# Handout Five

• View
7

0

Embed Size (px)

DESCRIPTION

Num 5

### Text of 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.

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

Documents
Documents
Documents
Education
Documents
Documents
Art & Photos
Technology
Documents
Documents
Documents
Documents
Documents
Education
Documents
Documents
Documents
Education
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents