5 some important discrete probability distributions

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Section 5.1 The Probability Distribution for a Discrete Random Variable January 9, 2014 1 5 Some Important Discrete Probability Distributions 5.1 The Probability Distribution for a Discrete Random Variable What are some continuous random variables? Discrete distributions Example: (Discrete case) Roll a (not necessarily fair) six-sided die once. The possible outcomes are the faces (i.e., dots) on the die. A different die might have six colors for the six sides (like a Rubik’s cube). Example: (Discrete case) Toss a (not necessarily fair) coin once.

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Page 1: 5 Some Important Discrete Probability Distributions

Section 5.1 The Probability Distribution for a Discrete Random Variable January 9, 2014 1

5 Some Important Discrete

Probability Distributions

5.1 The Probability Distribution for a

Discrete Random Variable

What are some continuous random variables?

Discrete distributions

Example: (Discrete case) Roll a (not necessarily fair) six-sided die once. The

possible outcomes are the faces (i.e., dots) on the die.

⑦ ⑦

⑦ ⑦

⑦ ⑦

⑦ ⑦

⑦ ⑦

⑦ ⑦ ⑦

⑦ ⑦ ⑦

A different die might have six colors for the six sides (like a Rubik’s cube).

Example: (Discrete case) Toss a (not necessarily fair) coin once.

Page 2: 5 Some Important Discrete Probability Distributions

Section 5.1 The Probability Distribution for a Discrete Random Variable January 9, 2014 2

Definition: The probability distribution of a discrete random variable

X consists of the possible values of X along with their associated probabilities.

We sometimes use the terminology population distribution.

Example: Fair (six-sided) die. Let X be the numerical outcome. Determine the

probability distribution of X, graph this probability distribution, and compute the

mean of the probability distribution.

Example: Toss a fair coin 3 times. Let X = number of heads.

Let Y = number of matches among the three tosses.

Page 3: 5 Some Important Discrete Probability Distributions

Section 5.1 The Probability Distribution for a Discrete Random Variable January 9, 2014 3

Terminology: The expected value of X is the mean of a random variable X.

Example: (hypothetical) Suppose an airline often overbooks flights, because past

experience shows that some passengers fail to show.

Let the random variable X be the number of passengers who cannot be boarded

because there are more passengers than seats.

x P (x)

0 0.6

1 0.2

2 0.1

3 0.1

sum 1

(a) Compute the expected value of X.

(b) Suppose each unboarded passenger costs the airline $100 in a ticket voucher.

Compute the average cost to the airline per flight in ticket vouchers.

Brief review of means and standard deviations

The mean, µ, of a random variable is the average of all outcomes in the population,

and is the limiting value of X̄ as n gets large.

The standard deviation, σ, of a random variable measures the spread of all

outcomes in the population, and is the limiting value of s as n gets large.

Page 4: 5 Some Important Discrete Probability Distributions

Section 5.3 Binomial Distribution January 9, 2014 4

The variance, σ2, of a random variable also measures the spread in the population,

and is the limiting value of s2 as n gets large.

σ is more intuitive than σ2, partly because σ has the same units as the original data.

5.3 Binomial Distribution

Example: Toss an unfair coin 5 times, where π = P (heads) = 0.4.

Let X be the number of heads.

Suppose we want to determine P (X = 2).

Factorials: m! = m(m− 1)(m− 2) · · · 1 for positive integers m.

Example: Compute 4!, 5!, 1!, and 0!.

A Bernoulli trial can have two possible outcomes, success or failure.

Definition of a binomial random variable X.

1. Let n, the number of Bernoulli trials, be fixed in advance.

2. The Bernoulli trials are independent.

3. The probability of success of a Bernoulli trial is π, which is the same for all

observations.

Page 5: 5 Some Important Discrete Probability Distributions

Section 5.3 Binomial Distribution January 9, 2014 5

Let X be the number of successes. Then, X is a binomial(n, π) random variable.

P (X = x) =n!

x! (n− x)!πx (1− π)n−x, for x = 0, 1, 2, . . . , n

Example: n = 5 tosses of an unfair coin.

Assume π = P (heads) = 0.4 (OR 40% Democrats from a huge population).

Let X be the number of heads.

Determine the probability distribution of X and construct the line graph.

x P (x)

0 0.0776

1 0.2592

2 0.3456

3 0.2304

4 0.0768

5 0.01024

0 1 2 3 4 5

00.0

50.1

0.150

.20.2

50.3

0.35

X

PROB

ABILIT

Y

Determine the probability of obtaining at least one heads among the five coin tosses.

In 100 tosses of this coin, on average, how many heads do you expect?

Mean and standard deviation of a binomial random

Page 6: 5 Some Important Discrete Probability Distributions

Section 5.3 Binomial Distribution January 9, 2014 6

variable

µ = nπ, σ2 = nπ(1− π), σ =√

nπ(1− π)

Example: Revisit. Let X ∼ Binomial(n = 5, π = 0.4). Compute the mean and

standard deviation of X.

Example: Consider a huge population where 30% of the people are Democrats.

Let X be the number of Democrats in a sample of size 1000. Compute the mean

and standard deviation of X.

For large sample sizes (i.e., nπ ≥ 5 and n(1− π) ≥ 5), a binomial random variable and

a sample proportion are approximately normally distributed by the Central

Limit Theorem.

Example: Viewing the Central Limit Theorem.

(a) Consider the graphs below for binomial random variables, using π = 0.3

and n = 1, 2, 3, 4, 5, 10, 15, 20, and 30.

Page 7: 5 Some Important Discrete Probability Distributions

Section

5.3Binom

ialDistribu

tionJanuary

9,2014

7

01

0.0 0.2 0.4 0.6

X

PROBABILITY

01

2

0.0 0.1 0.2 0.3 0.4 0.5

X

PROBABILITY

01

23

0.0 0.1 0.2 0.3 0.4

X

PROBABILITY

01

23

4

0.0 0.1 0.2 0.3 0.4

X

PROBABILITY

01

23

45

0.0 0.1 0.2 0.3X

PROBABILITY0

24

68

10

0 0.05 0.1 0.15 0.2 0.25

X

PROBABILITY

05

10

15

0 0.05 0.1 0.15 0.2

X

PROBABILITY

05

10

15

20

0 0.05 0.1 0.15

X

PROBABILITY

05

10

15

20

25

30

0.00 0.05 0.10 0.15

X

PROBABILITY

(b)Con

sider

thegrap

hsbelow

forsample

pro

portio

ns,p,

usin

gπ=

0.3

andn=

1,2,

3,4,

5,10,

15,20,

and30.

Page 8: 5 Some Important Discrete Probability Distributions

Section 5.3 Binomial Distribution January 9, 2014 8

0 1

0.00.2

0.40.6

n = 1, and Pi = 0.3

p

PROB

ABILI

TY

0 0.5 1

0.00.1

0.20.3

0.40.5

n = 2, and Pi = 0.3

p

PROB

ABILI

TY

0 0.2 0.4 0.6 0.8 1

0.00.1

0.20.3

0.4

n = 3, and Pi = 0.3

p

PROB

ABILI

TY

0 0.2 0.4 0.6 0.8 1

0.00.1

0.20.3

0.4

n = 4, and Pi = 0.3

p

PROB

ABILI

TY

0 0.2 0.4 0.6 0.8 1

0.00.1

0.20.3

n = 5, and Pi = 0.3

p

PROB

ABILI

TY0 0.2 0.4 0.6 0.8 1

00.0

50.1

0.15

0.20.2

5

n = 10, and Pi = 0.3

p

PROB

ABILI

TY

0 0.2 0.4 0.6 0.8 1

00.0

50.1

0.15

0.2

n = 15, and Pi = 0.3

p

PROB

ABILI

TY

0 0.2 0.4 0.6 0.8 1

00.0

50.1

0.15

n = 20, and Pi = 0.3

p

PROB

ABILI

TY

0 0.2 0.4 0.6 0.8 1

00.0

50.1

0.15

n = 30, and Pi = 0.3

p

PROB

ABILI

TY

Example: The Democrats.

(a) Use the 95% part of the empirical rule on the binomial random variable.

00.0

050.0

10.0

150.0

20.0

25

Probability is 0.95

X

probab

ility de

nsity

functio

n

271 300 329

(b) Use the 95% part of the empirical rule on the sample proportion.

Page 9: 5 Some Important Discrete Probability Distributions

Section 5.4 Poisson Distribution January 9, 2014 9

05

1015

2025

Probability is 0.95

p

probab

ility de

nsity

functio

n0.271 0.3 0.329

Read p. 194, Microsoft Excel.

5.4 Poisson Distribution

Consider the Binomial(n, π) distribution, such that n is huge, π is small, but nπ is

moderate (i.e., neither huge nor small).

Example: Radioactive decay. Consider a radioactive substance containing

3,000,000 atoms, such that decaying atoms are independent of each other, and

π = P (A particular atom decays in the next day) = 1/1, 000, 000.

Compute the mean number of atomic decays in the next day.

Consider letting n → ∞ and π → 0 such that nπ → λ, a positive constant, where

X ∼ Binomial(n, π).

In this limit, X ∼ Poisson(λ).

If X ∼ Poisson(λ) for λ > 0, then

P (X = x) =1

x!λx e−λ, for x = 0, 1, 2, . . .

Page 10: 5 Some Important Discrete Probability Distributions

Section 5.4 Poisson Distribution January 9, 2014 10

Example: Revisit radioactive decay. Consider a radioactive substance containing

3,000,000 atoms, such decaying atoms are independent of each other, and

π = P (A particular atom decays in the next day) = 1/1, 000, 000.

(a) Let X1 be the number of decays in one day. Determine the probability that

at least one atom decays in the next day.

(b) Let X2 be the number of decays in two days. Determine the probability that

at least one atom decays in the next two days.

0 2 4 6 8

00.0

50.1

0.15

0.2

Poisson( λ1 = 3 )

X1

prob

abilit

y mas

s fun

ction

0 5 10 15

00.0

50.1

0.15

Poisson( λ2 = 6 )

X 2

prob

abilit

y mas

s fun

ction

Remark: Typically, when a Binomial(n, π) distribution is reasonably

approximated by a Poisson(λ) distribution, n and π are difficult to determine (or

estimate), but λ can be estimated from the data. How?

The Poisson Process is explained by the following:

(a) The probability of a success (such as a radioactive decay) in the next day is

independent of its past.

(b) The mean of a Poisson process based on two days is twice as large as the

mean of the same Poisson process based on one day.

Page 11: 5 Some Important Discrete Probability Distributions

Section 5.4 Poisson Distribution January 9, 2014 11

Example: Consider the number of recombinations (breaks) in DNA (chromosome

pairs) when DNA strands are passed to offspring.

Read p. 200, Microsoft Excel.

Read pp. 212–214, Appendix E5: Using Microsoft Excel for

Discrete Probability Distributions.