2011 Pearson Education, Inc

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2 Statistics for Business and Economics Chapter 4 Random Variables & Probability Distributions

3 Content 1. Two Types of Random Variables 2. Probability Distributions for Discrete Random Variables 3. The Binomial Distribution 4. Poisson and Hypergeometric Distributions 5. Probability Distributions for Continuous Random Variables 6. The Normal Distribution

4 Content (continued) 7. Approximating a Binomial Distribution with a Normal Distribution 8. Sampling Distributions 9. The Sampling Distribution of a Sample Mean and the Central Limit Theorem

5 Learning Objectives 1. Develop the notion of a random variable 2. Learn that numerical data are observed values of either discrete or continuous random variables 3. Study two important types of random variables and their probability models: the binomial and normal model 4. Define a sampling distribution as the probability of a sample statistic 5. Learn that the sampling distribution of x follows a normal model

6 Thinking Challenge You re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What s the chance you ll get it right? If you guessed on all 33 questions, what would be your grade? Would you pass?

7 4.1 Two Types of Random Variables

8 Random Variable A random variable is a variable that assumes numerical values associated with the random outcomes of an experiment, where one (and only one) numerical value is assigned to each sample point.

9 Discrete Random Variable Random variables that can assume a countable number (finite or infinite) of values are called discrete.

10 Discrete Random Variable Examples Experiment Random Variable Possible Values Make 100 Sales Calls # Sales 0, 1, 2,..., 100 Inspect 70 Radios # Defective 0, 1, 2,..., 70 Answer 33 Questions # Correct 0, 1, 2,..., 33 Count Cars at Toll Between 11:00 & 1:00 # Cars Arriving 0, 1, 2,...,

11 Continuous Random Variable Random variables that can assume values corresponding to any of the points contained in one or more intervals (i.e., values that are infinite and uncountable) are called continuous.

12 Continuous Random Variable Examples Experiment Random Variable Possible Values Weight 100 People Weight 45.1, 78,... Measure Part Life Hours 900, 875.9,... Amount spent on food $ amount 54.12, 42,... Measure Time Between Arrivals Inter-Arrival Time 0, 1.3, 2.78,...

13 4.2 Probability Distributions for Discrete Random Variables

14 Discrete Probability Distribution The probability distribution of a discrete random variable is a graph, table, or formula that specifies the probability associated with each possible value the random variable can assume.

15 Requirements for the Probability Distribution of a Discrete Random Variable x 1. p(x) 0 for all values of x 2. p(x) = 1 where the summation of p(x) is over all possible values of x.

16 Discrete Probability Distribution Example Experiment: Toss 2 coins. Count number of tails. Probability Distribution Values, x Probabilities, p(x) 0 1/4 = /4 = /4 = T/Maker Co.

17 Visualizing Discrete Probability Distributions Listing { (0,.25), (1,.50), (2,.25) } Graph p(x) x p ( x ) # Tails Table f(x) Count Formula p(x) n! = p x!(n x)! x (1 p) n x

18 Summary Measures 1. Expected Value (Mean of probability distribution) Weighted average of all possible values = E(x) = x p(x) 2. Variance Weighted average of squared deviation about mean 2 = E[(x 2 (x 2 p(x) 3. Standard Deviation 2

19 Summary Measures Calculation Table x p(x) x p(x) x (x 2 (x 2 p(x) Total x p(x) (x 2 p(x)

20 Thinking Challenge You toss 2 coins. You re interested in the number of tails. What are the expected value, variance, and standard deviation of this random variable, number of tails? T/Maker Co.

21 Expected Value & Variance Solution* x p(x) x p(x) x (x 2 (x 2 p(x) =

22 Probability Rules for Discrete Random Variables Let x be a discrete random variable with probability distribution p(x), mean µ, and standard deviation. Then, depending on the shape of p(x), the following probability statements can be made: Chebyshev s Rule Empirical Rule P x x µ 0.68 P x 2 x µ P x 3 x µ

23 4.3 The Binomial Distribution

24 Binomial Distribution Number of successes in a sample of n observations (trials) Number of reds in 15 spins of roulette wheel Number of defective items in a batch of 5 items Number correct on a 33 question exam Number of customers who purchase out of 100 customers who enter store (each customer is equally likely to pyrchase)

25 Binomial Probability Characteristics of a Binomial Experiment 1. The experiment consists of n identical trials. 2. There are only two possible outcomes on each trial. We will denote one outcome by S (for success) and the other by F (for failure). 3. The probability of S remains the same from trial to trial. This probability is denoted by p, and the probability of F is denoted by q. Note that q = 1 p. 4. The trials are independent. 5. The binomial random variable x is the number of S s in n trials.

26 Binomial Probability Distribution n n! p( x) p q p (1 p) x x! ( n x)! x n x x n x p(x) = Probability of x Successes p = Probability of a Success on a single trial q = 1 p n = Number of trials x = Number of Successes in n trials (x = 0, 1, 2,..., n) n x = Number of failures in n trials

27 Binomial Probability Distribution Example Experiment: Toss 1 coin 5 times in a row. Note number of tails. What s the probability of 3 tails? n! x p( x) p (1 p) x!( n x)! n x T/Maker Co. 5! p(3).5 (1.5) 3!(5 3)!

28 Binomial Probability Table n = 5 (Portion) k Cumulative Probabilities p(x 3) p(x 2) = =.312 p

29 Binomial Distribution Characteristics Mean E(x) np P(X) n = 5 p = 0.1 Standard Deviation npq P(X) n = 5 p = X X

30 Binomial Distribution Thinking Challenge You re a telemarketer selling service contracts for Macy s. You ve sold 20 in your last 100 calls (p =.20). If you call 12 people tonight, what s the probability of A. No sales? B. Exactly 2 sales? C. At most 2 sales? D. At least 2 sales?

31 Binomial Distribution Solution* n = 12, p =.20 A. p(0) =.0687 B. p(2) =.2835 C. p(at most 2) = p(0) + p(1) + p(2) = =.5584 D. p(at least 2) = p(2) + p(3)...+ p(12) = 1 [p(0) + p(1)] = =.7251

32 4.4 Other Discrete Distributions: Poisson and Hypergeometric

33 Poisson Distribution 1. Number of events that occur in an interval events per unit Time, Length, Area, Space 2. Examples Number of customers arriving in 20 minutes Number of strikes per year in the U.S. Number of defects per lot (group) of DVD s

34 Characteristics of a Poisson Random Variable 1. Consists of counting number of times an event occurs during a given unit of time or in a given area or volume (any unit of measurement). 2. The probability that an event occurs in a given unit of time, area, or volume is the same for all units. 3. The number of events that occur in one unit of time, area, or volume is independent of the number that occur in any other mutually exclusive unit. 4. The mean number of events in each unit is denoted by.

35 Poisson Probability Distribution Function 2 p(x) = Probability of x given = Mean (expected) number of events in unit e x p ( x ) x x e (x = 0, 1, 2, 3,...) = (base of natural logarithm) = Number of events per unit!

36 Poisson Probability Distribution Function = 0.5 Mean E(x) P(X) X Standard Deviation P(X) 2 4 = X

37 Poisson Distribution Example Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes? 1995 Corel Corp.

38 Poisson Distribution Solution 72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval px ( ) x e - x! e p(4) !

39 Poisson Probability Table (Portion) x : : : : : : : : : : : : : : Cumulative Probabilities p(x 4) p(x 3) = =.191

40 Thinking Challenge You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction? T/Maker Co.

41 Poisson Distribution Solution: Finding * 75 words/min = (75 words/min)(60 min/hr) = 4500 words/hr 6 errors/hr = 6 errors/4500 words = errors/word In a 255-word transaction (interval): = ( errors/word )(255 words) =.34 errors/255-word transaction

42 Poisson Distribution Solution: Finding p(0)* px ( ) x e - x! e p(0) !

43 Characteristics of a Hypergeometric Random Variable 1. The experiment consists of randomly drawing n elements without replacement from a set of N elements, r of which are S s (for success) and (N r) of which are F s (for failure). 2. The hypergeometric random variable x is the number of S s in the draw of n elements.

44 Hypergeometric Probability Distribution Function p x r N r x n x N n [x = Maximum [0, n (N r),, Minimum (r, n)] µ nr N 2 r N r N 2 n N n N 1 where...

45 Hypergeometric Probability Distribution Function N = Total number of elements r = Number of S s in the N elements n = Number of elements drawn x = Number of S s drawn in the n elements

46 4.5 Probability Distributions for Continuous Random Variables

47 Continuous Probability Density Function The graphical form of the probability distribution for a continuous random variable x is a smooth curve

48 Continuous Probability Density Function This curve, a function of x, is denoted by the symbol f(x) and is variously called a probability density function (pdf), a frequency function, or a probability distribution. The areas under a probability distribution correspond to probabilities for x. The area A beneath the curve between two points a and b is the probability that x assumes a value between a and b.

49 4.6 The Normal Distribution

50 Importance of Normal Distribution 1. Describes many random processes or continuous phenomena 2. Can be used to approximate discrete probability distributions Example: binomial 3. Basis for classical statistical inference

51 Normal Distribution 1. Bell-shaped & symmetrical f ( x ) 2. Mean, median, mode are equal x Mean Median Mode

52 Probability Density Function f (x) 1 2 e 1 2 x 2 where µ = Mean of the normal random variable x = Standard deviation π = e = P(x < a) is obtained from a table of normal probabilities

53 Effect of Varying Parameters ( & )

54 Normal Distribution Probability Probability is area under curve! P(c x d) c d f (x) dx? f ( x ) c d x

55 Standard Normal Distribution The standard normal distribution is a normal distribution with µ = 0 and = 1. A random variable with a standard normal distribution, denoted by the symbol z, is called a standard normal random variable.

56 The Standard Normal Table: P(0 < z < 1.96) Standardized Normal Probability Table (Portion) Z = 0 Probabilities = z Shaded area exaggerated

57 The Standard Normal Table: P( 1.26 z 1.26) Standardized Normal Distribution = z = 0 Shaded area exaggerated P( 1.26 z 1.26) = =.7924

58 The Standard Normal Table: P(z > 1.26) Standardized Normal Distribution = = 0 z P(z > 1.26) = =.1038

59 The Standard Normal Table: P( 2.78 z 2.00) Standardized Normal Distribution = z = 0 P( 2.78 z 2.00) = =.0201 Shaded area exaggerated

60 The Standard Normal Table: P(z > 2.13) Standardized Normal Distribution = z = 0 Shaded area exaggerated P(z > 2.13) = =.9834

61 Non-standard Normal Distribution Normal distributions differ by mean & standard deviation. Each distribution would require its own table. f(x) x That s an infinite number of tables!

62 Property of Normal Distribution If x is a normal random variable with mean μ and standard deviation, then the random variable z, defined by the formula z x µ has a standard normal distribution. The value z describes the number of standard deviations between x and µ.

63 Standardize the Normal Distribution Normal Distribution z x Standardized Normal Distribution = 1 x = 0 z One table!

64 Finding a Probability Corresponding to a Normal Random Variable 1. Sketch normal distribution, indicate mean, and shade the area corresponding to the probability you want. 2. Convert the boundaries of the shaded area from x values to standard normal random variable z z x µ Show the z values under corresponding x values. 3. Use Table IV in Appendix A to find the areas corresponding to the z values. Use symmetry when necessary.

65 Non-standard Normal μ = 5, σ = 10: Normal Distribution = 10 P(5 < x < 6.2) z x Standardized Normal Distribution = = x = 0.12 z Shaded area exaggerated

66 Non-standard Normal μ = 5, σ = 10: Normal Distribution = 10 P(3.8 x 5) z x Standardized Normal Distribution = = 5 x Shaded area exaggerated -.12 = 0 z

67 Non-standard Normal μ = 5, σ = 10: z x Normal Distribution = 10 P(2.9 x 7.1).21 z x Standardized Normal Distribution = x z Shaded area exaggerated

68 Non-standard Normal μ = 5, σ = 10: Normal Distribution = 10 P(x 8) z x Standardized Normal Distribution = = 5 8 x = 0.30 z Shaded area exaggerated

69 Non-standard Normal μ = 5, σ = 10: z x Normal Distribution = 10 P(7.1 X 8).21 z x Standardized Normal Distribution = = x = z Shaded area exaggerated

70 Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours and = 200 hours. What s the probability that a bulb will last A. between 2000 and 2400 hours? B. less than 1470 hours?

71 Solution* P(2000 x 2400) Normal Distribution z x Standardized Normal Distribution = 200 = = x = z

72 Solution* P(x 1470) Normal Distribution = 200 z x Standardized Normal Distribution = = 2000 x 2.65 = 0 z

73 Finding z-values for Known Probabilities What is Z, given P(z) =.1217? Standardized Normal Probability Table (Portion).1217 = 1 Z Shaded area exaggerated = 0?.31 z

74 Finding x Values for Known Probabilities Normal Distribution = 10 Standardized Normal Distribution = = 5 8.1? x = 0.31 z x z Shaded areas exaggerated

75 4.8 Approximating a Binomial Distribution with a Normal Distribution

76 Normal Approximation of Binomial Distribution 1. Useful because not all binomial tables exist 2. Requires large sample size 3. Gives approximate probability only 4. Need correction for continuity n = 10 p = 0.50 p(x) x

77 Why Probability Is Approximate p(x) Binomial Probability: Bar Height Probability Added by Normal Curve Probability Lost by Normal Curve x Normal Probability: Area Under Curve from 3.5 to 4.5

78 Correction for Continuity 1. A 1/2 unit adjustment to discrete variable 2. Used when approximating a discrete distribution with a continuous distribution 3. Improves accuracy (4.5) 4.5 (4 +.5)

79 Using a Normal Distribution to Approximate Binomial Probabilities 1. Determine n and p for the binomial distribution, then calculate the interval: 3 np 3 np 1 p If interval lies in the range 0 to n, the normal distribution will provide a reasonable approximation to the probabilities of most binomial events.

80 Using a Normal Distribution to Approximate Binomial Probabilities 2. Express the binomial probability to be approximated by the form P x a For example, or P x b P x a P x 3 P x 2 P x 5 1 P x 4 P 7 x 10 P x 10 P x 6

81 Using a Normal Distribution to Approximate Binomial Probabilities 3. For each value of interest a, the correction for continuity is (a +.5), and the corresponding standard normal z-value is z a.5 µ

82 Using a Normal Distribution to Approximate Binomial Probabilities 4. Sketch the approximating normal distribution and shade the area corresponding to the event of interest. Using Table IV and the z-value (step 3), find the shaded area. This is the approximate probability of the binomial event.

83 Normal Approximation Example What is the normal approximation of p(x = 4) given n = 10, and p = 0.5? P(x) x

84 Normal Approximation Solution 1. Calculate the interval: np 3 np 1 p , 9.74 Interval lies in range 0 to 10, so normal approximation can be used 2. Express binomial probability in form: P x 4 P x 4 P x 3

85 Normal Approximation Solution 3. Compute standard normal z values: z a.5 n p n p 1 p z a.5 n p n p 1 p

86 Normal Approximation Solution 4. Sketch the approximate normal distribution: = 0 = z

87 Normal Approximation Solution 5. The exact probability from the binomial formula is (versus.2034) p(x) x

88 4.10 Sampling Distributions

89 Parameter & Statistic A parameter is a numerical descriptive measure of a population. Because it is based on all the observations in the population, its value is almost always unknown. A sample statistic is a numerical descriptive measure of a sample. It is calculated from the observations in the sample.

90 Common Statistics & Parameters Sample Statistic Population Parameter Mean x Standard Deviation s Variance s 2 2 Binomial Proportion ^ p p

91 Sampling Distribution The sampling distribution of a sample statistic calculated from a sample of n measurements is the probability distribution of the statistic.

92 Developing Sampling Distributions Suppose There s a Population... Population size, N = 4 Random variable, x Values of x: 1, 2, 3, 4 Uniform distribution T/Maker Co.

93 Population Characteristics Summary Measure Population Distribution N x i i 1 N 2.5 P(x) x

94 All Possible Samples of Size n = 2 16 Samples 16 Sample Means 1st Obs 2nd Observation st Obs 2nd Observation ,1 1,2 1,3 1, ,1 2,2 2,3 2, ,1 3,2 3,3 3, ,1 4,2 4,3 4, Sample with replacement

95 Sampling Distribution of All Sample Means 1st Obs Sample Means Sampling Distribution of the Sample Mean 2nd Observation P(x) x

96 Summary Measure of All Sample Means X N x i i 1 N

97 Comparison Population Sampling Distribution P(x) x P(x) x 2.5 x 2.5

98 4.11 The Sampling Distribution of a Sample Mean and the Central Limit Theorem

99 Properties of the Sampling Distribution of x 1. Mean of the sampling distribution equals mean of sampled population*, that is, x E x. 2. Standard deviation of the sampling distribution equals Standard deviation of sampled population Square root of sample size That is, x n.

100 Standard Error of the Mean The standard deviation x is often referred to as the standard error of the mean.

101 Theorem 4.1 If a random sample of n observations is selected from a population with a normal distribution, the sampling distribution of x will be a normal distribution.

102 Sampling from Normal Populations Central Tendency Dispersion x x n Sampling with replacement n = 4 x = 5 Population Distribution = 50 = 10 Sampling Distribution x n =16 x = 2.5 x - = 50 x

103 Standardizing the Sampling Distribution of x z x x x x Sampling Distribution x n Standardized Normal Distribution = 1 x x = 0 z

104 Thinking Challenge You re an operations analyst for AT&T. Long-distance telephone calls are normally distributed with = 8 min. and = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes? T/Maker Co.

105 Sampling Distribution Solution* Sampling Distribution x =.4 z x n 25 z x n Standardized Normal Distribution = x z

106 Sampling from Non-Normal Populations Central Tendency Dispersion x x n Sampling with replacement Population Distribution n = 4 x = 5 = 50 = 10 Sampling Distribution x n =30 x = 1.8 x - = 50 x

107 Central Limit Theorem Consider a random sample of n observations selected from a population (any probability distribution) with mean μ and standard deviation. Then, when n is sufficiently large, the sampling distribution of x will be approximately a normal distribution with mean x and standard deviation x n. The larger the sample size, the better will be the normal approximation to the sampling distribution of x.

108 Central Limit Theorem As sample size gets large enough (n 30)... x n sampling distribution becomes almost normal. x x

109 Central Limit Theorem Example The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of.2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than oz? SODA

110 Central Limit Theorem Solution* Sampling Distribution x =.03 z x n Standardized Normal Distribution.0384 = x z Shaded area exaggerated.4616

111 Key Ideas Properties of Probability Distributions Discrete Distributions 1. p(x) 0 2. p x 1 all x Continuous Distributions 1. P(x = a) = 0 2. P(a < x < b) = area under curve between a and b

112 Key Ideas Normal Approximation to Binomial x is binomial (n, p) P x a P z a.5 µ

113 Key Ideas Methods for Assessing Normality 1. Histogram

114 Key Ideas Methods for Assessing Normality 2. Stem-and-leaf display

115 Key Ideas Methods for Assessing Normality 3. (IQR)/S Normal probability plot

116 Key Ideas Generating the Sampling Distribution of x