P ROBABILITY & S TATISTICS FOR P-8 T EACHERS Chapter 5 Probability Distributions.

PROBABILITY & STATISTICS FOR P-8

TEACHERS

Chapter 5

Probability Distributions

PROBABILITY DISTRIBUTIONSWe will now combine the methods of descriptive statistics (Chapter 2 and 3) and those of probability (Chapter 4) to describe and analyze

probability distributions.

Probability Distributions describe what will probably happen instead of what actually did happen, and they are often given in the format of a graph, table, or formula.

In this chapter we will construct probability distributions by presenting possible outcomes along with the relative frequencies we expect.

COMBINING DESCRIPTIVE METHODS AND PROBABILITIES

PROBABILITY DISTRIBUTIONS A random variablerandom variable is a variable whose

values are determined by chance.

A discrete probability distributiondiscrete probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values.

The sum of the probabilitiesprobabilities of all events in a sample space add up to 1. Each probability is between 0 and 1, inclusively.

RANDOM VARIABLESTwo types of random variables:

A discrete random variablediscrete random variable can assume a countable number of values.Number of steps to the top of the Eiffel

Tower

A continuous random variablecontinuous random variable can assume any value along a given interval of a number line.The time a tourist stays at the top

once s/he gets there

RANDOM VARIABLES Discrete random variables

Number of salesNumber of callsShares of stockPeople in lineMistakes per page

Continuous random variablesLengthDepthVolumeTimeWeight

PROBABILITY DISTRIBUTIONS

The probability distributionprobability distribution of a discrete random variable is a graph, table or formula that specifies the probability associated with each possible outcome the random variable can assume.0 ≤ P(x) ≤ 1 for all values of xP(x) = 1

x P(x)

0 0.16

1 0.18

2 0.22

3 0.10

4 0.30

5 -0.01

Is the following a probability distribution?

PROBABILITY DISTRIBUTIONS

0.95

P(x) < 0

∑P(x) ≠ 1

Not a probability distribution

PROBABILITY DISTRIBUTIONConstruct a probability distribution for tossing a coin twice and recording the number of heads

# of Heads

x

ProbabilityP(x)

0 0.25

1 0.50

2 0.25

HH

H T

T H

T T

P(0 heads)=

P(1 head) =

P(2 heads)=

1/42/41/4

A probability histogramprobability histogram is a histogram in which the horizontal axis corresponds to the value of the random variable and the vertical axis represents the probability of that value of the random variable.

PROBABILITY HISTOGRAM

DVDs Rented at a Video Store

0

0.1

0.2

0.30.4

0.5

0.6

0.7

0 1 2 3 4 5

Number of DVDs Rented

Pro

ba

bili

ty

Draw a probability histogram of the probability distribution to the right, which represents the number of DVDs a person rents from a video store during a single visit.

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01

PROBABILITY HISTOGRAM

a. Determine the probability distribution of the random variable x.

b. Construct a probability histogram for the random variable x.

A survey was completed regarding how many siblings are in your family. Let X denote the number of siblings of a randomly selected student.

PROBABILITY DISTRIBUTION

Solution


Probability Distribution

Probability Histogram

Just like with any distribution, we would like to analyze the data. The mean and standard deviation are the most common measurements.

To find these measurements, we will treat the probability distribution just like a frequency distribution.

Since probability distributions represent theoretical data, we will treat the results as a true population.


MEAN AND STANDARD DEVIATION

Mean: µ = ∑ x P(x)

Variance: σ2 = ∑ (x – µ)2 P(x)

or σ2 = ∑ x2 P(x) – µ2

MEAN & STANDARD DEVIATIONFind the mean and standard deviaton of the number of spots that appear when a die is tossed.

Outcomex

Probability

P(x)

1 1/6

2 1/6

3 1/6

4 1/6

5 1/6

6 1/6

µ = ∑ x P(x)= 21/6= 3.5

x P(x)

1/6

2/6

3/6

4/6

5/6

6/6

21/6

MEAN & STANDARD DEVIATIONFind the mean and standard deviaton of the number of spots that appear when a die is tossed.

Outcomex

Probability

P(x)

1 1/6

2 1/6

3 1/6

4 1/6

5 1/6

6 1/6

σ2 = ∑ x2 P(x) – µ2

= 91/6 – (21/6)2

= 105/36 σ = √ 105/36 = 1.708

x2 P(x)

1/6

4/6

9/6

16/6

25/6

36/6

91/6

The probability distribution shown represents the number of trips of five nights or more that American adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean.

MEAN & STANDARD DEVIATION

MEAN & STANDARD DEVIATION

# Tripsx

Probability

P(x)

0 0.06

1 0.70

2 0.20

3 0.03

4 0.01

x P(x)

0.00

0.70

0.40

0.09

0.04

1.23

x2 P(x)

0.00

0.70

0.80

0.27

0.16

1.93µ = ∑ x P(x) = 1.23σ2 = ∑ x2 P(x) – µ2 = 1.93 – (1.23)2 = 0.4171

σ = √0.4171 = 0.6458

EXPECTATION The expected valueexpected value, or expectation expectation, of

a discrete random variable of a probability distribution is the theoretical average of the variable.

The expected value is, by definition, the mean of the probability distribution.

E(x) = µ = ∑ x P(x)

WINNING TICKETSOne thousand tickets are sold at $1 each for four prizes of $100, $50, $25, and $10. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if you purchase a ticket?Winnings

xProbabilit

yP(x)

99 1/1000

49 1/1000

24 1/1000

9 1/1000

-1 996/1000

x P(x)

99/1000

49/1000

24/1000

9/1000

-996/1000

-815/1000

E(x) = ∑ x P(x)

= –0.815On average, you will lose 82 cents for every dollar you spend

Winnings = prize amount – $1 ticket price

WINNING TICKETS (ALTERNATE APPROACH)

One thousand tickets are sold at $1 each for four prizes of $100, $50, $25, and $10. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if you purchase a ticket?Winnings

xProbabilit

yP(x)

100 1/1000

50 1/1000

25 1/1000

10 1/1000

0 996/1000

x P(x)

100/1000

50/1000

25/1000

10/1000

0/1000

185/1000

E(x) = ∑ x P(x) – initial cost

= 0.185 – $1

= –0.815

THE BINOMIAL DISTRIBUTION Many types of probability problems have

only two possible outcomes or they can be reduced to two outcomes.

Examples include: when a coin is tossed it can land on heads or tails, when a baby is born it is either a boy or girl, etc.

THE BINOMIAL DISTRIBUTION

The binomial experimentbinomial experiment is a probability experiment that satisfies these requirements:

1. Each trial can have only two possible outcomes—success or failure.

2. There must be a fixed number of trials.

3. The outcomes of each trial must be independent of each other.

4. The probability of success must remain the same for each trial.

NOTATION FOR THE BINOMIAL DISTRIBUTION

The numerical probability of success

The numerical probability of failure

The number of trials

The number of successes

p

q

n

x

Note that x = 0, 1, 2, 3,...,n


!

- ! ! X n Xn

P X p qn X X

In a binomial experiment, the probability of exactly X successes in n trials is

number of possible probability of adesired outcomes desired outcome

or

X n Xn xP X C p q


The Binomial Probability Distributionthe probability of exactly x successes in n trials is

P(x) = nCx px qn-x

If 40% of the class is female, what is the probability that 1 of the next 2 students walking in will be female?


P(x) = nCx px qn-x

P(1) = 2C1 p1 q2-1

= (2) (.4)1 (.6)1

= 0.48n = 2 (# trials)

x = 1 (# successes)

p = .4 (probability of success)

q = .6 (probability of failure)

A survey found that one out of five Americans say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month.

3 7

10! 1 43

7!3! 5 5

P 0.201


P(x) = nCx px qn-x

P(3) = 10C3 p3 q10-3

n = 10x = 3p = 1/5q = 4/5

Individual baseball cards, chosen at random from a set of 20, are given away inside cereal boxes. Stan needs one more card to complete his set so he buys five boxes of cereal. What is the probability that he will complete his set?

The probability of Stan completing his set is 20%.


P(x) = nCx px qn-x

P(1) = 5C1 p1 q5-1

= (5) (.05)1 (.95)4

= 0.204

n = 5x = 1p = 1/20q = 19/20

A test consists of 10 multiple choice questions, each with four possible answers. To pass the test, one must answer at least nine questions correctly. Find the probability of passing, if one were to guess the answer for each question.

P(x successes) = nCx px qn - x

P(x ≥ 9 successes) = P(9 successes) + P(10 successes)

10C9

1

4

9

3

4

1

10C10 1

4

10

3

4

0

= 0.000028610 + 0.000000954= 0.000029564 The probability of

passing is 0.003%.


n = 10

x = 9, 10

p = 1/4

q = 3/4

P(x ≥ 9)

A survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part-time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part-time jobs.

3 25!3 0.30 0.70

2!3! P 0.132

4 15!4 0.30 0.70

1!4! P 0.028

5 05!5 0.30 0.70

0!5! P 0.002


n = 5p = 0.3q = 0.7x = 3, 4,

5P(x ≥ 3) = 0.132 + 0.028 + 0.002 =

0.162

A family has nine children. What is the probability that there is at least one girl?

This can be best solved using the compliment, that is, the probability of zero girls:

P(x successes) = nCx px qn - x

P(x ≥ 1) = 1 – P(0) = 1 – 0.001953

= 0.998

P (0) = 9C0 p0 q9 - 0

= (1) (.5)0 (.5)9

= 0.001953


n = 9p = 0.5q = 0.5x = 0

BINOMIAL DISTRIBUTIONCreate a probability distribution table for tossing a coin 3 times

# Headsx

0

1

2

3

P(x) = nCx px qn-x

P(0) = 3C0 p0 q3-0

= (1) (.5)0 (.5)3 = 0.125

P(1) = (3) (.5)1 (.5)2 = 0.375

P(2) = (3) (.5)2 (.5)1 = 0.375

P(3) = (1) (.5)3 (.5)0 = 0.125

P(x)

0.125

0.375

0.375

0.125

TOSSING COINSA coin is tossed 3 times. Find the probability of getting exactly two heads, using Table B.

123, 0.5, 2 n p X 2 0.375 P

(a) Construct a binomial probability histogram with n = 8 and p = 0.15.

(b) Construct a binomial probability histogram with n = 8 and p = 0. 5.

(c) Construct a binomial probability histogram with n = 8 and p = 0.85.

For each histogram, comment on the shape of the distribution.

BINOMIAL DISTRIBUTION


Mean: np2Variance: npq

The mean, variance, and standard deviation of a variable that has the binomial distribution can be found by using the following formulas.

Standard Deviation: npq

The Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins.

8000 0.02 160 np

2 8000 0.02 0.98 156.8 157 npq

8000 0.02 0.98 12.5 13 npq


According to the Experian Automotive, 35% of all car-owning households have three or more cars. Find the mean and standard deviation for a random sample of 400 homes.


µ = np

= (400)(0.35)

= 140

σ = √npq

= √(400)(0.35)(0.65)

= 9.54

P ROBABILITY & S TATISTICS FOR P-8 T EACHERS Chapter 5 Probability Distributions.

Documents

x p x slide

heads x probability

trips x probability

outcome x probability

x p x variance

probability distribution

probability chapter

p robabilities slide