copyright ©2006 brooks/cole, a division of thomson learning, inc. random variables chapter 8

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

RandomVariables

Chapter 8

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 2

8.1 What is a Random Variable?

Random Variable: assigns a number to each outcome of a random circumstance, or, equivalently, to each unit in a population.

Two different broad classes of random variables:

1. A continuous random variable can take any value in an interval or collection of intervals.

2. A discrete random variable can take one of a countable list of distinct values.


Random factors that will determine how enjoyable the event is:

Temperature: continuous random variable

Number of airplanes that fly overhead:discrete random variable

Example 8.1 Random Variables at an Outdoor Graduation or Wedding


• What is the probability that three tosses of a fair coin will result in three heads?

• Assuming boys and girls are equally likely, what is the probability that 3 births will result in 3 girls?

• Assuming probability is 1/2 that a randomly selected individual will be taller than median height of a population, what is the probability that 3 randomly selected individuals will all be taller than the median?

Example 8.2 Probability an Event Occurs Three Times in Three Tries

Answer to all three questions = 1/8.

Discrete Random Variable X = number of times the “outcome of interest” occurs in three independent tries.


8.2 Discrete Random Variables

Discrete random variable: can only result in a countable set of possibilities – often a finite number of outcomes, but can be infinite.

X = the random variable.

k = a number the discrete r.v. could assume.

P(X = k) is the probability that X equals k.

Example 8.3 It’s Possible to Toss Forever

Repeatedly tossing a fair coin, and define:X = number of tosses until the first head occurs

Any number of flips is a possible outcome.

P(X = k) = (1/2)k


Probability Distribution of a Discrete R.V.

Using the sample space to find probabilities:

Step 1: List all simple events in sample space.

Step 2: Find probability for each simple event.

Step 3: List possible values for random variable X and identify the value for each simple event.

Step 4: Find all simple events for which X = k, for each possible value k.

Step 5: P(X = k) is the sum of the probabilities for all simple events for which X = k.

Probability distribution function (pdf) X is a table or rule that assigns probabilities to possible values of X.


Example 8.4 How Many Girls are Likely? Family has 3 children. Probability of a girl is ½.What are the probabilities of having 0, 1, 2, or 3 girls?

Sample Space: For each birth, write either B or G. There are eight possible arrangements of B and G for three births. These are the simple events.

Sample Space and Probabilities: The eight simple events are equally likely.

Random Variable X: number of girls in three births. For each simple event, the value of X is the number of G’s listed.


Example 8.4 How Many Girls? (cont)

Probability distribution function for Number of Girls X:

Value of X for each simple event:

Graph of the pdf of X:


Conditions for Probabilities for Discrete Random Variables

Condition 1 The sum of the probabilities over all possible values of a discrete random variable must equal 1.

Condition 2 The probability of any specific outcome for a discrete random variable must be between 0 and 1.


Cumulative Distribution Function of a Discrete Random Variable

Cumulative distribution function (cdf) for a random variable X is a rule or table that provides the probabilities P(X ≤ k) for any real number k.

Cumulative probability = probability that X is less than or equal to a particular value.

Example 8.4 Cumulative Distribution Function for the Number of Girls (cont)


Finding Probabilities for Complex Events

Example 8.4 A Mixture of Children

pdf for Number of Girls X:

What is the probability that a family with 3 children will have at least one child of each sex?

If X = Number of Girls then either family has one girl and two boys (X = 1) or two girls and one boy (X = 2).

P(X = 1 or X = 2) = P(X = 1) + P(X = 2) = 3/8 + 3/8 = 6/8 = 3/4


8.3 Expectations for Random Variables

The expected value of a random variable is the mean value of the variable X in the sample space, or population, of possible outcomes.

If X is a random variable with possible values x1, x2, x3, . . . , occurring with probabilities p1, p2, p3, . . . , then the expected value of X is calculated as

ii pxXE


Example 8.6 California Decco Lottery

How much would you win/lose per ticket over long run?

Player chooses one card from each of four suits. Winning card drawn from each suit. If one or more matches the winning cards => prize. It costs $1.00 for each play.

=> Lose an average of 35 cents per play.


Standard Deviation for a Discrete Random Variable

The standard deviation of a random variable is essentially the average distance the random variable falls from its mean over the long run.

If X is a random variable with possible values x1, x2, x3, . . . , occurring with probabilities p1, p2, p3, . . . , and expected value E(X) = , then

ii

ii

pxX

pxXVX

2

22

ofDeviation Standard

of Variance


Example 8.7 Stability or Excitement

Two plans for investing $100 – which would you choose?

Expected Value for each plan:

Plan 1:E(X ) = $5,000(.001) + $1,000(.005) + $0(.994) = $10.00

Plan 2: E(Y ) = $20(.3) + $10(.2) + $4(.5) = $10.00


Example 8.7 Stability or Excitement (cont)

Variability for each plan:

Plan 1: V(X ) = $29,900.00 and = $172.92

Plan 2: V(X ) = $48.00 and = $6.93

The possible outcomes for Plan 1 are much more variable. If you wanted to invest cautiously, you would choose Plan 2, but if you wanted to have the chance to gain a large amount of money, you would choose Plan 1.


8.3 Binomial Random Variables

1. There are n “trials” where n is determined in advance and is not a random value.

2. Two possible outcomes on each trial, called “success” and “failure” and denoted S and F.

3. Outcomes are independent from one trial to the next.

4. Probability of a “success”, denoted by p, remains same from one trial to the next. Probability of “failure” is 1 – p.

Class of discrete random variables = Binomial -- results from a binomial experiment.

Conditions for a binomial experiment:


Examples of Binomial Random Variables

A binomial random variable is defined as X=number

of successes in the n trials of a binomial experiment.


Finding Binomial Probabilities

p = probability win = 0.2; plays of game are independent.

X = number of wins in three plays.

What is P(X = 2)?

knk ppknk

nkXP

1

!!

!for k = 0, 1, 2, …, n

Example 8.9 Probability of Two Wins in Three Plays

096.0)8(.)2(.3

2.12.!23!2

!32

12

232

XP


Expected Value and Standard Deviation for a Binomial Random Variable

For a binomial random variable X based on n trials and success probability p,

pnp

npXE

1 deviation Standard

Mean


Example 8.12 Extraterrestrial Life? 50% of large population would say “yes” if asked, “Do you believe there is extraterrestrial life?”

Sample of n = 100 is taken.

X = number in the sample who say “yes” is approximately a binomial random variable.

In repeated samples of n=100, on average 50 people would say “yes”. The amount by which that number would differ from sample to sample is about 5.

55.)5(.100 deviation Standard

50)5(.100 Mean

XE


8.5 Continuous Random Variables

Continuous random variable: the outcome can be any value in an interval or collection of intervals.

Probability density function for a continuous random variable X is a curve such that the area under the curve over an interval equals the probability that X is in that interval.

P(a X b) = area under density curve over the interval between the values a and b.


Example 8.13 Time Spent Waiting for Bus

Bus arrives at stop every 10 minutes. Person arrives at stop at a random time, how long will s/he have to wait?

X = waiting time until next bus arrives.

X is a continuous random variable over 0 to 10 minutes.

Note: Height is 0.10 so total area under the curve is (0.10)(10) = 1

This is an example of a Uniform random variable


Example 8.13 Waiting for Bus (cont)

What is the probability the waiting time X was in the interval from 5 to 7 minutes?

Probability = area under curve between 5 and 7

= (base)(height) = (2)(.1) = .2


8.6 Normal Random Variables

If a population of measurements follows a normal curve, and if X is the measurement for a randomly selected individual from that population, then

X is said to be a normal random variable

X is also said to have a normal distribution

Any normal random variable can be completely characterized by its mean, , and standard deviation, .


Example 8.14 College Women’s Heights

Data suggest the distribution of heights of college women modeled by a normal curve with mean = 65 inches and standard deviation = 2.7 inches.

Note: Tick marks given at the mean and at 1, 2, 3 standard deviations above and below the mean.

Empirical Rule values are exact characteristics of a normal curve model


Standard Scores

The formula for converting any value x to a z-score is

x

zdeviation Standard

MeanValue

A z-score measures the number of standard deviations that a value falls from the mean.


Example 8.14 Height (cont)

For a population of college women, the z-score corresponding to a height of 62 inches is

This z-score tells us that 62 inches is 1.11 standard deviations below the mean height for this population.

11.17.2

6562

deviation Standard

MeanValue

z


Finding Probabilities for z-scores

Table A.1 = Standard Normal (z) Probabilities

• Body of table contains P(Z z*).

• Left-most column of table shows algebraic sign, digit before the decimal place, the first decimal place for z*.

• Second decimal place of z* is in column heading.


More Finding Probabilities for z-scores

Table A.1 = Standard Normal (z) Probabilities

P(Z -3.00) =.0013 (see in portion above)

P(Z −2.59) = .0048

P(Z 1.31) = .9049

P(Z 2.00) = .9772

P(Z -4.75) = .000001 (from in the extreme)


Example 8.15 Probability Z > 1.31

P(Z > 1.31) = 1 – P(Z 1.31)

= 1 – .9049 = .0951


Example 8.16 Probability Z is between –2.59 and 1.31

P(-2.59 Z 1.31)

= P(Z 1.31) – P(Z -2.59)

= .9049 – .0048 = .9001


Use z-scores to Solve General Problems


What is the probability that a randomly selected college woman is 62 inches or shorter?

1335.11.1

7.2

656262

ZP

ZPXP

About 13% of college women are 62 inches or shorter.


Use z-scores to Solve General Problems


What proportion of college woman are taller than 68 inches?

1335.8665.1

11.1111.17.2

658668

ZPZPZPXP

About 13% of college women are taller than

68 inches.


Finding PercentilesIf 25th percentile of pulse rates is 64 bpm, then 25% of pulse rates are below 64 and 75% are above 64. The percentile is 64 bpm, and the percentile ranking is 25%.

Step 1: Find z-score that has specified cumulative probability. Using Table A.1, find percentile rank in body of table and read off the z-score.

Step 2: Calculate the value of variable that has the z-score found in step 1: x = z* + .


Example 8.17 75th Percentile of Systolic Blood Pressure

Blood Pressures are normal with mean 120 and standard deviation 10. What is the 75th percentile?

Step 1: Find closest z* with area of 0.7500 in body of Table A.1.

z = 0.67

Step 2: Calculate x = z* +

x = (0.67)(10) + 120 = 126.7 or about 127.


8.7 Approximating Binomial Distribution ProbabilitiesIf X is a binomial random variable based on n trials with success probability p, and n is large, then the random variable X is also approximately normal, with mean and standard deviation given as:

Conditions: Both np and n(1 – p) are at least 10.

pnp

npXE

1 deviation Standard

Mean


Example 8.18 Number of H in 30 Flips

X = number of heads in n = 30 flips of fair coinBinomial distribution with n = 30 and p = .5.

Distribution is bell-shaped and can be approximated by a normal curve.

74.25.)5(.30 deviation Standard

15)5(.30 Mean

XE


Example 8.19 Political Woes

Poll: n = 500 adults; 240 supported position

If 50% of all adults support position, what is the probability that a random sample of 500 would find 240 or fewer holding this position? P(X 240)

2.115.)5(.100 deviation Standard

250)5(.500 Mean

XEX is approximately normal with

1867.89.2.11

250240240

ZPZPXP

Not unlikely to see 48% or less, even if 50% in population favor.


8.8 Sums, Differences, Combinations of R.V.s

A linear combination of random variables, X, Y, . . . is a combination of the form:

L = aX + bY + …

where a, b, etc. are numbers – positive or negative.

Most common:Sum = X + Y Difference = X – Y


Means of Linear Combinations

The mean of L is:

Mean(L) = a Mean(X) + b Mean(Y) + …

Most common:

Mean( X + Y) = Mean(X) + Mean(Y)

Mean(X – Y) = Mean(X) – Mean(Y)


Variances of Linear Combinations

If X, Y, . . . are independent random variables, then

Variance(L) = a2 Variance(X) + b2 Variance(Y) + …

Most common:

Variance( X + Y) = Variance(X) + Variance(Y)

Variance(X – Y) = Variance(X) + Variance(Y)


If X, Y, . . . are independent normal random variables, then L = aX + bY + … is normally distributed.

In particular:

X + Y is normal with

X – Y is normal with

Combining Independent Normal Random Variables

22 deviation standard

mean

YX

YX

22 deviation standard

mean

YX

YX


Example 8.20 Will Meg Miss Her Flight?

X = Meg’s travel time is normal with mean 25 minutes and std dev 3 minutes.

Y = Airport time is normal with mean 15 minutes and std dev 2 minutes.

X and Y independent so T = X + Y = total time is normal with

6.323 deviation standard

401525 mean

22

0823.39.16.3

404545

ZPZPTP

If Meg leaves with 45 minutes before flight takes off, what is probability she will miss her flight?


If X, Y, . . . are independent binomial random variables with nX, nY, etc trials and all have same success probability p, then the sum X + Y + … is a binomial random variable with n = nX + nY + … and success probability p.

If the success probabilities differ, the sum is not a binomial random variable.

Combining Independent Binomial Random Variables


Example 8.22 Donations Add Up

X = # donors by volunteer 1; binomial n = 10, p = .2

Y = # donors by volunteer 2; binomial n = 12, p = .2

W = # donors by volunteer 3; binomial n = 18, p = .2

T = X + Y + W is binomial n = 40, p = .2

2682.10 TP

Fund drive: Three volunteers call potential donors. About 20% called make a donation, independent of who called. If volunteers make 10, 12 and 18 calls, respectively, what is the probability that they get at least ten donors?

Using calculator or computer:

copyright ©2006 brooks/cole, a division of thomson learning, inc. random variables chapter 8

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