Chapter 1 Notes



Note: I will discuss this material during our class on September 20.

Chapter 5

Basic Definitions

• Experiment

• Outcome

• Sample Space

• Event

Probability

• Definition

• Axioms of Probability

• Mutually Exclusive/ Non- mutually Exclusive

• Independent / Dependent

Random Variables

• Definition

• Probability Distribution

• Discrete vs. Continuous RV

• Expected Value

• Probability Histogram

Discrete Distribution

• Binomial: Assumptions, Bernoulli Trials, Binomial Formula, mean and Variance

• Poisson Distribution

• Geometric Distribution

• Hypergeometric Distribution

Continuous Distributions

• Probability Density Function

• Cumulative Distribution Function

• Mean/ Expected Value of a Continuous Random Variable (RV)

• Variance/ Standard Deviation of a Continuous RV

• Other Continuous Distributions: Uniform, Exponential, Weibuil

• Example of a Joint Distribution

• The Normal Distribution

The Central Limit Theorem

INTRODUCTION TO PROBABILITY

Information in the Appendix. See if you can do any of these problems – we will discuss in our next class.

1. An experiment consists of drawing one car from a deck of 52 cards. What is the probability of

a) a red card

b) an ace

c) a king

d) a king or an ace

e) a red card or a king

2. An experiment consists or drawing two cards with replacement. What is the probability of:

a) a king on the first draw and a jack on the second

b) a three on the first and a nine on the second?

3. An experiment consists or drawing two cards without replacement. What is the probability of:

a) a king on the first draw and a jack on the second

b) a king on both draws?

4. A bin contains 5 aluminum, 2 steel and 3 brass parts. Three parts are selected. Find the probability that they are drawn in the order brass, aluminum, steel. What is the probability that 2 are aluminum and 1 is brass?

5. The probability that an integrated circuit chip will have a defective etching is 0.12; the probability that it will have a crack defect is 0.29, and the probability that it has both defects is 0.07.

a) What is the probability that a newly manufactured chip will have either an etching or a crack defect?

b) What is the probability that a new chip will have neither defect?

PROBABILITY DISTRIBUTIONS

Random Variable - Maps a real number to the outcome of an experiment

Two Types of Random Variables

1. Discrete the random variable assumes discrete (countable) values

2. Continuous the random variable can assume values represented by a continuous interval of numbers

Examples of Discrete Random Variable

• the number of automobile accidents in Houston.

• the number of building permits issued by the city during the last year

• the number of power failures per month

• the number of defective parts produced in a manufacturing operation

Examples of Continuous Random Variable

• force required to break a certain tensile specimen

• voltage

• distance

Probability Distribution for a Random Variable

Assigns probabilities to the possible outcomes as measures of the likelihood that the various numerical values will occur

Probability Function for a Discrete Random Variable

• For a discrete random variable X with possible outcomes of x1, x2, …., the probability function is a nonnegative function f(x) such that f(x) = P[X = x].

• Note that the probability function of a discrete random variable is often expressed using a table.

• Properties:

➢ f(x) assumes values between 0 and 1 (inclusive)

➢ the f(x) values sum to 1

Cumulative Probability Function

Assume X is a random variable. The function F(x) = P[X < x].

Note for a discrete random variable, F(x) = Σf(z) over z < x

Mean or Expected Value of a Discrete RV

E(X) = Σx * f(x)

E(X) = X1*P(X1) + …. + Xn*P(Xn)

Variance/Standard Deviation of a Discrete RV

Var (X) = Σ(x - EX)2 *f(x) = Σx2 *f(x) - EX2

Example of a Discrete Random Variable

Let X = the number of imperfections in a roll of sheet metal.

Suppose X has the following probability distribution.

Probability Distribution of the Number of Imperfections

|x |f(x) | |

|0 |0.40 | |

|1 |0.30 | |

|2 |0.15 | |

|3 |0.10 | |

|4 |0.05 | |

| |1.00 |Note that the sum of the probabilities is 1. Σf(x) = 1 |

The random variable X assumes the number of imperfection found, i.e. there can be 0,1,2,3, or 4 imperfections on the roll.

We formed a probability distribution by assigning probabilities to each of these outcomes. For example

f(2) = P[X = 2] = 0.15

Cumulative Probability Distribution of Number of Imperfections

|x |f(x) = P[X = x] |F(x) = P(X < x) |

|0 |0.40 |0.40 |

|1 |0.30 |0.70 |

|2 |0.15 |0.85 |

|3 |0.10 |0.95 |

|4 |0.05 |1.00 |

Note: F(3) = P[X < 3] = P[X < 2] = f(0) + f(1) + f(2) = 0.4 + 0.3 + 0.15 = 0.85

The expected value of a random variable is:

E(X) = Σx f(x) = 0*.4 + 1*.3 + 2*.15 + 3*.1 + 4*.05

= 0 + 0.3 + 0.3 + 0.3 + 0.2 = 1.1

On the average we expect the number of imperfections to be 1.1.

The variance of the random variable is:

Var (X) = Σ(x - EX)2 * f(x)

Var (X) = Σx2 * f(x) - EX2

Example: Probability Histogram

X = # of equipment failures in a one month period

In the following example, the number of equipment failures can take on a value from 0 to 9. The probability distribution on the left lists each possibility with the associated probability that it will occur. The cumulative function is also shown.

| |[pic] |

|[pic] | |

SOME DISCRETE PROBABILITY DISTRIBUTIONS

Binomial Distribution

The Binomial Distribution assumes the following

• An experiment is performed a finite number of times.

• Each outcome of the experiment can result in ‘success’ or ‘failure.’

• There is a constant probability of success (and we will label it p) and a probability of failure (q = 1 - p).

• The trials of the experiment are independent,

• X is the number of successes in n trial of the experiment

Examples

• number of defective parts

• number of projects that meet specifications

• number of employees that passed the training

• number of nonconforming transducers

• number of containers that are over filled

Probability Function for the Binomial

The binomial distribution has the following probability function

X = the number of successes in n independent Bernoulli trials of an experiment

f(x) = nCxpx * (1-p)n - x for x = 0,1,2….n

f(x) = 0 otherwise

Example of the Binomial

A manufacturer claims that only 10% of his machines require repair within one year. Find the probability of 5 repairs from 20 machines.

Use the Binomial formula to determine the probability of 5 repairs (i.e. successes) in 20 trials of the experiment.

|n = 20 |20C5 = 20!/5!* (20 – 5)! = 15,504 |

|x = 5 | |

| |px = (0.1)5 = 0.00001 |

|p = 0.10 | |

|q = 1 – 0.10 = 0.90 |(1- p)n - x = (0.9)15 = 0.20589 |

P(X = 5) = 15504*0.00001*0.20589= 0.0319

Poisson Distribution

The Poisson distribution counts the number of relatively rare events over a specified interval of space or time,

Examples of the Poisson

• the number of flaws in a length of wire

• the number of particles of contamination that occur on a storage disk

• the number of messages arriving for routing through a switching center in a communications network

• the number of imperfections in a bolt of cloth

• the number of arrivals at a retail outlet

Probability Function for the Poisson

X = # of success in an interval of time, space, distance

f(x) = e-λ λx/x! for x = 0,1,2,…...

f(x) = 0 otherwise

Example of Poisson

Tin plates that are produced by a continuous electrolytic process are inspected. The number of imperfections spotted per minute is 0.2. Find the probability of 1 imperfection in 3 minutes.

e = 2.718…

x = 1

λ = 0.2 * 3 = 0.6 (If there are 0.2 imperfections in 1 minute, we have 0.6 imperfections in 3 minutes.)

f(1) = (e-0.6 ∗ 0.61)/1! = 0.329287

Geometric Distribution

This distribution is similar to the Binomial, but it counts the number of trials to the first success.

Probability Function for the Geometric

X = # of trials until the first success

f(x) = px(1-p)n-x for x = 0,1,2….n

f(x) = 0 otherwise

Example of the Geometric Distribution

The probability that a measuring device will show excessive drift is 0.05. A series of devices is tested. What is the probability that the 6th device will show excessive drift? Find the probability of the 1st drift on the 6th trail.

P(X = 6) = (0.05)*(0.95)5 = 0.039

Discrete Probability Distributions

1. Human error is the reason for 75% of all accidents in a plant. Find the probability that human error will be reported as the reason for two of the next four accidents. [27/128]

2. During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of the chips receive a thick enough coating, find the probabilities that among 15 chips:

2.1 at least 12 will have a thick enough coating [0.2969]

2.2 at most 6 will have a thick enough coating [0.0152]

2.3 exactly 10 will have a thick enough coating [0.2061]

3. The probability that the noise level of a wide band amplifier will exceed 2 dB is 0.05. for a group of 12 amplifiers, find:

3.1 one will exceed 2 dB

3.2 at most two will exceed 2 dB

3.3 two or more will exceed 2 dB

1. Given that the switch board of a consultants office receives on the average 0.6 calls per minute, find the probability that:

4.1 in a given minute, there will be at least one call

4.2 in a 4-minute interval, there will be at least three calls

2. At a check out counter, customers arrive at an average rate of 1.5 per minute. Find the probability that:

1. at most four will arrive in any given minute [0.981]

2. at least three will arrive during an interval of 2 minutes [0.577]

3. at most 15 will arrive during an interval of 6 minutes. [0.978]

Binomial and Poisson / Minitab

Binomial

Use Calc/Probability Distributions/Binomial

MTB > # Binomial

MTB > #P = 0.05 and N = 16

MTB > CDF;

SUBC> Binomial 16 .05.

Cumulative Distribution Function

Binomial with n = 16 and p = 0.0500000

x P( X # Inverse Binomial

MTB >

MTB > INVCDF .1247;

SUBC> Binomial 16 .05.

Inverse Cumulative Distribution Function

Binomial with n = 16 and p = 0.0500000

x P( X

MTB > CDF;

SUBC> Poisson 5.

Cumulative Distribution Function

Poisson with mu = 5.00000

x P( X 0 for all x that are elements of R

(

( f(x)dx = 1

’ (

b

P(a < x < b) > ( f(x)dx

a

Note that consequence of X being a continuous random variable is that

P(X =x) = 0

and when evaluating the probability of an interval, it is not necessary to consider the equality sign i.e. P(a < X < b) = P(a < X < b).

Example

The lead concentration in gasoline ranges from 0.2 to 0.6 grams per liter. Define the random variable X.

X: grams of lead per liters

The density of the random variable is given by

f(x) = kx - 1 for 0.2 < x < 0.6

f(x) = 0 otherwise

a) Find the value of k.

b) What is the probability that a liter of gas will have between .3 and .5 grams of lead? [Ref. Milton and Arnold]

Solution

According to the definition

(

((kx-1)dx = 1

(

.2 . 6 (

( 0dx + ((kx-1)dx + ( 0dx = 1

−( ..2 .6

Cumulative Distribution Function

An alternate way of describe a continuous random variable is the cumulative distribution function (cdf).

Assume X is a random variable. The function

F(x) = P[X < x]

x

( f(x) dx for −( < x < (

−(

Example B

For the distribution function of Example a, find the cumulative density function.

Mean or Expected Value of a Continuous RV

(

E(X) = (x f(x)dx

−(

Variance/Standard Deviation of a Continuous RV

(

Var (X) = ((x - E(X))2 f(x)dx

−(

NORMAL DISTRIBUTIONS

Family of distributions, all with the same general shape.

Symmetric about the mean

The y-coordinate (height) specified in terms of the mean and the standard deviation of the distribution

Normal Probability Density

For all x

Note: e is the mathematical constant, 2.718282 ...

Standard Normal Distribution

The normal distribution with μ =0 and σ =1 is called the standard normal.

For all x:

Transformations

Normal distributions can be transformed to the standard normal.

We use what is called the z-score, which is a value that gives the number of standard deviations that X is from the mean.

Standard Normal Table

Use the table in the text to verify the following.

P(z < -2) = F(2) = 0.0228

F(2) = 0.9773

F(1.42) = 0.9222

F(-0.95) = 0.1711

Example of the Normal

The amount of instant coffee that is put into a 6 oz jar has a normal distribution with a standard deviation of 0.03. oz. What proportion of the jar contain:

a) Less than 6.06 oz?

b) More than 6.09 oz?

c) Less than 6 oz?

Normal Example - part a)

Assume μ = 6 and σ = 0.03.

The problem requires us to find

P(X < 6.06)

Convert x = 6.06 to a z-score

z = (6.06 - 6)/.03 = 2

and find: P(z < 2) = .9773

So 97.73% of the jars have less than 6.06 oz.

Normal Example - part b)

Again μ = 6 and σ = 0.03.

The problem requires us to find:

P(X > 6.09)

Convert x = 6.09 to a z-score

z = (6.09 - 6)/.03 = 3

and find

P(z > 3) = 1- P(x < 3) = 1- .9987= 0.0013

So 0.13% of the jars have more than 6.09oz.

Normal Distribution/ MTB

MTB > # Use Calc\Probability Distributions

MTB > # Normal Distribution

MTB > CDF -1.77;

SUBC> Normal 0.0 1.0.

Cumulative Distribution Function

Normal with mean = 0 and standard deviation = 1.00000

x P( X PDF -1.77;

SUBC> Normal 0.0 1.0.

Probability Density Function

Normal with mean = 0 and standard deviation = 1.00000

x f( x )

-1.7700 0.0833

MTB > InvCDF .0833;

SUBC> Normal 0.0 1.0.

Inverse Cumulative Distribution Function

Normal with mean = 0 and standard deviation = 1.00000

P( X

CENTRAL LIMIT THEOREM

specifies a theoretical distribution

formulated by the selection of all possible random samples of a fixed size n

a sample mean is calculated for each sample

Sampling Distribution Of The Mean

The mean of the sample means is equal to the mean of the population from which the samples were drawn.

The variance of the distribution is σ divided by the square root of n. (the standard error.)

Standard Error

Standard Deviation for the Distribution of Sample Means

[pic]

Central Limit Theorem

1. Consider a population with mean μ and standard deviation σ.

1. Draw a random sample of n observations from this population where n is a large number (n> 30).

2. Find the mean [pic] for each and every sample.

3. The distribution of the sample means [pic] will be approximately normal. This distribution is called the Sampling Distribution of the Means or the Distribution of Sample Means.

4. The mean and standard deviation (called the standard error) of the Distribution of Sample Means is:

| |[pic] |The mean of the Sampling Distribution equals the mean of the Population |

| |[pic] |The standard error equals the standard deviation of the population divided by the square root of the sample |

| | |size. |

6. The approximation becomes more accurate as n becomes large.

Example of CLT

A certain brand of tires has a mean life of 25,000 miles with a standard deviation of 1600 miles. What is the probability that the mean life of 64 tires is less than 24,600 miles?

Solution

The sampling distribution of the means has a mean of 25,000 miles (the population mean)

μ = 25,000 mi.

and a standard deviation (i.e. standard error) of

[pic]= 1600/8 = 200

Convert 24,600 mi. to a z-score and use the normal table to determine the required probability.

z = (24,600 – 25,000)/200 = -2

P(z < -2) = 0.0228

or 2.28% of the sample means will be less than 24,600 mi.

Distribution of Individual Values for 6 Samples from a Population with an Exponential Distribution

|[pic] |[pic] |

|[pic] |[pic] |

|[pic] |[pic] |

Distribution of the Means of 30 Samples [pic]

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[pic]

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