Chapter 4 Discrete Random Variables

1

• Chapter 4 Discrete Random Variables
• 4.1 Discrete and Continuous
• 4.2 Probability Distribution
• 4.3 Expectation and Variance
• 4.4 Binominal
• 4.5 Poisson
• 4.6 Hypergeometric
• Homework 3,5,7,9,11,13,15,21,23,27,31,
• 43,52,53,61,63,68,69,77

2

• Last chapter we discussed several useful
  concepts of dealing with probability problems.
  However, it is very difficult to write down
  sample spaces of some random experiments. In
  these cases the concept random variable is very
  useful. A random variable is a variable that
  assumes values associated with the random
  outcomes of a random experiment, where one and
  only one numerical values is assigned to each
  sample point. It is random because we can not
  predict the outcome of a random experiment. It
  is a variable because there are more than one
  possible sample points in a random experiment.

3

• ltExample 4.1gt (Basic)
• One hundred fair coins are tossed and the up
  faces are observed.
• (a) Is it convenient to write down the sample
  space for this random experiment?
• (b) Do we need to write down the sample space if
  we are interested in counting the number of heads
  in this random experiment?
• ltSolutionsgt
• (a) No, it takes vary long time to write the
  sample space of this random experiment.
• (b) No, we can use the concept of random variable
  to solve our problem.

4

• Section 4.1 Discrete and Continuous Random
  Variable
• Some random variable can assume values on
  countable many numbers (such as integers) and
  some random variable can assume values on one or
  more intervals. For example, the distance
  between you home and UCF is between 0 and 100
  miles that is an interval, i.e. the distance
  between your home and UCF is a continuous random
  variable. But the number of head in coin tossing
  experiment is a countable number, i.e. the number
  of heads in a coin tossing experiment is a
  discrete random variable.

5

• ltExample 4.2gt (Basic)
• List five discrete random variables and five
  continuous random variables.

6

• Sec 4.2 Probability Distributions for Discrete
  Random Variables
• This chapter will focus on the discussion of
  discrete random variable. A complete description
  of a discrete random variable requires that we
  specify all the possible values the random
  variable can assume and the probability
  associated with each value. Usually, we can use
  the following four steps to complete a
  probability table.
• Step 1 Find out the variable of interest.
• Step 2 List all the sample points in the sample
  space.
• Step 3 List all the possible values of this
  random variable.
• Step 4 Assign the probabilities to all the
  possible values.

7

• ltExample 4.3gt (Basic)
• A company has five applicants for two
  positions three from UCF and two from UF.
  Suppose that the five applicants are equally
  qualified and no preference is given for choosing
  either school. Let x be the number of UCF
  graduates chosen to fill the two positions.
• (a) What is the random variable of interest?
• (b) Write down the sample space.
• (c) Write the probability table.

8

• The probability distribution of a discrete
  random variable is a graph, a table, or a formula
  that specifies the probability associated with
  each possible value the random variable can
  assume. The probability distribution should not
  include values that have zero probabilities. The
  rules to assign probability discussed in Section
  3.2 should be followed as well. Thus, the
  probability of any value of a random variable is
  between 0 and 1 and the sum of the probabilities
  of all possible values of a random variable is
  equal to one.

9

• ltExample 4.4gt (Basic)
• A random variable has the following probability
  table
• x 0 1 2 3 4
• p(x) 0.1 0.2 0.3 ? 0.15
• (a) Find P(x3).
• (b) Is x a continuous random variable?

10

• ltExample 4.5gt (Advance)
• A lady claims that she can taste the difference
  between PEPSI and COKE. Therefore, we conduct an
  experiment to confirm her claim. Four cups of
  cola that some are COKE colas and some are PEPSI
  colas are displayed in front of her. After
  tasting these colas, she needs to identify the
  contents in each cups. We are interested in the
  correct decisions made by her in this experiment.
  
• (a) What is the random variable of interest?
• (b) Write down the sample space.
• (c) Write the probability table.

11

• Sec 4.3 Expectation and Variance of a Discrete
  Random Variable
• We discussed how to obtain the sample mean, the
  sample variance, and the sample standard
  deviation in chapter 2. Now, we introduce the
  formulas of getting the population mean, the
  population variance, and the population standard
  deviation of a discrete random variable.
  Suppose X is a discrete random variable with
  probability distribution p(x). The expectation
  of X is the population mean of X. Let m, ,
  and s be the population mean, the population
  variance, and the population standard deviation
  of X, respectively. Then
• m E(x) S xp(x),
• s2 E(x-m)2 S (x-m)2
  p(x), and

12

• ltExample 4.6gt (Basic)
• Consider the probability table of random
  variable x below.
• x p(x)
• 1 0.1
• 3 0.2
• 5 0.3
• 6 0.3
• 10 0.1

13

• (a) Find the expectation of this random variable.
• ltSolutiongt
• x p(x) xp(x)
• 1 0.1 0.1
• 3 0.2 0.6
• 5 0.3 1.5
• 6 0.3 1.8
• 10 0.1 1.0
• mean m Sxp(x) 0.1 0.6 1.5 1.8 1.0
  5

14

• (b) Find the standard deviation of this random
  variable.
• ltSolutiongt
• x-m (x-m)2 (x-m)2p(x)
• -4 16 1.6
• -2 4 0.8
• 0 0 0
• 1 1 0.3
• 5 25 2.5
• the variance S p(x) 1.60.800.32.55.
  2 and the standard deviation s 2.28.

15

• (c) What is the probability that x falls within
  the interval (m-2s, m2s)?
• (d) Does the result satisfy the Chebyshevs Rule?
• (e) Does the result satisfy the Empirical Rule?
  Explain.
• ltSolutionsgt
• (c) m-2s 5 - 22.28 0.44
• m2s 5 22.28 9.56
• Thus, the probability that x falls within the
  interval (m-2s, m2s) is 0.9 (0.90.10.20.30.3)
  .
• (d) Yes.
• (e) No, because the random variable x does not
  have a mound-shape distribution.

16

• ltExample 4.7gt (Basic)
• You need to pay one dollar to buy an instant
  lottery ticket. In this instant lottery game,
  you have 10 chance to win a one dollar bill and
  5 chance to win a five dollar bill. You are
  interested in the money which you can win in a
  single play.

17

• (a) Write down the probability table.
• ltSolutiongt
• x p(x)
• 1 0.1
• 5 0.05
• 0 0.85
• Note The lottery official does (not?) want you
  to know that you have 85 chance to win
  nothing.

18

• (b) Find the mean.
• ltSolutiongt
• x p(x) xp(x)
• 1 0.1 0.1
• 5 0.05 0.25
• 0 0.85 0
• mean m Sxp(x) 0.10.250 0.35

19

• (c) Find the standard deviation of the game.
• ltSolutiongt
• x-m (x-m)2 (x-m)2p(x)
• 0.65 0.4225 0.04225
• 4.65 21.6225 1.081125
• -0.35 0.1225 0.104125
• the variance s2 S (x-m)2 p(x)
  0.042251.0811250.104125 1.2275 and the
  standard deviation s 1.10793.

20

• (d) Do you believe the lottery officials claim
  the more you play the more you win?
• ltSolutiongt
• Clearly, I dont believe it because lottery
  revenue is another form of taxes for the lottery
  players.

21

• ltExample 4.8gt (Basic)
• A study selected a sample of fifth grade pupils
  and recorded how many years of school they
  eventually completed. Let X be the highest year
  of school that a randomly selected fifth grader
  completes. (Students who go on to college are
  included in the outcome of x12.) The
  probability is as follows
• x 4 5 6 7 8 9 10
  11 12
• p(x) .01 .007 .007 ? .032 .068 .070 .041
  .752
• (a) Find P(X 7).
• ltSolutiongt P(X7)1-(0.010.0070.0130.032
• 0.0680.0700.0410.752) 0.007.

22

• (b) Find the mean and standard deviation.
• ltSolutionsgt
• x p(x) x p(x) (x-m)2 (x-m)2p(x)
• 4 0.01 0.04 52.577001 0.52577001
• 5 0.007 0.035 39.075001 0.273525007
• 6 0.007 0.042 27.573001 0.193011007
• 7 0.013 0.091 18.071001 0.234923013
• 8 0.032 0.256 10.569001 0.338208032
• 9 0.068 0.612 5.067001 0.344556068
• 10 0.070 0.700 1.565001 0.10955007
• 11 0.041 0.451 0.063001 0.002583041
• 12 0.752 9.024 0.561001 0.421872752
• 11.251 2.44400
• Thus, m11.251, s2 2.44400, and s 1.563.

23

• (c) Find P(x gt 9).
• (d) Can you apply the Empirical rule to find the
  probability of X falls into the interval (m-2s,
  m2s)?
• ltSolutionsgt
• (c) P(X 9) .068.070.041.762 0.931
• (d) m-2s 11.251 - 2 1.563 8.124
• m2s 11.251 2 1.563 14.378
• Thus, the probability that x falls within the
  interval (m-2s, m2s) is 0. 934. Although the
  probability is close to 0.95, we can only apply
  the chebyshevs Rule because the empirical
  distribution of this random variable is not
  mound-shape.

24

• Sec 4.4 The Binomial Random Variable
• The responses of many experiments have only two
  alternatives such as "Yes or No, "True or
  False", "Male or Female, and "Failure or
  Success". These types of experiments have some
  characteristic in common. First, they consist of
  n identical and independent trials. Second,
  there are only two possible outcomes, denoted by
  S and F on each trail. Third, the possibility of
  each outcome remains unchanged from trial to
  trial, that is, the probability of S is p and
  probability of F is q(1-p) in each trial.

25

• Fourth, we are interested in the random variable
  x represented the number of S happened in n
  trails (n is a fixed number). Therefore, it is
  worth to develop a special probability model to
  deal with this kind of random variables. Any
  random variable that has these four
  characteristics is called binomial random
  variable and can be dealt with by using this
  special probability model.

26

• ltExample 4.9gt (Basic) List several random
  variables that have only two possible outcomes.
• ltSolutionsgt
• Gender of a student in STA 3023
• Win or Loss in a football game
• Pass or Fail in an exam
• Hit or Miss in a state lottery drawing
• True or False to answer a question

27

• ltExample 4.10gt (Basic) For each of the following
  situations, indicate whether a binomial
  distribution is a reasonable probability model
  for the random variable X.
• (a) A couple decides to continue to have children
  until their first girl is born X is the total
  number of children the couple has.
• (b) Fifty students are taught about binomial
  probabilities by a television program. After
  completing their study, all students take the
  same examination X is the number of students who
  passed this exam.
• (c) A chemist repeats a solubility test 10 times
  on the same substance. Each test is conducted at
  a temperature 10 degrees higher than the previous
  test.

28

• Suppose that X is a binomial random variable.
  The probability of success on any single trial is
  p and there are n trials in this random
  experiment. The probability density function of
  X is
• where
• p the probability of success on any single
  trial
• n total number of trials
• q 1 - p
• x number of successes in n trials.

29

• Let m and s be the mean and standard deviation
  of the binomial random variable X. In stead of
  using the expectation summation rules to
  calculate m and s, we can find m and s easily
  using the formulas
• m np,
• s2 npq np(1-p), and

30

• ltExample 4.11gt (Basic) To test the side effect
  of a newly developed medicine, we conduct an
  animal experiment. Five dogs are given this drug
  and each dog has 20 chance to develop certain
  symptoms. We are interested in the number of
  dogs that develop this symptom.
• (a) Is this a binomial random variable?

31

• (b) Write down the probability table of this
  random variable.
• ltsolution to part (b)gt
• Probability Table X P(Xx)
• 0 0.32768
• 1 0.4096
• 2 0.2048
• 3 0.0512
• 4 0.0064
• 5 0.00032

32

• (c) Find the mean and standard deviation of this
  random variable.
• ltsolution to part (c)gt
• m np 5 0.2 1
• s2 npq 5 0.2 0.8 0.8
• s 0.894.

33

• ltExample 4.12gt (Basic)
• A firm receives a shipment of 500 hi-fi
  speakers. For any randomly selected sample of 9
  speakers, if 2 or more of the speakers are
  defective then rejects this shipment. What is
  the probability that this firm will accept the
  shipment if the proportion of defective is
• (a) 0.20.
• (b) 0.10.
• (c) 0.05.

34

• ltExample 4.13gt (Basic) An oil exploration firm
  plans to drill six holes. Due to experience, the
  probability of each hole yielding oil is 0.12.
  Since the holes are in quite different locations,
  the outcome of drilling one hole is statistically
  independent of drilling of any other holes.
• (a) Give the expectation and standard deviation
  of the number of holes that results in oil.
• ltSolution to part (a)gt
• (a) n 6 and p 0.12
• m np 6 0.12 0.72
• s
  0.796

35

• (b) If the firm will be able to stay in business
  only if two or more holes produce oil, what is
  the probability that it can survive.
• ltSolution to part (b)gt
• P(X ³ 2) P(X2) (X3)P(X4)P(X5)
• P(X6) 0.129534197 0.023551672
• 0.002408693760.000131383296
• 0.000002985984 _at_ 0.156

36

• Note
• (1) We can not use the Binomial probabilities
  Table to obtain this probability because p 0.12
  is not in the Table.
• (2) We can obtain this probability much easier
  with the concept of complement event
• P(X ? 2) 1 - P(X?1)
• 1 - P(X0) - P(X1)
• 1 - 0.4644044086 - 0.37996698
• _at_ 0.156

37

• Section 4.5 The Poisson Random Variable
• The random variables produced by many random
  experiments can be well described by using
  Poisson probability model.
• Typical examples are as follows
• (1) the number of customers served per hour in a
  given restaurant,
• (2) the number of alcohol related traffic
  accidents per month at a busy intersection,
• (3) the number of diseased trees per acre in a
  certain national park,
• (4) the number of telephone calls received per
  minute during your lunch hour.

38

• Poisson random variable has the following common
  characteristics.
• (1). The experiment consists of counting the
  number of times that a certain event occurs
  during a given unit of time or in a given area or
  volume.
• (2). The probability of an event occurs in a
  given unit of time is same for all time units.
• (3). The number of events that occur in one unit
  of time, area, or volume is independent of the
  number that occur in other units.

39

• The probability density function of a Poisson
  random variable is
• Both the mean and the variance of a Poisson
  random variable equals to l, i.e. m l and s2
  l.

40

• ltExample 4.14gt (Basic)
• Suppose x is a Poisson random variable, use Table
  III on page 804 to find the following
  probabilities.
• (a) P(x 2) when l 1.
• ltSolution to part (a)gt
• part of Table III
• x 0 1 2 3 4 5 6
• l 1 .368 .736 .920 .981 .996 .999
  1.000
• Thus, P(X2) 0.920.

41

• (b) P(x ³ 2) when m 2.
• ltSolution to part (b)gt
• part of Table III
• x 0 1 2 3 4 5
  6 7 8
• l2 .135 .406 .677 .857 .956 .987 0.997
  0.999 1.000
• Since l m 2, P(x ³ 2) 1 - P(x 1) 1 -
  0.406 0.594.

42

• (c) P(x gt3) where s2 3.
• ltSolution to part (c)gt
• part of Table III
• x0 1 2 3 4 5
  6 7 8
• l3 0.050 0.199 0.423 0.647 0.815 0.916
  0.996 0.988 0.996
• Since s2 m 3, P(x gt 3) 1 - P(x 2) 1 -
  0.423 0.577.

43

• Note
• For Poisson random variable, we can find P(x
  a) from the table directly if a is an integer.
  We need to apply the concept of complement event
  to find the probability of P(x gt a) or P(x ³ a).
  We need to know that P(x gt a) 1 - P(x a) and
  P(x ³ a ) 1 - P(x a-1).

44

• ltExample 4.15gt (basic) We know that the mean of
  a Poisson random variable is equal to 2. Find
  the probabilities of x equal to 1, 2, and 3.
• ltSolutionsgt
• We can use the probability function to compute
  the probability of a Poisson random variable as
  well.

45

• ltExample 4.16gt (Intermediate)
• According to the records of an airline, the
  number of people who buy tickets but fail to show
  up for the early morning flight between Orlando
  and Washington D.C. is a Poisson random variable.
  We know that the standard deviation of this
  random variable is 2. Determine the
  probabilities that the number of no shows in an
  early flight
• (a) is equal to 5,
• (b) is less than or equal to 3,
• (c) is greater than or equal to 6.

46

• ltSolutions to EX 4.16gt
• l s2 2 2 4
• (a)
• (b) p(x 3) 0.433. (From Table III)
• (c ) p(x³6) 1 - p(x5) 1 - 0.785 0.215.

47

• Section 4.6 The Hypergeometric Random Variable
• Hypergeometric random variable is another
  popular discrete random variable. Suppose there
  are a total of N balls r red balls and (N-r)
  white balls, in a bag. And n balls are randomly
  selected from this bag without replacement. Let X
  denote the number of red balls in the n balls
  selected. Then the distribution of X is called a
  
• Hypergeometric distribution,
• with parameters N, r and n.

48
The probability density function of this
hypogeometric distribution is given by
49

• ltExample 4.17gt (Basic) Given that x is a
  hypergeometric random variable, compute p(x), m,
  and s2 for each of the following cases
• (a) N5, n3, r3, x1.
• ltSolutionsgt

50

• (b) N9, n5, r3, x3.
• ltSolution to part (b)gt

51

• Collection of Definitions
• Random Variable
• A random variable is a rule that assigns one and
  only one numerical value to each sample point in
  a random experiment.
• Discrete Random Variable
• A discrete random variable is a random variable
  that can assume only countable number of values.
• Continuous Random Variable
• A continuous random variable is a random
  variable that can assume values in one or more
  intervals.

52

• Probability Distribution
• The probability distribution of a discrete
  random variable is a way, such as a graph, a
  table, or a formula, that specifies the
  probability associated with each possible value
  the random can assume.
• Expectation of a Discrete Random Variable
• The expectation of a discrete random variable is
  the population mean of this random variable. We
  can use the following formula to compute the
  expectation of a discrete random variable
• m E(x) S xp(x).

53

• Variance of Discrete Random Variable
• The variance of a discrete random variable is
  the population variance of the random variable,
  given by formula
• s2 E(x-m)2 S (x-m)2 p(x).
• Standard Deviation of Discrete Random Variable
• The standard deviation of a discrete random
  variable is equal to the square root of the
  variance of this random variable, i.e.

54

• Binomial Distribution
• The probability density function of a binomial
  random variable is
• Here
• p the probability of success on any single
  trial
• n total number of trials
• x number of successes in n trials
• q 1 - p
• The mean of a binomial random variable is
• np, i.e. m np
• The variance of a binomial random variable is
• npq, i.e. s2 npq np(1-p).

55

• Poisson Random Variable
• The probability density function of a Poisson
  random variable is
• Both the mean and the variance of a Poisson
  random variable equals to l, i.e.
• m l and s2 l.

56

• Hypergeometric Random Variable
• The probability density function of a
  Hypergeometric random variable is
• where
• N total number of balls in the bag
• r the number of red balls in the bag
• n the number of balls drawn without
  replacement
• x the number of red balls in the n balls
  selected.

57

• The mean of a Hypergeometric random variable is
• and the variance of a Hypergeometric random
• variable is