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Num 5

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.

1

Chapter 6

Some Continuous Probability Distributions

2

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

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

2piexp{(x)2

22}, < x 0.

Gamma (, ) 1()x

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

(2,2

)1

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

Lognormal LogN(, 2) 1x

2pie 1

22[ln(x)]2

, x > 0

3

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

.

4

Probability density function for a continuous uniform random

variable.

5

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?

6

Solution:

7

Normal Distribution

8

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.

9

Probability density function for a normal random variable.

10

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

exp{x2

10}, if < x

Solution:

13

Normal curves with 1 < 2 and 1 = 2.

14

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).

15

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.

16

Mean of a Normal Random Variable

f(x;, 2) =12pi

e(x)2

22 , < x

Variance of a Normal Random Variable

f(x;, 2) =12pi

e(x)2

22 , < x

Areas Under Normal Curve

19

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

12pi

e(x)2

22 dx

is represented by the area of the shaded region.

20

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

21

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

22

corresponding values z1 =x1 and z2 =

x2 . Consequently,

P (x1 < X < x2) = x2x1

n(x;, )dx

= P(z1 =

x1

k) = 0.3015 and

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

26

Solution:

27

Area of Example 4.

28

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.

Solution:

29

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

to give x = z + .

30

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.

31

Solution:

32

Area of Example 6.

33

Applications of the Normal Distribution

34

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:

36

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?

37

Solution:

38

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:

39

Normal Approximation to the Binomial

40

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).

41

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

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

42

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

k=x1

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

npq

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

43

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:

44

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:

45