Chapter 6: Random Variables and the Normal Distribution. 6.1 Discrete Random Variables. 6.2 Binomial Probability Distribution

Transcription

1 Chapter 6: Random Variables and the Normal Distribution 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution

2 6.1 Discrete Random Variables Objectives: By the end of this section, I will be able to 1) Identify random variables. 2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs. 3) Calculate the mean, variance, and standard deviation of a discrete random variable.

3 Random Variables A variable whose values are determined by chance Chance in the definition of a random variable is crucial

4 Example Notation for random variables Suppose our experiment is to toss a single fair die, and we are interested in the number rolled. We define our random variable X to be the outcome of a single die roll. a. Why is the variable X a random variable? b. What are the possible values that the random variable X can take? c. What is the notation used for rolling a 5? d. Use random variable notation to express the probability of rolling a 5.

5 Example 6.2 continued Solution a) We don t know the value of X before we toss the die, which introduces an element of chance into the experiment b) Possible values for X: 1, 2, 3, 4, 5, and 6. c) When a 5 is rolled, then X equals the outcome 5, or X = 5. d) Probability of rolling a 5 for a fair die is 1/6, thus P(X = 5) = 1/6.

6 Types of Random Variables Discrete random variable - either a finite number of values or countable number of values, where countable refers to the fact that there might be infinitely many values, but they result from a counting process Continuous random variable infinitely many values, and those values can be associated with measurements on a continuous scale without gaps or interruptions

7 Example Identify each as a discrete or continuous random variable. (a) Total amount in ounces of soft drinks you consumed in the past year. (b) The number of cans of soft drinks that you consumed in the past year.

8 ANSWER: (a) continuous (b) discrete Example

9 Example Identify each as a discrete or continuous random variable. (a) The number of movies currently playing in U.S. theaters. (b) The running time of a randomly selected movie (c) The cost of making a randomly selected movie.

10 ANSWER (a) discrete (b) continuous (c) continuous Example

11 Discrete Probability Distributions Provides all the possible values that the random variable can assume Together with the probability associated with each value Can take the form of a table, graph, or formula Describe populations, not samples

12 Example Table 6.2 in your textbook The probability distribution table of the number of heads observed when tossing a fair coin twice

13 Probability Distribution of a Discrete Random Variable The sum of the probabilities of all the possible values of a discrete random variable must equal 1. That is, ΣP(X) = 1. The probability of each value of X must be between 0 and 1, inclusive. That is, 0 P(X ) 1.

14 Example Let the random variable x represent the number of girls in a family of four children. Construct a table describing the probability distribution.

15 Example Determine the outcomes with a tree diagram:

16 Example Total number of outcomes is 16 Total number of ways to have 0 girls is 1 P(0 girls) 1/ Total number of ways to have 1 girl is 4 P(1girl) 4/ Total number of ways to have 2 girls is 6 P(2 girls) 6/

17 Example Total number of ways to have 3 girls is 4 P(3 girls) 4/ Total number of ways to have 4 girls is 1 P(4 girls) 1/

18 Example Distribution is: x P(x) NOTE: P(x) 1

19 Mean of a Discrete Random Variable The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times Also called the expected value or expectation of the random variable X. Denoted as E(X ) Holds for discrete and continuous random variables

20 Finding the Mean of a Discrete Random Variable Multiply each possible value of X by its probability. Add the resulting products. X P X

21 Variability of a Discrete Random Variable Formulas for the Variance and Standard Deviation of a Discrete Random Variable Definition Formulas 2 X 2 P X X 2 P X Computational Formulas X P X X P X 2 2

22 Example x P(x) x P(x) 2 2 x x P( x) xp(x) 2.0

23 Example x P(x) x P(x) 2 2 x x P( x) x 2 P( x)

24 Discrete Probability Distribution as a Graph Graphs show all the information contained in probability distribution tables Identify patterns more quickly FIGURE 6.1 Graph of probability distribution for Kristin s financial gain.

25 Page 270 Example

26 Example Probability distribution (table) x P(x) Omit graph

27 Page 270 Example

28 ANSWER Example X number of goals scored (a) Probability X is fewer than 3 P( X 0 X 1 X 2) P( X 0) P( X 1) P( X 2)

29 Example ANSWER (b) The most likely number of goals is the expected value (or mean) of X x P(x) x P(x) xp(x) She will most likely score one goal

30 Example ANSWER (c) Probability X is at least one P( X 1 X 2 X 3) P( X 1) P( X 2) P( X 3)

31 Summary Section 6.1 introduces the idea of random variables, a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text. Random variables are variables whose value is determined at least partly by chance. Discrete random variables take values that are either finite or countable and may be put in a list. Continuous random variables take an infinite number of possible values, represented by an interval on the number line.

32 Summary Discrete random variables can be described using a probability distribution, which specifies the probability of observing each value of the random variable. Such a distribution can take the form of a table, graph or formula. Probability distributions describe populations, not samples. We can find the mean μ, standard deviation σ, and variance σ 2 of a discrete random variable using formulas.

33 6.2 Binomial Probability Distribution

34 6.2 Binomial Probability Distribution Objectives: By the end of this section, I will be able to 1) Explain what constitutes a binomial experiment. 2) Compute probabilities using the binomial probability formula. 3) Find probabilities using the binomial tables. 4) Calculate and interpret the mean, variance, and standard deviation of the binomial random variable.

35 Factorial symbol For any integer n 0, the factorial symbol n! is defined as follows: 0! = 1 1! = 1 n! = n(n - 1)(n - 2) 3 2 1

36 Example Find each of the following 1. 4! 2. 7!

37 Example ANSWER 1. 4! !

38 Factorial on Calculator Calculator 7 MATH PRB 4:! which is 7! Enter gives the result 5040

39 Combinations An arrangement of items in which r items are chosen from n distinct items. repetition of items is not allowed (each item is distinct). the order of the items is not important.

40 Example of a Combination The number of different possible 5 card poker hands. Verify this is a combination by checking each of the three properties. Identify r and n.

41 Example Five cards will be drawn at random from a deck of cards is depicted below

42 Example An arrangement of items in which 5 cards are chosen from 52 distinct items. repetition of cards is not allowed (each card is distinct). the order of the cards is not important.

43 Combination Formula The number of combinations of r items chosen from n different items is denoted as n C r and given by the formula: n C r n! r! n r!

44 Example Find the value of 7C 4

45 Example ANSWER: 7C 4 7! 4!(7 4)! 7! 4!3! 7 6 ( ) 3 ( )

46 Combinations on Calculator Calculator 7 MATH PRB 3:nCr 4 To get: 7C 4 Then Enter gives 35

47 Example of a Combination Determine the number of different possible 5 card poker hands.

48 Example ANSWER: 52C 5 2,598,960

49 Motivational Example Genetics In mice an allele A for agouti (gray-brown, grizzled fur) is dominant over the allele a, which determines a non-agouti color. Suppose each parent has the genotype Aa and 4 offspring are produced. What is the probability that exactly 3 of these have agouti fur?

50 Motivational Example A single offspring has genotypes: A a A AA Aa Sample Space { AA, Aa, aa, aa} a aa aa

51 Motivational Example Agouti genotype is dominant Event that offspring is agouti: { AA, Aa, aa} Therefore, for any one birth: P(agouti genotype) 3/ 4 P(not agouti genotype) 1/ 4

52 Motivational Example Let G represent the event of an agouti offspring and N represent the event of a non-agouti Exactly three agouti offspring may occur in four different ways (in order of birth): NGGG, GNGG, GGNG, GGGN

53 Motivational Example Consecutive events (birth of a mouse) are independent and using multiplication rule: ) ( ) ( ) ( ) ( ) ( G P G P G P N P G G G N P ) ( ) ( ) ( ) ( ) ( G P G P N P G P G G N G P

54 Motivational Example ) ( ) ( ) ( ) ( ) ( G P N P G P G P G N G G P ) ( ) ( ) ( ) ( ) ( N P G P G P G P N G G G P

55 Motivational Example P(exactly 3 offspring has agouti fur) birth N OR secondbirth N OR third birth N OR fourth birth N) (first 3 P

56 Binomial Experiment Agouti fur example may be considered a binomial experiment

57 Binomial Experiment Four Requirements: 1) Each trial of the experiment has only two possible outcomes (success or failure) 2) Fixed number of trials 3) Experimental outcomes are independent of each other 4) Probability of observing a success remains the same from trial to trial

58 Binomial Experiment Agouti fur example may be considered a binomial experiment 1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur) 2) Fixed number of trials (4 births) 3) Experimental outcomes are independent of each other 4) Probability of observing a success remains the same from trial to trial (¾)

59 Binomial Probability Distribution When a binomial experiment is performed, the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution.

60 Binomial Probability Distribution Formula For a binomial experiment, the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is X n P ( X ) ncx p (1 p) X where n C X X! n! n X!

61 Binomial Probability Distribution Formula Let q=1-p P ( X ) n C X p X q n X

62 Rationale for the Binomial Probability Formula P(x) = n! (n x )!x! p x q n-x The number of outcomes with exactly x successes among n trials

63 Binomial Probability Formula P(x) = n! (n x )!x! p x q n-x Number of outcomes with exactly x successes among n trials The probability of x successes among n trials for any one particular order

64 Agouti Fur Genotype Example X event of a birth with agouti fur P( X ) (4 4! 3)!3!

65 Binomial Probability Distribution Formula: Calculator 2 ND VARS A:binompdf(4,.75, 3) n, p, x Enter gives the result

66 Binomial Distribution Tables n is the number of trials X is the number of successes p is the probability of observing a success See Example 6.16 on page 278 for more information FIGURE 6.7 Excerpt from the binomial tables.

67 Page 284 Example

68 Example ANSWER X number of heads P( X 5) 20C 5 (0.5) 5 15,504 (0.5) (0.5) 5 20 (0.5) 5 15

69 Binomial Mean, Variance, and Standard Deviation Mean (or expected value): μ = n p Variance: 2 np(1 p) Use q 1 p, then 2 npq Standard deviation: np (1 p) npq

70 Example 20 coin tosses The expected number of heads: np (20)(0.50) 10 Variance and standard deviation: 2 npq (20)(0.50)(0.50)

71 Page 284 Example

72 Is this a Binomial Distribution? Four Requirements: 1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket) 2) Fixed number of trials (50) 3) Experimental outcomes are independent of each other 4) Probability of observing a success remains the same from trial to trial (assumed to be 58.4%=0.584)

73 Example ANSWER (a) X number of baskets P( X 25) 50 C (0.584) 25 (0.416) 25

74 Example ANSWER (b) The expected value of X np (50)(0.584) 29.2 The most likely number of baskets is 29

75 Example ANSWER (c) In a random sample of 50 of O Neal s shots he is expected to make 29.2 of them.

76 Page 285 Example

77 Is this a Binomial Distribution? Four Requirements: 1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not) 2) Fixed number of trials (120) 3) Experimental outcomes are independent of each other 4) Probability of observing a success remains the same from trial to trial (assumed to be 11%=0.11)

78 Example ANSWER (a) X=number of white males who contracted AIDS through injected drug use P( X 10) 120 C (0.11) 10 (0.89) 110

79 Example ANSWER (b) At most 3 men is the same as less than or equal to 3 men: P( X 3) P( X 0) P( X 1) P( X 2) P( X 3) Why do probabilities add?

80 Example Use TI-83+ calculator 2 ND VARS to get A:binompdf(n, p, X) P( X 0) P( X 1) P( X 2) P( X 3) = binompdf(120,.11, 0) + binompdf(120,.11, 1) + binompdf(120,.11, 2) + binompdf(120,.11,3) E

81 Example ANSWER (c) Most likely number of white males is the expected value of X np (120)(0.11) 13.2

82 Example ANSWER (d) In a random sample of 120 white males with AIDS, it is expected that approximately 13 of them will have contracted AIDS by injected drug use

83 Page 286 Example

84 Example ANSWER (a) 2 npq (120)(0.11)(0.89)

85 RECALL: Outliers and z Scores Data values are not unusual if 2 z -score 2 Otherwise, they are moderately unusual or an outlier (see page 131)

86 Z score Formulas Sample Population z x x z x s

87 Z-score for 20 white males who contracted AIDS through injected drug use: z Example It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS.

88 Summary The most important discrete distribution is the binomial distribution, where there are two possible outcomes, each with probability of success p, and n independent trials. The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula.

89 Summary Binomial probabilities can also be found using the binomial tables or using technology. There are formulas for finding the mean, variance, and standard deviation of a binomial random variable.

90 6.3 Continuous Random Variables and the Normal Probability Distribution Objectives: By the end of this section, I will be able to 1) Identify a continuous probability distribution. 2) Explain the properties of the normal probability distribution.

91 FIGURE Histograms (a) Relatively small sample (n = 100) with large class widths (0.5 lb). (b) Large sample (n = 200) with smaller class widths (0.2 lb).

92 Figure 6.15 continued (c) Very large sample (n = 400) with very small class widths (0.1 lb). (d) Eventually, theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small.

93 Continuous Probability Distributions A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take Density curve is drawn above the horizontal axis Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

94 Requirements for Probability Distribution of a Continuous Random Variable 1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables). 2) The vertical height of the density curve can never be negative. That is, the density curve never goes below the horizontal axis.

95 Probability for a Continuous Random Variable Probability for Continuous Distributions is represented by area under the curve above an interval.

96 The Normal Probability Distribution Most important probability distribution in the world Population is said to be normally distributed, the data values follow a normal probability distribution population mean is μ population standard deviation is σ μ and σ are parameters of the normal distribution

97 FIGURE 6.19 The normal distribution is symmetric about its mean μ (bell-shaped).

98 Properties of the Normal Density Curve (Normal Curve) 1) It is symmetric about the mean μ. 2) The highest point occurs at X = μ, because symmetry implies that the mean equals the median, which equals the mode of the distribution. 3) It has inflection points at μ-σ and μ+σ. 4) The total area under the curve equals 1.

99 Properties of the Normal Density Curve (Normal Curve) continued 5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 0.5 (Figure 6.19). 6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions. As X moves farther from the mean, the density curve approaches but never quite touches the horizontal axis.

100 The Empirical Rule For data sets having a distribution that is approximately bell shaped, the following properties apply: About 68% of all values fall within 1 standard deviation of the mean. About 95% of all values fall within 2 standard deviations of the mean. About 99.7% of all values fall within 3 standard deviations of the mean.

101 FIGURE 6.23 The Empirical Rule.

102 Drawing a Graph to Solve Normal Probability Problems 1. Draw a generic bell-shaped curve, with a horizontal number line under it that is labeled as the random variable X. Insert the mean μ in the center of the number line.

103 Steps in Drawing a Graph to Help You Solve Normal Probability Problems 2) Mark on the number line the value of X indicated in the problem. Shade in the desired area under the normal curve. This part will depend on what values of X the problem is asking about. 3) Proceed to find the desired area or probability using the empirical rule.

104 Page 295 Example

105 ANSWER Example

106 Page 296 Example

107 ANSWER Example

108 Page 296 Example

109 ANSWER Example

110 Summary Continuous random variables assume infinitely many possible values, with no gap between the values. Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve.

111 Summary The normal distribution is the most important continuous probability distribution. It is symmetric about its mean μ and has standard deviation σ. One should always sketch a picture of a normal probability problem to help solve it.