Probability Distributions

1
Chapter 6
6-1

• Probability Distributions

2
Outline
6-2

• 6-1 Introduction
• 6-2 Probability Distributions
• 6-3 Mean, Variance, and Expectation
• 6-4 The Binomial Distribution

3
Outline
6-3

• 6-5 Other Types of Distributions
• 4-6 Summary

4
Objectives
6-4

• Construct a probability distribution for a random
  variable.
• Find the mean, variance, and expected value for a
  discrete random variable.
• Find the exact probability for X successes in n
  trials of a binomial experiment.

5
Objectives
6-5

• Find the mean, variance, and standard deviation
  for the variable of a binomial distribution.
• Find the probabilities for outcomes of variables
  using the Poisson, hypergeometric, and
  multinomial distributions.

6
6-2 Probability Distributions
6-6

• A variable is defined as a characteristic or
  attribute that can assume different values.
• A variable whose values are determined by chance
  is called a random variable.

7
6-2 Probability Distributions
6-7

• If a variable can assume only a specific number
  of values, such as the outcomes for the roll of a
  die or the outcomes for the toss of a coin, the
  the variable is called a discrete variable.
• Discrete variables have values that can be
  counted.

8
6-2 Probability Distributions
6-8

• If a variable can assume all values in the
  interval between two given values then the
  variable is called a continuous variable.
  Example - temperature between 680 to 780.
• Continuous random variables are obtained from
  data that can be measured rather than counted.

9
6-2 Probability Distributions - Tossing Two
Coins
6-9

First Toss
T
10
6-2 Probability Distributions - Tossing Two
Coins
6-10

• From the tree diagram, the sample space will be
  represented by HH, HT, TH, TT.
• If X is the random variable for the number of
  heads, then X assumes the value 0, 1, or 2.

11
6-2 Probability Distributions - Tossing Two
Coins
6-11
Sample Space
Number of Heads
0 1 2
TT TH HT HH
12
6-2 Probability Distributions - Tossing Two
Coins
6-12
13
6-2 Probability Distributions
6-13

• A probability distribution consists of the values
  a random variable can assume and the
  corresponding probabilities of the values. The
  probabilities are determined theoretically or by
  observation.

14
6-2 Probability Distributions --
Graphical Representation
6-14
15
6-3 Mean, Variance, and Expectation for Discrete
Variable
6-15
16
6-3 Mean for Discrete Variable - Example
6-16

• Find the mean of the number of spots that appear
  when a die is tossed. The probability
  distribution is given below.

17
6-3 Mean for Discrete Variable - Example
6-17
That is, when a die is tossed many times, the
theoretical mean will be 3.5.
18
6-3 Mean for Discrete Variable - Example
6-18

• In a family with two children, find the mean
  number of children who will be girls. The
  probability distribution is given below.

19
6-3 Mean for Discrete Variable - Example
6-19
That is, the average number of girls in a
two-child family is 1.
20
6-3 Formula for the Variance of a
Probability Distribution
6-20

• The variance of a probability distribution is
  found by multiplying the square of each outcome
  by its corresponding probability, summing these
  products, and subtracting the square of the mean.

21
6-3 Formula for the Variance of a
Probability Distribution
6-21
22
6-3 Variance of a Probability
Distribution - Example
6-22

• The probability that 0, 1, 2, 3, or 4 people will
  be placed on hold when they call a radio talk
  show with four phone lines is shown in the
  distribution below. Find the variance and
  standard deviation for the data.

23
6-3 Variance of a Probability
Distribution - Example
6-23
24
6-3 Variance of a Probability
Distribution - Example
6-24

• Now, m (0)(0.18) (1)(0.34)
  (2)(0.23) (3)(0.21) (4)(0.04)
  1.59.
• s2 (02)(0.18) (12)(0.34)
  (22)(0.23) (32)(0..21) (42)(0.04) -
  1.592 1.26 (one decimal place).
• s Ö(1.26) 1.12.

25
6-3 Variance of a Probability
Distribution - Example
6-25
s2 3.79 - 1.592 1.26
26
6-3 Expectation
6-26
27
6-3 Expectation - Example
6-27

• A ski resort loses 70,000 per season when it
  does not snow very much and makes 250,000 when
  it snows a lot. The probability of it snowing at
  least 75 inches (i.e., a good season) is 40.
  Find the expected profit.

28
6-3 Expectation - Example
6-28

• The expected profit (250,000)(0.40)
  (-70,000)(0.60) 58,000.

29
6-4 The Binomial Distribution
6-29

• A binomial experiment is a probability experiment
  that satisfies the following four requirements
• Each trial can have only two outcomes or outcomes
  that can be reduced to two outcomes. These
  outcomes can be considered as either success or
  failure.

30
6-4 The Binomial Distribution
6-30

• There must be a fixed number of trials.
• The outcomes of each trial must be independent of
  each other.
• The probability of success must remain the same
  for each trial.

31
6-4 The Binomial Distribution
6-31

• The outcomes of a binomial experiment and the
  corresponding probabilities of these outcomes
  are called a binomial distribution.

32
6-4 The Binomial Distribution
6-32

• Notation for the Binomial Distribution
• P(S) p, probability of success
• P(F) 1 - p q, probability of failure
• n number of trials
• X number of successes.

33
6-4 Binomial Probability Formula
6-33
34
6-4 Binomial Probability - Example
6-34

• If a student randomly guesses at five
  multiple-choice questions, find the probability
  that the student gets exactly three correct.
  Each question has five possible choices.
• Solution n 5, X 3, and p 1/5. Then, P(3)
  5!/(5 - 3)!3!(1/5)3(4/5)2 0.05.

35
6-4 Binomial Probability - Example
6-35

• A survey from Teenage Research Unlimited
  (Northbrook, Ill.) found that 30 of teenage
  consumers received their spending money from
  part-time jobs. If five teenagers are selected
  at random, find the probability that at least
  three of them will have part-time jobs.

36
6-4 Binomial Probability - Example
6-36

• Solution n 5, X 3, 4, and 5, and p 0.3.
  Then, P(X ³ 3) P(3) P(4) P(5) 0.1323
  0.0284 0.0024 0.1631.
• NOTE You can use Table B to find the Binomial
  probabilities as well.

37
6-4 Binomial Probability - Example
6-37

• A report from the Secretary of Health and Human
  Services stated that 70 of single-vehicle
  traffic fatalities that occur at night on
  weekends involve an intoxicated driver. In a
  sample of 15 single-vehicle traffic fatalities
  that occur at night on a weekend is selected,
  find the probability that exactly 12 involve a
  driver who is intoxicated.

38
6-4 Binomial Probability - Example
6-38

• Solution n 15, X 12, and p 0.7. From
  Table B, P(X 12) 0.170

39
6-4 Mean, Variance, Standard Deviation for the
Binomial Distribution - Example
6-39

• A coin is tossed four times. Find the mean,
  variance, and standard deviation of the number of
  heads that will be obtained.
• Solution n 4, p 1/2, and q 1/2.
• m np (4)(1/2) 2.
• s2 npq (4)(1/2)(1/2) 1.
• s Ö1 1.

40
6-5 Multinomial Distribution
6-40
41
6-5 Multinomial Distribution - Example
6-41

• In a large city, 50 of the people choose a
  movie, 30 choose dinner, and 20 choose shopping
  as a leisure activity. If a sample of five
  people is randomly selected, find the probability
  that three are planning to go to a movie, one to
  dinner, and one to a shopping mall.

42
6-5 Multinomial Distribution - Example
6-42

• Solution n 5, X1 3, X2 1, X3 1,
  p1 0.5, p2 0.3, and p3 0.2.
• Thus P(X) 5!/3!1!1!(0.5)3(0.3)1(0.2)1
  0.15.

43
6-5 Formula for the Poisson Distribution
6-43

• The probability of X occurrences in an interval
  of time, area, volume, etc., for a variable where
  l (lambda) is the mean number of occurrences per
  unit (time, area, volume etc.) is

44
6-5 Formula for the Poisson Distribution
6-44
(e 2.7183)
45
6-5 Poisson Distribution - Example
6-45

• If there are 200 typographical errors randomly
  distributed in a 500-page manuscript, find the
  probability that a given page contains exactly
  three errors.
• Solution l (200/500) 0.4 error per page and
  X 3. P(3) e-l lx/X! (2.7183)-0.4(0.4)3/3!
  0.0072.

46
6-5 Poisson Distribution - Example
6-46

• A sales firm receives, on the average, three
  calls per hour on its toll-free number. For any
  given hour, find the probability that it will
  receive the following. (Use Table C).
• At most three calls.

47
6-5 The Hypergeometric Distribution
6-47

• When sampling is done without replacement, the
  binomial distribution does not give exact
  probabilities, since the trials are not
  independent. The smaller the size of the
  population, the less accurate the binomial
  probabilities will be.

48
6-5 The Hypergeometric Distribution
6-48

• The hypergeometric distribution is a distribution
  of a variable that has two outcomes when sampling
  is done without replacement.
• The probabilities for the hypergeometric
  distribution can be calculated by using the
  formula on the next slide.

49
6-5 Formula for the Hypergeometric
Distribution
6-49

• Given a population with only two types of objects
  (females and males, defective and nondefective
  etc.), such that there are a items of one kind
  and b items of another kind and a b equals the
  total population, the probability P(X) of
  selecting without replacement a sample of size n
  with X items of type a and (n - X) items of type
  b is

50
6-5 Formula for the Hypergeometric
Distribution
6-50
51
6-5 Hypergeometric Distribution -
Example
6-51

• Ten people apply for a job as assistant manager
  of a restaurant. Five have completed college and
  5 have not. If the manager selects three
  applicants at random, find the probability that
  all three are college graduates.

52
6-5 Hypergeometric Distribution -
Example
6-52

• Solution a 5 college graduates b 5
  nongraduates n 3 X 3 and n - X 0.
• Substituting in the formula gives P(X) 5C3
  5C0/10C3 10/120 1/12.

53
6-5 Hypergeometric Distribution -
Example
6-53

• A recent study found that four out of nine houses
  were underinsured. If five houses are selected
  from the nine houses, find the probability that
  two are underinsured.

54
6-5 Hypergeometric Distribution -
Example
6-54

• Solution a 4 , b 5, n 5, X 2, and n - X
  3.
• Substituting in the formula gives P(X) 4C2
  5C3/9C5 60/126 10/21.