Learning Objectives for Ch. 5

Transcription

1 Chapter : Probabilit Distributions Hildebrand, Ott and Gra Basic Statistical Ideas for Managers Second Edition Learning Objectives for Ch. Understanding the counting techniques needed for sequences and combinations. Understanding that a binomial random variable counts the number of successes in a fixed number of trials with each trial being a success or failure. Assumptions needed to use the binomial. Understanding that a Poisson random variable counts the number of occurrences of an event in a unit of time, area or volume. Assumptions needed to use the Poisson. 2 Learning Objectives for Ch. Understanding that a normal random variable measures a characteristic of interest and has a bell-shaped distribution. Learning how to calculate probabilities for binomial, Poisson and normal random variables. Understanding how to use a normal probabilit plot to determine if data is from a normal distribution. 3

2 Section. Counting Possible Outcomes 4. Counting Possible Outcomes Under the classical interpretation of probabilit: P(Event) = Number of favorable outcomes Total number of outcomes We need was to count the number of outcomes.. Counting Possible Outcomes Preliminar Concept Factorials The factorial smbol is! Definition of n! n! = n (n - ) (n - 2) Example: 3! = (3)(2)() = 6 B definition,! =. 6

3 . Counting Possible Outcomes One consideration in counting techniques Order matters sequences Order doesn t matter subsets Example: Consider the letters a and b. If order matters, there are 2 sequences: (a,b) and (b,a) If order does not matter, there is onl subset: {a,b} 7. Counting Possible Outcomes Number of sequences Rule: The number of sequences of k objects that can be formed from a set of r distinct objects, denoted r P k, is: rp k = (r) (r - ) (r k + ) Example: The number of sequences of 2 letters formed from the 4 letters a, b, c, d, is: (4) (3) = 2 The sequences are: (a,b) (a,c) (a,d) (b,c) (b,d) (c,d) (b,a) (c,a) (d,a) (c,b) (d,b) (d,c) 8. Counting Possible Outcomes Number of subsets or combinations Rule: The number of subsets of k objects that can be formed from a set of r distinct objects, denoted r C k, is: rc k = r! k! (r k)! r k Notation: Use r C k or ( ) 9

4 . Counting Possible Outcomes Example: The number of subsets of 2 letters formed from the 4 letters a, b, c, d is: r k 4! 2! (4-2)! r C k = ( ) = = 6 The subsets are: {a,b} {a,c} {a,d} {b,c} {b,d} {c,d}. Counting Possible Outcomes Exercise.67: Several states now have a Lotto game. A plaer chooses 6 distinct integers in the range to 4. If exactl those 6 numbers are selected as the winning numbers, the plaer receives a ver large prize. What is the probabilit that a particular set of 6 numbers will be drawn? You ma wish to think of the 6 numbers drawn as the success numbers. First approach: Order matters (even though it doesn t) Total number of outcomes = 4 P 6 = (4)(39)(38)(37)(36)(3) Number of favorable outcomes = 6 P 6 = (6)()(4)(3)(2)() = 6! P(Winning) = 6!/ [(4) (3)] = Counting Possible Outcomes Another perspective of the first approach P(Winning) = 6! / [(4) (3)] = (6)()(4)(3)(2)() (4)(39)(38)(37)(36)(3) 6 = = P(W ) P(W 2 /W ) P(W 6 /W and W ) Where W i {The i th number is a winning number} 2

5 . Counting Possible Outcomes Second approach: Order doesn t matter (and it reall doesn t) Total number of outcomes = 4 C 6 = 4! 6! 34! Number of favorable outcomes = 6C 6 34 C P (Winning) = 6!/[(4) (3)] = Moral: You can t lose if ou don t pla! = 6!. 34! = 6!!! 34! 3 Section.2 The Binomial Distribution 4.2 The Binomial Distribution Examples of a Bernoulli Trial. A coin toss results in a head (H) or a tail (T). 2. A bit sent through a digital communications channel is entered as either or and received either correctl or incorrectl. 3. An audited account is either current (C) or delinquent (D). 4. A consumer is either aware (A) of a particular product or not aware (N).. A flight reservation is either a show (S) or no-show (N).

6 .2 The Binomial Distribution Features of a Bernoulli Trial: Onl 2 possible outcomes for each trial, characterized as: Success (S) or Failure (F) π denotes P(S) ( π) denotes P(F). 6.2 The Binomial Distribution Bernoulli R.V. and Probabilit Distribution Let Y =, if trial results in S =, if trial results in F PY ( ) π π 7.2 The Binomial Distribution Graphical representation of a Bernoulli probabilit distribution P Y () π π The distribution is skewed when π. 8

7 .2 The Binomial Distribution E(Y) = ( - π) + (π ) = π V(Y) = Σ ( - µ) 2 P Y () = [ π] 2 ( π) + [ π] 2 π = π ( π) [π + ( π) ] = π ( π) 9.2 The Binomial Distribution Examples of Binomial Random Variable:. Toss a coin times. Let Y denote the number of heads in the tosses. 2. For the next 3 bits transmitted through a digital communications channel, let Y be the number of bits received that are in error accounts are randoml selected from a population of several thousand accounts and are audited. Let Y be the number of delinquent accounts in the sample. [The sampling has to be with replacement for the probabilit of success to remain constant. In realit, the sampling is done without replacement.] 4. randoml selected consumers are surveed as part of a market research stud. Let Y denote the number of these consumers who are aware of a particular product.. Out of flight reservations made, let Y be the number of passengers who show. 2.2 The Binomial Distribution Features of a Binomial Experiment: There are n Bernoulli trials [each one results in S or F]. The probabilit of a success, π = P(S), remains constant over the n trials; [P(F) = - π ]. The trials are independent. The binomial random variable is the total number of successes in n trials, where the ordering is unimportant. 2

8 .2 The Binomial Distribution Binomial Probabilit Distribution n = π π = n - P Y() ( - ),,,..., n n n! =! (n - )! The expression for P Y () can be used to calculate probabilities for a binomial random variable. What is the basis for the expression for P Y ()? 22.2 The Binomial Distribution Example: Y denotes number of bits in error in next 3 transmitted where P(Error) = π Outcomes Probabilit From P Y () E, E, E 3 π 3 3 π 3 (-π) = π 3 3 E, E, O E, O, E O, E, E 2 3 π 2 (- π) 3 π 2 (- π) = 3 π 2 (- π) 2 E, O, O O, E, O O, O, E 3 π (- π) 2 3 π (- π) 3 = 3 π (- π ) 2 O, O, O (- π) 3 (- π ) 3 Found b using principles of Chapter The Binomial Distribution Calculation of Probabilities Use the binomial probabilit distribution formula Instead of actuall calculating the probabilities, we can look them up in a table. Table at the end of Hildebrand, Ott & Gra gives the probabilities for n = 2() (2) 2,, and π =.(.).. We can also use software (MINITAB; EXCEL s BINOMDIST function) Two obvious cases n n n P[ successes] = π (- π) = (- π) n n n P[n successes] = π ( - π ) = π n 24

9 .2 The Binomial Distribution Mean and Variance of a Binomial Random Variable E[Y] = nπ V(Y) = σ 2 = nπ ( - π) 2.2 The Binomial Distribution An eas wa to find E(Y) and V(Y) Y = total number of successes in n trials = Number of successes on st trial + Number of successes on 2nd trial + + Number of successes on nth trial E(Y) = π + π +. + π = nπ V(Y) = π ( π) + π ( π)+ + π ( π) = nπ ( π) 26.2 The Binomial Distribution Exercise.6 [Revised so that number of potential customers is.] Executives at a soft drink compan wish to test a new formulation of their chief product. The new drink is tested in comparison to the current one. Each of potential customers is given a cup of the current formulation and a cup of the new one. The cups are labeled H and K to avoid bias. Each customer indicates a preference. Assume that, in fact, the customers can't detect a difference and are, in effect, guessing. Define Y to be the number (out of ) indicating preference for the new formulation. 27

10 .2 The Binomial Distribution a. What probabilit distribution should appl to Y? Do the assumptions underling that distribution seem plausible in this context? Each of the customers is a Bernoulli trial (either prefers new product or does not). If customers are guessing, the probabilit of preference for new product is.. Reasonable to assume trials are independent. Let Y be the number of customers who indicate a preference for the new product. Then Y is binomial with n = and π = The Binomial Distribution A graph of the probabilit distribution of Y follows..2 Probabilit Distribution of Y..8 P(Y=Y) The graph is smmetric because π = The Binomial Distribution b. Find the mean and standard deviation of Y. µ = nπ = ()(.) = 2 σ 2 = nπ (-π) = 2. σ = 3.4 3

11 .2 The Binomial Distribution c. (Cont d) Find the probabilit that the number of customers preferring the new brand is within 2 standard deviations of the mean. P[µ 2σ Y µ + 2σ ] = P[2 2(3.4) Y 2 + 2(3.4)] = P[ 7.93 Y 32.8] = P[ 8 Y 32] = P[Y=8] + P[Y=9] + + P[Y=32] = (From Table ) =.9672 Most of the time (97%), we should observe between 8 and 32 customers indicating a preference for the new product if, in fact, the are guessing. 3.2 The Binomial Distribution d. (Cont d) In one such test, 2 people preferred the new formulation. Find the probabilit that 2 or fewer would prefer the new formulation if the customers can t detect a difference. What, if anthing, can ou infer about consumer preferences from the results of the taste test. P(Y 2) =. (from Table ) If the hpothesis that the people can t detect a difference is correct, P(Y 2) is ver small [ <.]. Since this probabilit is ver small, it implies the hpothesis that the people can t detect a difference is incorrect! Or, π. Wh were the cups labeled H and K? Studies have shown that people have no preference for either of these letters, as opposed to the letters A and B. 32 Section.3 The Poisson Distribution 33

12 .3 The Poisson Distribution Named for Simeon D. Poisson (78-84) Examples of a Poisson random variable The number of work-related injuries per month at a manufacturing plant. The number of messages arriving at a personal computer in one hour. The number of network errors per da on a local area network. The Poisson random variable is the number of occurrences in a given unit The Poisson Distribution Features of a Poisson Experiment For a unit of time, area or volume Probabilit that an event occurs in a given unit is the same for all units. Probabilit of two or more events occurring at same time is. The occurrence of the event in one unit is independent of the number that occur in other units. The expected number of occurrences in each unit is denoted b µ. 3.3 The Poisson Distribution Poisson Probabilit Distribution µ e µ PY ( ) = =,,2... ( )! Calculation of Probabilities Use formula for p Y () Use Table 2 for µ =.(.) and.(.) and ()2 Use software 36

13 .3 The Poisson Distribution Mean and Variance for a Poisson Random Variable E(Y) = µ Var(Y) = µ 37.3 The Poisson Distribution Exercise.29: Suppose that the number of defaults on home mortgage loans at National Mortgage Compan follows a Poisson distribution with an average of 8.2 defaults per month. a. Compute the probabilit of exactl 2 defaults at NMC next month (8.2) P(Y = 2) = P (2) = e Y (2)! =.2992 { From Minitab 38.3 The Poisson Distribution A graph of the probabilit distribution of Y, the number of defaults per month follows..4 Probabilit Distribution of Y.2. P(Y=) The probabilit distribution quickl tapers off to. or less for 6. 39

14 .3 The Poisson Distribution b. What is the chance of at least one default next week? P(Y ) = P(Y = ) = -.27 = c. Because of poor economic times, NMC believes that the average number of defaults ma have increased from 8.2 per month. Last month, there were defaults. If the average number of defaults has not changed from 8.2, find P(Y ). P(Y ) = P(Y 4) = =.29 Since P(Y ) is small, this implies µ has changed from Section.4 The Normal Distribution 4.4 The Normal Distribution Continuous Random Variables in General Examples of continuous random variables: Stock market returns Qualit characteristics of finished products (such as net contents) Heights of males; heights of females Age at time of death 42

15 .4 The Normal Distribution Continuous Random Variables in General (Cont d) Features of a continuous random variable: The possible values are uncountable. The probabilit that the random variable takes on a specific value is. Onl an interval of values has a nonzero probabilit The Normal Distribution Continuous Random Variables in General (Cont d) The probabilit for an interval of values will be shown as the area under the pdf. () f Y P(a< Y < b) a b 44.4 The Normal Distribution Continuous Random Variables in General (Cont d) Details: It doesn t matter whether endpoints are included in the interval: P[a < Y < b] = P[a Y < b] = P[a < Y b] = P[a Y b] Wh? P[Y = a] = P[Y = b] =. Data are never continuous! 4

16 .4 The Normal Distribution The Standard Normal Random Variable The probabilit distribution of a standard normal random variable Z is shown below: f z (z) z 46.4 The Normal Distribution E(Z) = µ z = {The curve is smmetric around } V(Z) = σ z2 = Other Properties: Total area under the curve is. The curve is smmetric around. P(Z > ) = The Normal Distribution Determination of probabilities for a standard normal random variable: Use Table 3 (area from to a right-hand value z) Use software 48

17 .4 The Normal Distribution Exercise.3: Suppose that Z represents a standard normal random variable. i. Find P(Z -2.42). fz ( z) P(Z -2.42) =. - P( Z 2.42) = (from Table 3) =.78 z The Normal Distribution g. Find P(-.7 Z 2.33) P(-.7 Z 2.33) fz ( z) = P(-.7 Z ) + P( Z 2.33) z = (from Table 3) = The Normal Distribution Exercise.3: For the standard normal random variable Z, solve the following equation for k. a. P(Z k) =. fz ( z) From Table 3, P( Z 2.33) =.49 P(Z 2.33) =. k = k z

18 .4 The Normal Distribution Normal Random Variables in General fy( ) σ µ The probabilit distribution is mound-shaped. µ is the expected value of the distribution. σ is the standard deviation of the distribution. 2.4 The Normal Distribution Standardize Y to find areas under the normal curve of Y. Z Y µ Y = σ Y {Procedure for standardizing Y} Now use Table 3. The standardized variable Z measures how man standard deviations Y is above or below its mean. 3.4 The Normal Distribution Exercise.4: A potato chip packaging plant has a process line that fills 2 ounce bags of potato chips. At the current setting of the machine, the qualit control engineer knows that the actual distribution of weights in the bags follows a normal distribution with a mean of 2. ounces and a standard deviation of.8 ounces. a. What percentage of all bags filled contain exactl 2 ounces? P(Y = 2) =, since the probabilit at a point is. 4

19 .4 The Normal Distribution b. What percentage of all bags filled contain more than 2.4 ounces? () f Y P(Y > 2.4) = P(Z > ).8 = P( Z > 2.22) = = is 2.22 standard deviations from The Normal Distribution c. Find the 6th percentile of the actual weights of 2- ounce bags of potato chips. () f Y.6 2 Find k so that P(Y< k) =.6 Standardizing k 2 P(Z< ) =.6.8 From Table 3, P(Z <.23) =.6, k 2 Set =.23.8 k = 2 +(.8)(.23) k = 2.46 ounces 6.4 The Normal Distribution d. Management is concerned when 2-ounce bags of potato chips contain less than.7 ounces. The qualit control engineer can set the filling machine so that actual mean filling weight is whatever he chooses, but the standard deviation alwas remains at.8 ounces. What mean filling weight should he set the machine to if he wants onl % of all bags to contain less than.7 ounces? Find µ so that P (Y <.7) =..7 µ. = P(Y <.7) = P(Z < ).8. = P(Z < -2.33) from Table 3.7 µ Set = µ =.7 + (2.33)(.8) = 2.7 ounces 7

20 Section. Checking Normalit 8. Checking Normalit Man of the statistical techniques in later chapters assume that the data is from a normal distribution. Chapter 2 presented several graphical techniques that could be useful in assessing whether or not the data is from a normal distribution. For example, is a histogram mound-shaped? The answer to this question is facilitated b superimposing a normal distribution over the histogram. 9. Checking Normalit Example: Consider the returns for ^DJI first presented in Chapter. The histogram with a normal distribution superimposed follows. Histogram for R^DJI with Normal Distribution Superimposed Normal Mean StDev.287 N 3 8 Frequenc R^DJI 6

21 . Checking Normalit Conclusion: At first glance, it appears that the normal distribution is not a good fit. However, the shape of the histogram is determined b the number of class intervals and their width. So, this ma not be the best approach. Histogram for R^DJI with Normal Distribution Superimposed Normal Mean StDev.287 N 3 8 Frequenc R^DJI 6. Checking Normalit Another approach for assessing normalit is the Normal Probabilit Plot. The data are arranged in ascending order. Each data value, (i), is assigned a cumulative relative frequenc, p i : ( i.) pi = n Think of. as a correction factor. Other correction factors are sometimes used. 62. Checking Normalit For example, if the data set has 2 observations, then p = 2., p 2 = 6.,, p 2 = 98. The percentage of the observations less than or equal to () is 2.%. The percentage of the observations less than or equal to (2) is 6.%. ( (i), p i ) are plotted on a graph where the vertical axis is scaled so that if the data is from a normal distribution, the resulting plot should be approximatel linear. 63

22 . Checking Normalit Appearance of NPP s for data from a distribution that is not normal. Right-skewed data plot as a curve, with the slope getting flatter as one moves to the right. Left-skewed data plot as a curve, with the slope getting steeper as one moves to the right. Data from smmetric distributions with more tail area than the normal plot as an S-shape, with the slope steepest at both ends. The straight line drawn through the points can assist in assessing linearit. It can also be misleading if a few of the points are outliers. In the following examples, the sample size is fixed at 2. This value for n was arbitraril chosen. 64. Checking Normalit Example: What does the NPP look like for data from a standard normal distribution? Probabilit Plot of z Normal Mean.799 StDev.28 N 2 A D.96 P-Value Percent z 6. Checking Normalit Conclusion: Since the plotted points are nearl linear, conclude that the data came from a normal distribution. Probabilit Plot of z Normal Mean.799 StDev.28 N 2 A D.96 P-Value Percent z 66

23 . Checking Normalit Example: What does the NPP look like for data from a normal distribution with µ = and σ =? Probabilit Plot of Normal Mean 3.8 StDev. N 2 A D.23 P-Value Percent Checking Normalit Conclusion: Since the plotted points are nearl linear, conclude that the data came from a normal distribution. Probabilit Plot of Normal Mean 3.8 StDev. N 2 A D.23 P-Value Percent Checking Normalit Example: A uniform distribution is one that is of uniform or constant height for the range of values. For the interval from 3 to +3, a uniform distribution has height of (/6). What does the NPP look like for data from a uniform distribution that ranges from 3 to +3? Probabilit Plot of Normal Mean. S tdev.944 N 2 A D.844 P-Value Percent

24 . Checking Normalit Conclusion: Because the plot is S-shaped with the slope steepest at both ends, conclude that the data came from a smmetric distribution with more probabilit in each tail than the normal distribution. Probabilit Plot of Normal Mean. StDev.944 N 2 A D.844 P-Value.2 8 Percent Checking Normalit Example: What does the NPP look like for data from a distribution that is skewed to the right with E(Y) = 927 and σ Y = 87? Probabilit Plot of Normal Mean 94.9 StDev 8. N 2 A D.229 P-Value <. 8 7 Percent Checking Normalit Conclusion: Since the plot is curved with the slope getting flatter as one moves to the right, conclude that the data came from a right-skewed distribution. Probabilit Plot of Normal Mean 94.9 StDev 8. N 2 A D.229 P-Value <. 8 7 Percent

25 . Checking Normalit Example: Consider the returns for R^DJI. What does the NPP tell us? NPP for R^DJI Normal Mean StDev.287 N 3 AD.24 P-Value Percent R^DJI 73. Checking Normalit Conclusion: Because the NPP is linear, conclude that the R^DJI are normall distributed. However, it s a different stor for the RIBM data. The NPP for RIBM follows. NPP for RIBM Normal Mean.4368 StDev 2.3 N 3 A D.784 P-Value Percent RIBM 74. Checking Normalit Procedure to obtain a Normal Probabilit Plot using Minitab: Suppose the data to be analzed are stored in C Click on Stat Basic Statistics Normalit Test Enter C in box for Variable Select Percentile Lines option. The default option is None Select Tests for Normalit option. The default option is Anderson-Darling Enter Title for plot Click on OK 7

26 Kewords: Chapter Factorial Sequences Combinations Bernoulli trials Binomial random variable Binomial probabilit distribution Poisson random variable Poisson probabilit distribution Normal random variable Standard normal probabilit distribution Normal probabilit distribution Normal probabilit plot 76 Summar of Chapter The counting techniques needed for sequences and combinations. A binomial random variable counts the number of successes in n trials, with each trial being a success or failure. A Poisson random variable counts the number of occurrences of an event over a specified length of time. A normal random variable measures the characteristic of interest and the probabilit distribution is bell-shaped. 77 Summar of Chapter Computing probabilities for the binomial, Poisson and normal random variables. Assessing normalit of data b the normal probabilit plot. 78