handout five

Introduction to Probability and Statistics

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

2009

Handout #5

Lingzhou Xue

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

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Chapter 6

Some Continuous Probability Distributions

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Name Notation P.D.F. f(x)

Uniform U[a, b] 1b−a, a ≤ x ≤ b.

Normal N (µ, σ2) 1√2πσ

exp{−(x−µ)2

2σ2 }, −∞ < x <∞.

Exponential Exp(β) 1βe−x/β, x > 0.

Gamma Γ(α, β) 1βαΓ(α)x

α−1e−x/β, x > 0.

Chi-Squared χ2υ = Γ

(υ2,2

)1

2υ/2Γ(υ/2)xυ/2−1e−x/2, x > 0.

Lognormal LogN(µ, σ2) 1xσ√

2πe− 1

2σ2[ln(x)−µ]2, x > 0

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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) =

{1b−a, a ≤ x ≤ b;0, elsewhere.

E(X) =a+ b

2, and Var(X) =

(b− a)2

12.

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Probability density function for a continuous uniform random

variable.

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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?

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Solution:

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Normal Distribution

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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.

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Probability density function for a normal random variable.

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Normal Distribution

The density function of the normal random variable X, with

mean µ and variance σ2, is

f(x;µ, σ2) =1√2πσ

exp{−(x− µ)2

2σ2}, −∞ < x <∞.

where π = 3.12159 . . ., and e = 2.71828 . . . .

E(X) = µ, and Var(X) = σ2.

Standard Normal Distribution

The distribution of a normal random variable with mean 0 and

variance 1 is called a standard normal distribution.

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Example 2

Let the probability density function of a random variable X be

f(x) =1√10π

exp{−x2

10}, if −∞ < x <∞.

1. Find the mean and variance of the random variable X.

2. Use Chebyshev’s inequality to show that

P (|X| ≥√

10) ≤1

2.

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Solution:

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Normal curves with µ1 < µ2 and σ1 = σ2.

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Normal curves with µ1 = µ2 and σ1 < σ2.

Normal curves with µ1 < µ2 and σ1 < σ2.

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 µ

at x-axis if

f(µ− x) = f(µ+ x).

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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 = µ.

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

3. The curve has its points of inflection at x = µ± σ: concavedownward if µ−σ < X < µ+σ, 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.

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Mean of a Normal Random Variable

f(x;µ, σ2) =1√2πσ

e−(x−µ)2

2σ2 , −∞ < x <∞.

Prove that E(X) = µ.

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Variance of a Normal Random Variable

f(x;µ, σ2) =1√2πσ

e−(x−µ)2

2σ2 , −∞ < x <∞.

Prove that Var(X) = σ2.

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Areas Under Normal Curve

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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) =∫ x2

x1

1√2πσ

e−(x−µ)2

2σ2 dx

is represented by the area of the shaded region.

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

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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 µ and

σ. 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 − µσ

.

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

is given by z = x−µσ . Therefore, if X falls between the values

x = x1 and x = x2, the random variable Z will fall between the

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corresponding values z1 = x1−µσ and z2 = x2−µ

σ . Consequently,

P (x1 < X < x2) =∫ x2

x1

n(x;µ, σ)dx

= P

(z1 =

x1 − µσ

<X − µσ

< z2 =x2 − µσ

)= P

(z1 =

x1 − µσ

< Z < z2 =x2 − µσ

)=∫ z2

z1

n(x; 0,1)dz

where Z is seen to be a normal random variable with mean 0

and variance 1.

Example 3

Given a standard normal distribution, find the area under the

curve that lies

1. to the right of z = 1.84, and

2. between z = −1.97 and z = 0.86.

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Solution:

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Area of Example 3.

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Example 4

Given a standard normal distribution, find the value of k such

that

1. P (Z > k) = 0.3015 and

2. P (k < Z < −0.18) = 0.4917.

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Solution:

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Area of Example 4.

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Example 5

Given a random variable X having a normal distribution with

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

between 45 and 62.

Solution:

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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

formula

z =x− µσ

to give x = σz + µ.

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Example 6

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

value of x that has

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

2. 14% of area to the right.

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Solution:

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Area of Example 6.

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Applications of the Normal Distribution

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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:

35

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:

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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 falling

outside 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 µ = 3.0 and standard deviation σ = 0.005. On

average, how many manufactured ball bearings will be scrapped?

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Solution:

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Example 10

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

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Normal Approximation to the Binomial

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Theorem

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

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

T =X − np√npq

as n→∞, is the standard normal distribution.

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

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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 µ = np and

σ2 = npq = np(1− p) and

P (X ≤ x) =x∑

k=0

b(k;n, p) (1)

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

≈ P(Z ≤

x+ 0.5− np√npq

)(3)

where Z ∼ N(0,1), and the approximation will be good if np and

n(1− p) are greater than or equal to 5.

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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 µ = np and

σ2 = npq = np(1− p) and

P (x1 ≤ X ≤ x2) =x2∑

k=x1

b(k;n, p)

≈ area under normal curve to

the right of x1 − 0.5 and left of x2 + 0.5.

≈ P(x1 − 0.5− np√npq

≤ Z ≤x2 + 0.5− np√npq

)where Z ∼ N(0,1), and the approximation will be good if np and

n(1− p) are greater than or equal to 5.

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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?

Solution:

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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:

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Gamma and Exponential Distributions

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Exponential and Gamma Distributions

Exponential Distribution Describing the time until the occur-

rence of a Poisson event (or the time between Poisson events).

Gamma Distribution Describing the time (or space) occurring

until a specified number of Poisson events occur.

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Gamma and Exponential Distributions

Although the normal distribution can be used to solve many

problems in engineering and science, there are still numerous

situations that require different types of density functions. Two

such density functions, the gamma and exponential distributions.

The exponential and gamma distributions play an important role

in both queuing theory and reliability problems. Time between

arrivals at service facilities, and time to failure of component

parts and electrical systems.

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Gamma function

The gamma function is defined by

Γ(α) =∫ ∞

0xα−1e−xdx, for α > 0.

Prove: Γ(α) = (α− 1)Γ(α− 1) for α > 1.

Integrating by parts with u = xα−1 and dv = e−xdx, we obtain

Γ(α) = xα−1e−x|∞0 +∫ ∞

0(α−1)xα−2e−xdx = (α−1)

∫ ∞0

xα−2e−xdx

for α > 1 which yields the recursion formula

Γ(α) = (α− 1)Γ(α− 1).

Property

Γ(

1

2

)=√

2π.

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Gamma Distribution

The continuous random variable X has a gamma distribution,

with parameters α and β, if the density function given by

f(x;α, β) =

{ 1βαΓ(α)x

α−1e−x/β, x > 0;

0, elsewhere.

where α > 0 and β > 0.

E(X) = αβ and Var(X) = αβ2.

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Gamma Distribution

Gamma distributions.

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Exponential Distribution

The continuous random variable X has an exponential distribu-

tion, with parameter β, if the density function given by

f(x;β) =

{1βe−x/β, x > 0;

0, elsewhere.

where β > 0.

E(X) = β and Var(X) = β2.

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Example 13

What is the probability of an item surviving until t = 100 units ifthe item is exponentially distributed with a mean time betweenfailure of 80 units? Given that the item survived to 200 units,what is the probability of survival until t = 300units?Solution:

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Relationship to the Poisson Process

Let X be the amount of time until the first arrival in a Poisson

process with rate λ. Then X ∼ Exp(λ).

Proof:

Note that the number of arrivals in [0, t] is Poi(λt).

F (x) = P (X ≤ t) (4)

= 1− P (no arrivals in [0, t]) (5)

= 1−eλt(λt)0

0!(6)

= 1− eλt. (7)

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Relationship to the Poisson Process

The Poisson distribution is used to compute the probability ofspecific numbers of ”events” during a particular period of time orspace. In many applications, the time period or span of space isthe random variable. For example, an industrial engineer may beinterested in modeling the time T between arrivals at a congestedintersection during rush hour in a large city. An arrival representsthe Poisson event.

The poisson distribution was developed as a single-parameterdistribution with parameter λ, where λ may be interpreted as themean number of events per unit time. Consider now the randomvariable described by the time required for the first event tooccur. Using the Poisson distribution, we find that the probabilityof no events occurring in the span up to time t is given by

p(0;λt) =e−λt(λt)0

0!= e−λt.

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We can now make use of the above and let X be the time tothe first Poisson event. The probability that the length of timeuntil the first event will exceed x is the same as the probabilitythat no Poisson events will occur in x. The latter, of course, isgiven by e−λt. As a result,

P (X > x) = e−λx.

Thus the cumulative distribution function for X is given by

F (x) = P (0 ≤ X ≤ x) = 1− e−λx.

Now, in order that we recognize the presence of the exponen-tial distribution, we may differentiate the cumulative distributionfunction above to obtain the density function

f(x) =d

dxF (x) = λe−λx,

which is the density function of the exponential distribution withλ = 1/β.

Example 14

Let X equal the number of bad records in each 50 feet of a used

computer tape. Assume that X has a Poisson distribution with

mean 1. Let W equal the number of feet before the first bad

record is found.

1. Give the mean and variance of the number of flaws per foot.

2. How is W distributed?

3. Give the mean and variance of W .

4. Find P (W > 1001|W > 951).

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Solution:

Application of Gamma and Exponential Distributions

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Application of Gamma and Exponential Distributions

The application of the exponential distribution are in ”time to

arrival” or time to Poisson event problems. Notice that the mean

of the exponential distribution is the parameter β, the reciprocal

of the parameter in the Poisson distribution. Recall: the Poisson

distribution has no memory, implying the occurrences in succes-

sive time periods are independent. The important parameter β

is the mean time between events. In reliability theory, where

equipment failure often conforms to this Poisson process, β is

called mean time between failures. Many equipment breakdowns

do follow the Poisson process, and thus the exponential distri-

bution does apply. Other applications include survival times in

biomedical experiments and computer response time.

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Example 15

Suppose that a system contains a certain type of component

whose time, in years, to failure is given by T . The random

variable T is modeled nicely by the exponential distribution with

mean time to failure β = 5. If 5 of these components are in-

stalled in different systems, what is the probability that at least

2 are still functioning at the end of 8 years?

Solution:

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Memoryless Property for Exponential Distribution

A nonnegative random variable X is called memoryless if for all

s, t ≥ 0,

P (X ≥ t) = P (X ≥ s+ t|X ≥ s)

The probability of lasting an additional t time units is the same

as the probability of lasting t time units. The fact that it hasn’t

happened yet, tells us nothing about how much longer it will

take before it does happen.

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To show that an exponential distribution is memoryless.

Solution:

61

Example 16

The lifetime of a TV tube (in years) is an exponential random

variable with mean 10. If Jim bought his TV set 10 years ago,

what is the probability that its tube will last another 10 years?

Solution:

62

Example 17

Suppose that telephone calls arriving at a particular switchboardfollow a Poisson process with an average of 5 calls coming perminute. What is the probability that up to a minute will elapseuntil 2 calls have come in to the switchboard?Solution:

63

Example 18

Based on extensive testing it is determined that the time Y

in years before a major repair is required for a certain washing

machine is characterized by the density function

f(x) =

{14, y ≥ 0;0, elsewhere.

Note this ia na exponential with µ = 4 years. The machine is

considered a bargain if it is unlikely to require a major repair

before the sixth year. Thus, what is the probability P (Y > 6)?

Also, what is the probability that a major repair occurs in the

first year?

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Solution:

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Chi-Squared Distribution

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Chi-Squared Distribution

The continuous random variable X has a chi-squared distribution,

with υ degrees of freedom, if its density function is given by

f(x;α, β) =

1

2υ/2Γ(υ/2)xυ/2−1e−x/2, x > 0;

0, elsewhere.

where υ is a positive integer.

This is very important special case of the gamma distribution

is obtained by letting α = υ/2 and β = 2, where υ is a positive

integer.

E(X) = υ and Var(X) = 2υ.

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Lognormal Distribution

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Lognormal Distribution

The continuous random variable X has a lognormal distribution

if the random variable Y = ln(X) has a normal distribution with

mean µ and standard deviation σ. The resulting density function

of X is

f(x;µ, σ) =

1σx√

2πe− 1

2σ2[ln(x)−µ]2, x ≥ 0;

0, x < 0.

E(X) = eµ+σ2/2 and Var(X) = e2µ+σ2(eσ

2− 1).

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Example 19

Concentration of pollutants produced by chemical plants histor-

ically are known to exhibit that resembles a lognormal distri-

bution. This is important when one considers issues regarding

compliance to government regulations. Suppose it is assumed

that the concentration of a certain pollutant, in parts per million,

has a lognormal distribution with parameters µ = 3.2 and σ = 1.

What is the probability that the concentration exceeds 8 parts

per million?

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Solution:

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Example 20

The life, in thousands of miles, of a certain type of electroniccontrol for locomotives has an approximate lognormal distribu-tion with µ = 5.149 and σ = 0.737. Find the 5th percentile ofthe life of such locomotive.Solution:

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