionThe 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) iscalled standard gamma distribution.

    The cdf F(x) does not admit closed form. However, for the the standardgamma 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, theprobabilities of the gamma distribution can be calculated, using the TableA .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 10 150

    0.1

    0.2

    0.3

    α = 5 β = 1α = 4 β = 0.9α = 3 β = 0.8α = 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 mouseexposed of 240 rads of gamma radiation has a gamma distribution withα = 8 and β = 15. What is the probability that a mouse survives between60 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 selectedstudent who uses a terminal connected to a local time-sharing computerfacility 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 24min?(c) What is the probability that a student spends between 20 and 40 minusing 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

    (244 |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 DistributionThe gamma distribution with α = 1 and β = 1/λ is also called anexponential distribution with parameter λ > 0. It has the pdf

    f(x |λ) ={λe−λx , x ≥ 00, otherwise.

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

    E(X) =1λ

    and V(X) =1λ2.

    The cdf is F(x) =∫ x

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

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

    The Chi-squared DistributionThe distribution G(γ/2, 2) is called χ2 distribution with γ degrees offreedom. 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 = 0.5λ = 1λ = 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)( 12√

    y

    )− fX (−

    √y)

    (− 1

    2√

    y

    )=

    1√

    2πe−y/2

    (12

    y−1/2)

    +1√

    2πe−y/2

    (12

    y−1/2)

    =1√

    2πe−y/2y−1/2 =

    1

    21/2Γ( 12 )e−y/2y−1/2,

    which is G( 12 , 2) = χ21. Note we have used the fact fX (x) =

    1√2π

    e−x2/2.

    (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 componentsconnected in series as shown:

    1 2 3 4 5

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

    Let Xi denote the life-time of the i-th component. Then X1, . . . ,X5 areindependent rvs. Also, (Xi > t) denotes the event survives till time t . NoteX = 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 ofdistribution does X have?(b) Suppose there are n components, each having exponential lifetimewith parameter λ. What type of distribution does X have?

    Solution: Note first that X = min{X1,X2, . . . ,Xn} denotes the life-time ofthe 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 exponentialdistribution 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−(xβ )

    α

    xα−1, x > 00, otherwise

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

    µ = E(X) =β

    αΓ(1/α);σ2 = V(X) =

    2β2

    αΓ(2α

    ).

    (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 14 160

    0.1

    0.2

    0.3

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

    0f(x |α, β)dx

    βα

    ∫ t0

    e−(xβ )

    α

    xα−1dx

    =

    ∫ ( tβ )α0

    e−ydy,

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

    α−1dx. Thus, we get

    P(X ≤ t) =[− e−y

    ]( tβ )α0

    = 1 − e−(tβ )

    α

    .

    Therefore, P(X > t) = e(−tβ )

    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−(

    1010

    )2= 1 − e−1 � 0.632.

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

    1 − e−(

    c10

    )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 ageneral 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.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 16 / 43

  • Continuous Distributions and Probability Plots

    Nest, we discuss a variation of the normal distribution.The Lognormal DistributionDefinition: A variable X > 0 is said to follow LN(µ, σ) if ln(X) ∼ N(µ, σ).Its density is

    f(x; µ, σ) =

    1√2πσ 1x e1

    2σ2(ln x−µ)2

    , x > 00, otherwise.

    Note that ln(X) (and not X ) follows normal distribution with mean µ andvariance σ2.The graphs of some lognormal densities are given below.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 17 / 43

  • Large Sample Tests

    0 5 10 15 200

    5 · 10−2

    0.1

    0.15

    0.2

    0.25 µ = 1.0 σ = 1.0µ = 1.5 σ = 0.5µ = 2.0 σ = 1.0µ = 3.0 σ = 1.5

    Figure: Lognormal densities

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 18 / 43

  • Continuous Distributions and Probability Plots

    Facts:(i) It is easy to check ∫ ∞

    −∞f(y |µ, σ)dy = 1,

    using the substitution y = ln(x).

    (ii) If X ∼ LN(µ, σ), then Y = ln(X) ∼ N(µ, σ). Hence,

    P(X ≤ x) = P[ln(X) ≤ ln(x)] = P(Z ≤ ln(x) − µ

    σ

    )= Φ

    ( ln(x) − µσ

    ).

    (iii) E(X) = eµ+σ22 ; V(X) = e2µ+σ

    2(eσ

    2 − 1).

    The above facts are used to compute probabilities.

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

  • Continuous Distributions and Probability Plots

    Example 7 (Ex 79): Let the rv X denote thethe hourly median power (indecibels) of received radio signals transmitted between two cities. Theauthors of an article argue that the lognormal distribution provides areasonable probability model for X .If the parameter values are µ = 3.5 and σ = 1.2, calculate the following:

    (a) The mean value and standard deviation of received power.

    (b) The probability that received power is between 50 and 250dB.

    (c) The probability that X is less than its mean value. Why is thisprobability not 0.5?

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 20 / 43

  • Continuous Distributions and Probability Plots

    Solution: Given that X ∼ LN(3.5, 1.2). Then(a):

    E(X) = e3.5+(1.2)2/2 = 68.0335;

    V(X) = e2(3.5)+(1.2)2.(e(1.2)

    2−1)

    = 14907.168

    σx = 122.0949.

    (b) : P(50 ≤ X ≤ 250) = P(Z ≤ ln(250) − 3.5

    1.2

    )− P

    (Z ≤ ln(50) − 3.5

    1.2

    )= P(Z ≤ 1.68) − P(Z ≤ .34)= 0.9535 − 0.6331 = 0.3204. (use table)

    (c) P(X ≤ 68.0335) = P(Z ≤ ln(68.0335)−3.51.2

    )= P(Z ≤ .60) = 0.7257. The

    lognormal distribution is not a symmetric distribution.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 21 / 43

  • Continuous Distributions and Probability Plots

    Example 8. The r.v. X = modulus of elasticity (MOE) of a certain system isobserved, in a study, to follow LN(0.375, 0.25). Find(i) P(X < 3)(ii) P(1 < X < 2)(iii) For what value of c, only 1% of systems have X > c? Note, c is nothingbut the 99th percentile.

    Solution:(i) Note X < 3⇔ Y = ln(X) < ln(3) = 1.0986, where Y ∼ N(0.375, 0.25)Therefore,

    Y < 1.098⇔ Z = Y − .375.25

    <1.098 − 0.375

    .25= 2.89,

    where Z ∼ N(0, 1).From normal tables, P(Z < 2.89) = 0.998. Hence, P(X < 3) = 0.998 (or99.8%).

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 22 / 43

  • Continuous Distributions and Probability Plots

    (ii) Similarly,1 < X < 2⇔ ln(1) < ln(X) < ln(2)

    ⇔ ln(1) < Y < ln(2)

    ⇔ ln(1)−0.3750.25 <Y−0.375

    0.25 <ln(2)−0.375

    0.25

    Hence, P(1 < X < 2) = P(−1.512 < Z < 1.272).NowP(Z < 1.272) = 0.8984; P(Z < −1.512) = 0.0653. (see normal table)

    P(−1.512 < Z < 1.272) = 0.8331.

    Hence,P(1 < X < 2) = 83.3%

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 23 / 43

  • Continuous Distributions and Probability Plots

    (iii) Need to find c such that

    P(X > c) = 0.01⇔ P(X ≤ c) = 0.99.

    That is, need to find c such that

    P[ ln(X) − .375

    .25≤ ln(c) − .375

    .25

    ]= 0.99

    That is, c is such that

    P(Z ≤ ln(c) − .375

    .25

    )= 0.99

    From normal table, 99th percentile of Z is 2.33.Therefore,

    ln(c) − .375.25

    = 2.33

    Solving ln(c) = .9575, we get c = 2.605.(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 24 / 43

  • Continuous Distributions and Probability Plots

    Quantile (or Probability) Plots

    The probability plots can be used to check if the data x1, . . . , xn is from aparticular distribution.

    The basic idea is to compare sample quantiles (percentiles) withpopulation quantiles (percentiles).

    Definition:

    (i) Order the observations as x(1) ≤ x(2) ≤ . . . ≤ x(n). Then,

    (ii) x(i) may be called (in )-th or technically (

    i−0.5n )-th sample quantile.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 25 / 43

  • Continuous Distributions and Probability Plots

    Example 9Let n = 10 and x1, x2, . . . , x10 be a random sample andx(1) ≤ x(2) ≤ . . . ≤ x(n) be their ordered values. Then, sample quantiles are

    x(1) is1 − 0.5

    10=

    0.510

    = 0.5th quantile (or 5th pecrcentile)

    x(2) is2 − 0.5

    10=

    1.510

    = 0.15th quantile (or 15th pecrcentile)

    .

    .

    .

    x(10) is10 − 0.5

    10=

    0.9510

    = 0.95th quantile (or 95th pecrcentile)

    The basic ideas of a quantile plot are summarized below.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 26 / 43

  • Continuous Distributions and Probability Plots

    Procedure of Probability (Quantile) Plot

    Data: x1, x2, . . . , xn.Order: x(1) ≤ x(2) ≤ . . . ≤ x(n).Compute:

    (.5n

    ),(

    1.5n

    ), . . . ,

    (n−.5

    n

    )Population quantiles: η .5

    n, η 1.5

    n, . . . , η n−.5

    n,

    If the observations are basically from the population with density f(x), thenx(i) ≈ η( i−.5n ). This is the basic idea.

    Plot the points :(η .5

    n, x(1)

    ),(η 1.5

    n, x(2)

    ), . . . ,

    (η n−.5

    n, x(n)

    )If the graph is approximately linear, then the data x1, . . . , xn are fromdensity f(x).

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 27 / 43

  • Continuous Distributions and Probability Plots

    1. Normal Probability Plot

    Normal probability plot is used to check if the data y1, . . . , yn are fromN(µ, σ) :

    (i) Order: y(1) ≤ . . . ≤ y(n).

    (ii) Standardize: z(1) =y(1)−µσ , . . . , z(n) =

    y(n)−µσ

    (iii) Let p1 = 1−0.5n , p2 =2−0.5

    n , . . . , pn =n−0.5

    n .

    (iv) Find population percentiles of N(0, 1) : ηp1 , ηp2 , . . . , ηpn .

    (v) Plot the points: (ηp1 , z(1)), (ηp2 , z(2)), . . . , (ηpn , z(n)).

    If the graph is approximately linear, then sample is from the normaldistribution.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 28 / 43

  • Continuous Distributions and Probability Plots

    Example 10.

    Check if the data 9.9, 4.8, 8.1, 15.2, 12.1 are from N(10, 4).

    Ordered data x(i): 4.8, 8.1, 9.9, 12.1, 15.2

    Normalized data: z(i) =x(i)−10

    4 : −1.3,−0.475,−0.025, 0.525, 1.3.Also, here n = 5 andp1 = .55 = .1; p2 =

    1.55 = .3; p3 =

    2.55 = .5; p4 =

    3.55 = .7; p5 =

    4.55 = .9;

    From normal table (without interpolation)η.1 = −1.28; η.3 = −0.52; η.5 = 0; η.7 = 0.53; η.9 = 1.29Draw or graph the points (ηpi , z(i)):(−1.28,−1.3), (−0.52,−0.475), (0,−0.025), (0.53, 0.525), (1.29, 1.3).

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 29 / 43

  • Continuous Distributions and Probability Plots

    The plot is given below:

    Figure: Scatter plot of z(i) vs ηpi .

    −1.5 −1 −0.5 0 0.5 1 1.5

    −1

    0

    1

    ηpi

    z (i)

    Since the graph is approximately linear, we conclude that the sample isfrom N(10, 4).(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 30 / 43

  • Continuous Distributions and Probability Plots

    Example 11. The following data refers to the coating thickness of a paint:.83, .88, .88, 1.04, 1.09, 1.12, 1.29, 1.31, 1.48, 1.49, 1.59, 1.62, 1.65, 1.71,1.76, 1.83

    Can you assume that the above data comes from normal distribution?

    Note also that the mean and the standard deviation may be taken to be thecorresponding sample quantities, as they are not mentioned in theproblem.

    The quantile plot for the above-data obtained from Minitab Package isgiven below:Since the graph is roughly linear, we conclude that the coating thickness ofa paint follows normal.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 31 / 43

  • (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 32 / 43

  • Continuous Distributions and Probability Plots

    Example 12: Consider the following normal probability plot.

    It is clear that the sample is not from the normal distribution.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 33 / 43

  • Continuous Distributions and Probability Plots

    2. Lognormal Quantile Plot

    If X ∼ LN(µ, σ), then ln(X1), . . . , ln(Xn) are from N(µ, σ). Therefore, checkif the data ln(X1), . . . , ln(Xn) are from N(µ, σ), using normal quantile plot.

    Example 12. Consider the following sample of hourly median powerreadings of radio signals:2.7, 5.4, 22.8, 30.5, 55.7, 66.2, 97.3, 186.5, 240.0.

    Are the above values from a normal or a lognormal distribution?

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 34 / 43

  • Continuous Distributions and Probability Plots

    Solution. A normal quantile plot of the observed values follows. Clearly,the variable x, the hourly median power, is not normally distributed, as thenormal quantile plot is curvilinear.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 35 / 43

  • Continuous Distributions and Probability Plots

    (b) By taking the natural logarithm of the variable and constructing anormal quantile plot of the transformed data, we obtain:

    This plot looks quite linear indicating that it is plausible that theseobservations were sampled from a lognormal distribution.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 36 / 43

  • Continuous Distributions and Probability Plots

    3. Weibull Quantile Plot.

    If X ∼ W(α, β), then area to the left of ηp , the p-th quantile, satisfies1 − e−(

    ηpβ )

    α

    = p⇒ e−(

    ηpβ )

    α

    = (1 − p)⇒ ( ηpβ )α = −ln(1 − p).

    Taking log again we get,

    α[ln(ηp) − ln(β)] = ln[−ln(1 − p)]⇒ αln(ηp) + γ = ln[− ln(1 − p)].

    Therefore,ln(−ln(1 − p)) = αln(ηp) + γ

    ⇒ ln(−ln(1 − p)) and ln(ηp) are linearly related.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 37 / 43

  • Continuous Distributions and Probability Plots

    Let pi = i−0.5n , 1 ≤ i ≤ n. Then if data x1, . . . , xn are from Weibull, thenln(ηpi ) is approximately equal to ln(x(i)).

    Hence, ln[−ln(1 − pi)] and ln(x(i)) are linearly related.

    Plot the points (ln(−ln(1 − pi)), ln(x(i))), 1 ≤ i ≤ n, where pi = i−0.5n .

    If the graph is linear (not necessarily the line passing through the origin),then x1, . . . , xn are from W(α, β) for some α and β.

    Note that we do not have to compute the population quantiles ηpi .

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 38 / 43

  • Continuous Distributions and Probability Plots

    Example 13 The following are modulus of elasticity observations forcylinders.

    37.0, 37.5, 38.1, 40.0, 40.2, 40.8, 41.0,42.0, 43.1, 43.9, 44.1, 44.6, 45.0, 46.1, 47.0,62.0, 64.3, 68.8, 70.1, 74.5.

    Use the quantiles for a sample of size 20 given above to construct anormal quantile plot, and comment on the plausibility of a normalpopulation distribution.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 39 / 43

  • Continuous Distributions and Probability Plots

    Solution:The normal quantiles are easy to generate using Matlab. The Quantile plotis:

    The graph shows some nonlinearity and hence the data may not comefrom a normal distribution.(P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 40 / 43

  • Continuous Distributions and Probability Plots

    Example 14The following observations are on lifetime (hr) of power apparatusinsulation when thermal and electric stress acceleration were fixed atparticular values:

    282, 501, 741, 851, 1072, 1122, 1202, 1585, 1905, 2138

    .

    Construct a Weibull plot and comment.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 41 / 43

  • Continuous Distributions and Probability Plots

    Solution:This Weibull quantile plot was created in Minitab. The graph is givenbelow:

    -3 -2 -1 0 1

    5.5

    6.5

    7.5

    Weibull plotting points

    ln (

    lifetim

    e)

    The plot does not show any marked deviation from linearity. So, a Weibulldistribution would be plausible for this data.

    (P. Vellaisamy: MSU-STT-351-Sum-19A) Probability & Statistics for Engineers 42 / 43

  • Continuous Distributions and Probability Plots

    HomeworkSect 4.4: 61, 66, 70Sect 4.5: 72, 76, 81, 86Sect 4.6: 88, 92

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