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  • Introduction to Probability and Statistics

    Probability & Statistics for Engineers & Scientists, 9th Ed.


    Handout #5

    Lingzhou Xue

    The pdf file for this class is available on the class web page.


  • Chapter 6

    Some Continuous Probability Distributions


  • Name Notation P.D.F. f(x)

    Uniform U[a, b] 1ba, a x b.Normal N (, 2) 1


    22}, < x 0.

    Gamma (, ) 1()x

    1ex/, x > 0.Chi-Squared 2 =



    2/2(/2)x/21ex/2, x > 0.

    Lognormal LogN(, 2) 1x

    2pie 1


    , x > 0


  • Continuous Uniform Distribution (Rectangular Distribution)

    The density function of the continuous uniform random variable

    X on the interval [a, b] is

    f(x; a, b) =

    {1ba, a x b;0, elsewhere.

    E(X) =a+ b

    2, and Var(X) =

    (b a)212



  • Probability density function for a continuous uniform random



  • Example 1

    Suppose that a large conference room for a certain company can

    be reserved for no more than 4 hours. However, the use of the

    conference room is such that both long and short conferences

    occur quite often. In fact, it can assumed that length X of a

    conference has a uniform distribution on the interval [0, 4].

    1. What is the probability density?

    2. What is the probability that any given conference lasts at

    least 3 hours?


  • Solution:


  • Normal Distribution


  • The most important continuous probability distribution in the

    entire field of statistics is the normal distribution. Its graph,

    called the normal curve, is the bell-shaped curve, which de-

    scribes approximately many phenomena that occur in nature,

    industry, and research.

    Physical measurements in areas such as meteorological ex-periments, rainfall studies, and measurements of manufac-

    tured parts are often more than adequately explained with a

    normal distribution.

    Errors in scientific measurements are extremely well approx-imated by a normal distribution.


  • Probability density function for a normal random variable.


  • Normal Distribution

    The density function of the normal random variable X, with

    mean and variance 2, is

    f(x;, 2) =12pi

    exp{(x )2

    22}, < x

  • Example 2

    Let the probability density function of a random variable X be

    f(x) =110pi


    10}, if < x

  • Solution:


  • Normal curves with 1 < 2 and 1 = 2.


  • Normal curves with 1 = 2 and 1 < 2.

  • Normal curves with 1 < 2 and 1 < 2.

  • Symmetry

    A function f(x) is sometimes said to be symmetric about the

    origin at x-axis if

    f(x) = f(x).A function f(x) is sometimes said to be symmetric about the

    at x-axis if

    f( x) = f(+ x).


  • Properties of Normal Distribution

    1. The mode (the mode is the value that occurs the most fre-quently in a probability distribution.) occurs at x = .

    2. The curve is symmetric about a vertical axis through x = .

    3. The curve has its points of inflection at x = : concavedownward if < X < +, and concave upward otherwise.

    4. The normal curve approaches the horizontal axis asymptoti-cally as we proceed in either direction away from the mean.

    5. The total area under the curve and above the horizontal axisis equal to 1.


  • Mean of a Normal Random Variable

    f(x;, 2) =12pi


    22 , < x

  • Variance of a Normal Random Variable

    f(x;, 2) =12pi


    22 , < x

  • Areas Under Normal Curve


  • Areas Under the Normal Curve

    The curve of any continuous probability distribution or density

    function is constructed so that the area under the curve bounded

    by the two ordinates x = x1 and x = x2 equals the probability

    that the random variable X assumes a value between x1 and x2.

    Thus, for the normal curve

    P (x1 < X < x2) = x2x1



    22 dx

    is represented by the area of the shaded region.


  • P (x1 < X < x2)= area of the shaded region.


  • The difficulty encountered in solving integrals of normal density

    functions necessitates the tabulation of normal curve areas for

    quick reference. However, it would be a hopeless task to attempt

    to set up separate tables for every conceivable values of and

    . Fortunately, we are able to transform all the observations of

    any normal random variable X to a new set of observation of a

    normal random variable Z with mean 0 and variance 1. This can

    be done by means of the transformation

    Z =X


    Whenever X assumes a values x, the corresponding value of Z

    is given by z = x . Therefore, if X falls between the valuesx = x1 and x = x2, the random variable Z will fall between the


  • corresponding values z1 =x1 and z2 =

    x2 . Consequently,

    P (x1 < X < x2) = x2x1

    n(x;, )dx

    = P(z1 =


    k) = 0.3015 and

    2. P (k < Z < 0.18) = 0.4917.


  • Solution:


  • Area of Example 4.


  • Example 5

    Given a random variable X having a normal distribution with

    = 50 and = 10, find the probability that X assumes a value

    between 45 and 62.



  • Using the Normal Curve in Reverse

    Occasionally, we are required to find the value of z corresponding

    to a specified probability that falls between values. For conve-

    nience, we shall always choose the z value corresponding to the

    tabular probability that comes closest to the specified probability.

    We reverse the process and begin with a known area or proba-

    bility, find the z value, and then determine x by rearranging the


    z =x

    to give x = z + .


  • Example 6

    Given a normal distribution with = 40 and = 6, find the

    value of x that has

    1. 45% of the area to the left, and

    2. 14% of area to the right.


  • Solution:


  • Area of Example 6.


  • Applications of the Normal Distribution


  • Example 7

    A certain type of storage battery lasts, on average, 3.0 years witha standard deviation of 0.5 year. Assuming that the battery livesare normally distributed, find the probability that a given batterywill last less than 2.3 years.Solution:


  • Example 8

    An electrical firm manufactures light bulbs that have a life, be-fore burn-out, that is normally distributed with mean equal to800 hours and a standard deviation of 40 hours. Find the prob-ability that a bulb burns between 778 and 834 hours.Solution:


  • Example 9

    In an industrial process the diameter of a ball bearing is an impor-

    tant component part. The buyer sets specifications on the diam-

    eter to be 3.0 0.01 cm. The implication is that no part fallingoutside these specifications will be accepted. It is known that in

    the process the diameter of a ball bearing has a normal distribu-

    tion with mean = 3.0 and standard deviation = 0.005. On

    average, how many manufactured ball bearings will be scrapped?


  • Solution:


  • Example 10

    The average grade for an exam is 74, and the standard deviationis 7. If 25% of the class is given As, and the grades are curvedto follow a normal distribution what is the lowest possible A andthe height possible B?Solution:


  • Normal Approximation to the Binomial


  • Theorem

    If X is a binomial random variable with mean = np and variance

    = npq = np(1p), then the limiting form of the distribution of

    T =X npnpq

    as n, is the standard normal distribution.

    Normal approximation of b(x; 15,0.4).


  • Normal Approximation to the Binomial-I

    Let X be a binomial random variable with parameters n and p.

    Then X has approximately a normal distribution with = np and

    2 = npq = np(1 p) and

    P (X x) =x


    b(k;n, p) (1)

    area under normal curve to the left of x+ 0.5 (2) P

    (Z x+ 0.5 np



    where Z N(0,1), and the approximation will be good if np andn(1 p) are greater than or equal to 5.


  • Normal Approximation to the Binomial-II

    Let X be a binomial random variable with parameters n and p.

    Then X has approximately a normal distribution with = np and

    2 = npq = np(1 p) and

    P (x1 X x2) =x2


    b(k;n, p)

    area under normal curve tothe right of x1 0.5 and left of x2 + 0.5.

    P(x1 0.5 np

    npq Z x2 + 0.5 np


    )where Z N(0,1), and the approximation will be good if np andn(1 p) are greater than or equal to 5.


  • Example 11

    The probability that a patient recovers from a rare blood disease

    is 0.4. If 100 people are known to have contracted this disease,

    what is the probability that less than 30 survive?



  • Example 12

    A multiple-choice quiz has 200 questions each with 4 possibleanswers of which only 1 is the correct answer. What is theprobability that sheer guesswork yields from 25 to 30 correct an-swers for 80 of the 200 problems about which the student hasno knowledge?Solution: