Business Statistics Probability Distributions Random Variables A random

Business Statistics Probability Distributions

Random Variables… A random variable is a function or rule that assigns a number to each outcome of an experiment. Alternatively, the value of a random variable is a numerical event. Instead of talking about the coin flipping event as {heads, tails} think of it as {1, 0} “the number of heads when flipping a coin” (numerical events) 2

Two Types of Random Variables… Discrete Random Variable – one that takes on a countable number of values – E. g. values on the roll of dice: 2, 3, 4, …, 12 Continuous Random Variable – one whose values are not discrete, not countable – E. g. time (30. 1 minutes? 30. 10000001 minutes? ) Analogy: Integers are Discrete, while Real Numbers are Continuous 3

Probability Distributions… A probability distribution is a table, formula, or graph that describes the values of a random variable and the probability associated with these values. Since we’re describing a random variable (which can be discrete or continuous) we have two types of probability distributions: – Discrete Probability Distribution and – Continuous Probability Distribution 4

Probability Notation… An upper-case letter will represent the name of the random variable, usually X. Its lower-case counterpart will represent the value of the random variable. The probability that the random variable X will equal x is: P(X = x) or more simply P(x) 5

Discrete Probability Distributions… The probabilities of the values of a discrete random variable may be derived by means of probability tools such as tree diagrams or by applying one of the definitions of probability, so long as these two conditions apply: 6

Example 7. 1… Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data… 1, 218 ÷ 101, 501 = 0. 012 e. g. P(X=4) = P(4) = 0. 076 = 7. 6% 7

Example 7. 1… E. g. what is the probability there is at least one television but no more than three in any given household? “at least one television but no more than three” P(1 ≤ X ≤ 3) = P(1) + P(2) + P(3) =. 319 +. 374 +. 191 =. 884 8

Example 7. 2… Developing a probability distribution… Probability calculation techniques can be used to develop probability distributions, for example, a mutual fund sales person knows that there is 20% chance of closing a sale on each call she makes. What is the probability distribution of the number of sales if she plans to call three customers? Let S denote success, i. e. closing a sale P(S)=. 20 Thus SC is not closing a sale, and P(SC)=. 80 9

Example 7. 2… Developing a Probability Distribution… Sales Call 1 Sales Call 2 Sales Call 3 (. 2)(. 8)=. 032 P(S)=. 2 P(SC)=. 8 P(S)=. 2 SSS P(SC)=. 8 P(S)=. 2 S S SC S P(SC)=. 8 P(S)=. 2 S S C SC SC S S P(SC)=. 8 P(S)=. 2 SC S S C SC SC S P(SC)=. 8 SC SC SC P(SC)=. 8 P(S)=. 2 P(SC)=. 8 X 3 2 1 0 P(x). 23 =. 008 3(. 032)=. 096 3(. 128)=. 384. 83 =. 512 P(X=2) is illustrated here… 10

Population/Probability Distribution… The discrete probability distribution represents a population Example 7. 1 the population of number of TVs per household Example 7. 2 the population of sales call outcomes Since we have populations, we can describe them by computing various parameters. E. g. the population mean and population variance. 11

Population Mean (Expected Value) The population mean is the weighted average of all of its values. The weights are the probabilities. This parameter is also called the expected value of X and is represented by E(X). 12

Population Variance… The population variance is calculated similarly. It is the weighted average of the squared deviations from the mean. As before, there is a “short-cut” formulation… The standard deviation is the same as before: 13

Example 7. 3… Find the mean, variance, and standard deviation for the population of the number of color televisions per household… (from Example 7. 1) = 0(. 012) + 1(. 319) + 2(. 374) + 3(. 191) + 4(. 076) + 5(. 028) = 2. 084 14

Example 7. 3… Find the mean, variance, and standard deviation for the population of the number of color televisions per household… (from Example 7. 1) = (0 – 2. 084)2(. 012) + (1 – 2. 084)2(. 319)+…+(5 – 2. 084)2(. 028) = 1. 107 15

Example 7. 3… Find the mean, variance, and standard deviation for the population of the number of color televisions per household… (from Example 7. 1) = 1. 052 16

Laws of Expected Value… 1. E(c) = c The expected value of a constant (c) is just the value of the constant. 2. E(X + c) = E(X) + c 3. E(c. X) = c. E(X) We can “pull” a constant out of the expected value expression (either as part of a sum with a random variable X or as a coefficient of random variable X). 17

Example 7. 4… Monthly sales have a mean of $25, 000 and a standard deviation of $4, 000. Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6, 000. Find the mean monthly profit. 1) Describe the problem statement in algebraic terms: – sales have a mean of $25, 000 E(Sales) = 25, 000 – profits are calculated by… Profit =. 30(Sales) – 6, 000 18

Example 7. 4… Monthly sales have a mean of $25, 000 and a standard deviation of $4, 000. Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6, 000. Find the mean monthly profit. E(Profit) =E[. 30(Sales) – 6, 000] =E[. 30(Sales)] – 6, 000 [by rule #2] =. 30 E(Sales) – 6, 000 [by rule #3] =. 30(25, 000) – 6, 000 = 1, 500 Thus, the mean monthly profit is $1, 500 19

Laws of Variance… 1. V(c) = 0 The variance of a constant (c) is zero. 2. V(X + c) = V(X) The variance of a random variable and a constant is just the variance of the random variable (per 1 above). 3. V(c. X) = c 2 V(X) The variance of a random variable and a constant coefficient is the coefficient squared times the variance of the random variable. 20

Example 7. 4… Monthly sales have a mean of $25, 000 and a standard deviation of $4, 000. Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6, 000. Find the standard deviation of monthly profits. 1) Describe the problem statement in algebraic terms: sales have a standard deviation of $4, 000 V(Sales) = 4, 0002 = 16, 000 (remember the relationship between standard deviation and variance ) profits are calculated by… Profit =. 30(Sales) – 6, 000 21

Example 7. 4… Monthly sales have a mean of $25, 000 and a standard deviation of $4, 000. Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6, 000. Find the standard deviation of monthly profits. 2) The variance of profit is = V(Profit) =V[. 30(Sales) – 6, 000] =V[. 30(Sales)] [by rule #2] =(. 30)2 V(Sales) [by rule #3] =(. 30)2(16, 000) = 1, 440, 000 Again, standard deviation is the square root of variance, so standard deviation of Profit = (1, 440, 000)1/2 = $1, 200 22

Example 7. 4 (summary) Monthly sales have a mean of $25, 000 and a standard deviation of $4, 000. Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6, 000. Find the mean and standard deviation of monthly profits. The mean monthly profit is $1, 500 The standard deviation of monthly profit is $1, 200 23

Random Variables • Random Variable (RV): A numeric outcome that results from an experiment • For each element of an experiment’s sample space, the random variable can take on exactly one value • Discrete Random Variable: An RV that can take on only a finite or countably infinite set of outcomes • Continuous Random Variable: An RV that can take on any value along a continuum (but may be reported “discretely” • Random Variables are denoted by upper case letters (Y) • Individual outcomes for RV are denoted by lower case letters (y) 24

Probability Distributions • Probability Distribution: Table, Graph, or Formula that describes values a random variable can take on, and its corresponding probability (discrete RV) or density (continuous RV) • Discrete Probability Distribution: Assigns probabilities (masses) to the individual outcomes • Continuous Probability Distribution: Assigns density at individual points, probability of ranges can be obtained by integrating density function • Discrete Probabilities denoted by: p(y) = P(Y=y) • Continuous Densities denoted by: f(y) • Cumulative Distribution Function: F(y) = P(Y≤y) 25

Discrete Probability Distributions 26

Example – Rolling 2 Dice (Red/Green) Y = Sum of the up faces of the two die. Table gives value of y for all elements in S RedGreen 1 2 3 4 5 6 7 8 9 4 5 6 7 8 9 10 11 12 27

Rolling 2 Dice – Probability Mass Function & CDF y p(y) F(y) 2 1/36 3 2/36 3/36 4 3/36 6/36 5 4/36 10/36 6 5/36 15/36 7 6/36 21/36 8 5/36 26/36 9 4/36 30/36 10 3/36 33/36 11 2/36 35/36 12 1/36 36/36 28

Rolling 2 Dice – Probability Mass Function 29

Rolling 2 Dice – Cumulative Distribution Function 30

Expected Values of Discrete RV’s • Mean (aka Expected Value) – Long-Run average value an RV (or function of RV) will take on • Variance – Average squared deviation between a realization of an RV (or function of RV) and its mean • Standard Deviation – Positive Square Root of Variance (in same units as the data) • Notation: – Mean: E(Y) = m – Variance: V(Y) = 2 – Standard Deviation: 31

Expected Values of Discrete RV’s 32

Expected Values of Linear Functions of Discrete RVs 33

Example – Rolling 2 Dice y p(y) y 2 p(y) 2 1/36 2/36 4/36 3 2/36 6/36 18/36 4 3/36 12/36 48/36 5 4/36 20/36 100/36 6 5/36 30/36 180/36 7 6/36 42/36 294/36 8 5/36 40/36 320/36 9 4/36 36/36 324/36 10 3/36 300/36 11 2/36 242/36 12 1/36 12/36 144/36 Sum 36/36= 252/36= 1974/36=5 1. 00 7. 00 4. 833 34

Case Study 1. 1 • Case of a fruit seller who sells strawberries. Product has a very limited shelf life & is useless unless sold on the day of delivery. 1 Case of strawberries costs $20 and the wholesaler receives $50 for it. Cannot specify the number of cases that will be sold on any day, but analysis of past records produced this information. • What is the optimal stock? Days Sales No. of Days Sold Prob. of each number being sold 10 15 0. 15 11 20 0. 20 12 40 0. 40 13 25 0. 25 100 1. 00

Case Study 1. 2 An airlines needs to make a decision about its Flight 105. Currently there are 3 seats reserved for last minute customers, but the airlines does not know if anyone will buy them. If they release the seats now, they have to sell them for $250 each. Last minute customers must pay $475 per seat. The company also counts a $150 loss of goodwill for every last minute customer who is turned away. 1. 2. 3. 4. # last min customers requesting seats 0 1 2 3 Probability 0. 45 0. 30 0. 15 0. 10 How much revenue will be generated by releasing all 3 seats now? What is the company’s expected net revenue (revenue less loss of goodwill) if 3 seats are released now? What is the company’s expected net revenue if 2 seats are released now? How many seats should be released to maximize expected revenue?

Case Study 1. 3 In a lottery the total instant winnings of $34. 8 million was available in 70 million $1 tickets, with the ticket prizes ranging from $1 to $1, 000. Below are various prizes along with the probability of winning. Prize (x) & e lu me. a v ga d cte the e xp n of e he iatio t v ute de p m ard o C nd sta Probability P(x) $1, 000 0. 00002 $100 0. 00063 $20 0. 00400 $10 0. 00601 $4 0. 02403 $2 0. 08877 $1 0. 10479 0 0. 77176

Probability Distributions Discrete Probability Distributions Continuous Probability Distributions Binomial Uniform Poisson Normal 38

Discrete Probability Distributions • A discrete random variable is a variable that can assume only a countable number of values Many possible outcomes: – number of complaints per day – number of TV’s in a household – number of rings before the phone is answered Only two possible outcomes: – gender: male or female – defective: yes or no Co t n u 39

Continuous Probability Distributions • A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) – thickness of an item – time required to complete a task – temperature of a solution – height, in inches ea r u s e M • These can potentially take on any value, depending only on the ability to measure accurately. 40

The Binomial Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson 41

The Binomial Distribution • Characteristics of the Binomial Distribution: – A trial has only two possible outcomes – “success” or “failure”, “head” or “tail”, “win” or “lose”, “even” or “odd” – There is a fixed number, n, of identical trials – The trials of the experiment are independent of each other – The probability of a success, p, remains constant from trial to trial – If p represents the probability of a success, then (1 -p) = q is the probability of a failure 42

Binomial Experiment • Experiment consists of a series of n identical trials • Each trial can end in one of 2 outcomes: Success or Failure, Head or Tail, Win or Lose, Even or Odd • Trials are independent (outcome of one has no bearing on outcomes of others) • Probability of Success, p, is constant for all trials • Random Variable Y, is the number of Successes in the n trials is said to follow Binomial Distribution with parameters n and p • Y can take on the values y=0, 1, …, n • Notation: Y~Bin(n, p) 43

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Binomial Distribution Settings • A manufacturing plant labels items as either defective or acceptable • A firm bidding for a contract will either get the contract or not • A marketing research firm receives survey responses of “yes I will buy” or “no I will not” • New job applicants either accept the offer or reject it 47

Counting Rule for Combinations • A combination is an outcome of an experiment where x objects are selected from a group of n objects where: n! =n(n - 1)(n - 2). . . (2)(1) x! = x(x - 1)(x - 2). . . (2)(1) 0! = 1 (by definition) 48

Binomial Distribution Formula n! x n x P(x) = p q x ! (n - x )! P(x) = probability of x successes in n trials, with probability of success p on each trial x = number of ‘successes’ in sample, (x = 0, 1, 2, . . . , n) p = probability of “success” per trial q = probability of “failure” = (1 – p) n = number of trials (sample size) Example: Flip a coin four times, let x = # heads: n=4 p = 0. 5 q = (1 -. 5) =. 5 x = 0, 1, 2, 3, 4 49

Binomial Distribution • The shape of the binomial distribution depends on the values of p and n Mean – Here, n = 5 and p =. 1 . 6. 4. 2 0 P(X) X 0 – Here, n = 5 and p =. 5 . 6. 4. 2 0 n = 5 p = 0. 1 P(X) 1 2 3 4 5 n = 5 p = 0. 5 X 0 1 2 3 4 5 50

Binomial Distribution Characteristics • Mean • Variance and Standard Deviation Where n = sample size p = probability of success q = (1 – p) = probability of failure 51

Binomial Characteristics Examples Mean . 6. 4. 2 0 P(X) X 0 . 6. 4. 2 0 n = 5 p = 0. 1 P(X) 1 2 3 4 5 n = 5 p = 0. 5 X 0 1 2 3 4 5 52

Using Binomial Tables n = 10 x p=. 15 p=. 20 p=. 25 p=. 30 p=. 35 p=. 40 p=. 45 p=. 50 0 1 2 3 4 5 6 7 8 9 10 0. 1969 0. 3474 0. 2759 0. 1298 0. 0401 0. 0085 0. 0012 0. 0001 0. 0000 0. 1074 0. 2684 0. 3020 0. 2013 0. 0881 0. 0264 0. 0055 0. 0008 0. 0001 0. 0000 0. 0563 0. 1877 0. 2816 0. 2503 0. 1460 0. 0584 0. 0162 0. 0031 0. 0004 0. 0000 0. 0282 0. 1211 0. 2335 0. 2668 0. 2001 0. 1029 0. 0368 0. 0090 0. 0014 0. 0001 0. 0000 0. 0135 0. 0725 0. 1757 0. 2522 0. 2377 0. 1536 0. 0689 0. 0212 0. 0043 0. 0005 0. 0000 0. 0060 0. 0403 0. 1209 0. 2150 0. 2508 0. 2007 0. 1115 0. 0425 0. 0106 0. 0016 0. 0001 0. 0025 0. 0207 0. 0763 0. 1665 0. 2384 0. 2340 0. 1596 0. 0746 0. 0229 0. 0042 0. 0003 0. 0010 0. 0098 0. 0439 0. 1172 0. 2051 0. 2461 0. 2051 0. 1172 0. 0439 0. 0098 0. 0010 10 9 8 7 6 5 4 3 2 1 0 p=. 85 p=. 80 p=. 75 p=. 70 p=. 65 p=. 60 p=. 55 p=. 50 x Examples: n = 10, p =. 35, x = 3: P(x = 3|n =10, p =. 35) =. 2522 n = 10, p =. 75, x = 2: P(x = 2|n =10, p =. 75) =. 0004 53

Pat… Pat is a student taking a statistics course whose exam strategy is to rely on luck for the next quiz. The quiz consists of 10 multiple-choice questions. Each question has five possible answers, only one of which is correct. Pat plans to guess the answer to each question. • What is the probability that Pat gets no answers correct? • What is the probability that Pat gets two answers correct? • Algebraically then: n=10, and P(success) = 1/5 =. 20 54

Pat … Is this a binomial experiment? Check the conditions: There is a fixed finite number of trials (n=10). An answer can be either correct or incorrect. The probability of a correct answer (P(success)=. 20) does not change from question to question. Each answer is independent of the others. 55

Pat … • n=10, and P(success) =. 20 • What is the probability that Pat gets no answers correct? • I. e. # success, x, = 0; hence we want to know P(x=0) Pat has about an 11% chance of getting no answers correct using the guessing strategy. 56

Pat … • n=10, and P(success) =. 20 • What is the probability that Pat gets two answers correct? • I. e. # success, x, = 2; hence we want to know P(x=2) Pat has about a 30% chance of getting exactly two answers correct using the guessing strategy. 57

Cumulative Probability… Thus far, we have been using the binomial probability distribution to find probabilities for individual values of x. To answer the question: • “Find the probability that Pat fails the quiz” • requires a cumulative probability, that is, P(X ≤ x) • If a grade on the quiz is less than 50% (i. e. 5 questions • out of 10), that’s considered a failed quiz. • Thus, we want to know what is: P(X ≤ 4) to answer 58

Pat … • P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4) • We already know P(0) =. 1074 and P(2) =. 3020. Using the binomial formula to calculate the others: • P(1) =. 2684 , P(3) =. 2013, and P(4) =. 0881 • We have P(X ≤ 4) =. 1074 +. 2684 + … +. 0881 =. 9672 • Thus, its about 97% probable that Pat will fail the test using the luck strategy and guessing at answers… 59

Binomial Table… • Calculating binomial probabilities by hand is tedious and error prone. So we use the Binomial Tables. For the Pat example, n=10, so the first important step is to get the correct table! 60

Binomial Table… • The probabilities listed in the tables are cumulative, • i. e. P(X ≤ k) – k is the row index; the columns of the table are organized by P(success) = p 61

Binomial Table… • “What is the probability that Pat fails the quiz”? • i. e. what is P(X ≤ 4), given P(success) =. 20 and n=10 ? P(X ≤ 4) =. 967 62

Binomial Table… • “What is the probability that Pat gets no answers correct? ” • i. e. what is P(X = 0), given P(success) =. 20 and n=10 ? P(X = 0) = P(X ≤ 0) =. 107 63

Binomial Table… • “What is the probability that Pat gets two answers correct? ” • i. e. what is P(X = 2), given P(success) =. 20 and n=10 ? P(X = 2) = P(X≤ 2) – P(X≤ 1) =. 678 –. 376 =. 302 remember, the table shows cumulative probabilities… 64

Binomial Table… • The binomial table gives cumulative probabilities for • P(X ≤ k), but as we’ve seen in the last example, • P(X = k) = P(X ≤ k) – P(X ≤ [k– 1]) • Likewise, for probabilities given as P(X ≥ k), we have: • P(X ≥ k) = 1 – P(X ≤ [k– 1]) 65

Case Study 1. 4 In the electronics section of a large departmental store, it has been observed that the probability of a customer who is just browsing will buy something is 0. 3. Suppose that 15 customers browse products in the electronics section every hour. 1. 2. 3. 4. What is the probability that at least one browsing customer will buy something during a specified hour? What is the probability that at least four browsing customers will buy something during a specified hour? What is the probability that no browsing customers will buy anything during a specified hour? What is the probability that at not more than 4 browsing customer will buy something during a specified hour?

Case Study 1. 5 The mayor of a large city is concerned that a large number of people drawing unemployment cheques are secretly employed. The mayor’s assistants estimate that 40% of unemployment beneficiaries fall in this category, but the mayor is not convinced and asks one of his aides to conduct an investigation of 10 randomly selected unemployment beneficiaries. 1. 2. If the mayor’s assistants are correct, what is the probability that more than 8 of the individuals investigated have jobs? If the mayor’s assistants are correct, what is the probability that only 3 of the individuals investigated have jobs?

The Poisson Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson 68

The Poisson Distribution • Characteristics of the Poisson Distribution: – The outcomes of interest are relative to the possible outcomes – The average number of outcomes of interest per time or space interval is – The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest – The probability that an outcome of interest occurs in a given segment is the same for all segments 69

Poisson Distribution • Distribution often used to model the number of incidences of some characteristic in time or space: – Arrivals of customers in a queue – Numbers of flaws in a roll of fabric – Number of typos per page of text. Count of success (or failure) in a number of experiments, n where p is very small; n is very large; • Distribution obtained as follows: – – – Break down the “area” into many small “pieces” (n pieces) Each “piece” can have only 0 or 1 occurrences (p=P(1)) Let l=np ≡ Average number of occurrences over “area” Y ≡ # occurrences in “area” is sum of 0 s & 1 s over “pieces” Y ~ Bin(n, p) with p = l/n Take limit of Binomial Distribution as n with p = l/n 70

Poisson Distribution - Derivation 71

Poisson Distribution - Expectations 72

Poisson Distribution Formula where: t = size of the segment of interest x = number of successes in segment of interest = expected number of successes in a segment of unit size e = base of the natural logarithm system (2. 71828. . . ) 73

Poisson as an approximation of Binomial The Poisson distribution is a reasonable approximation of the Binomial when • n, the number of trials, is large • p, the binomial probability of success is small • np is finite • Since n is very large we can assume that the number of possible value of the Poisson variable is countably infinite • Thumb rule • n >= 20 • p <= 0. 05 In that case, we can substitute the mean of the binomial distribution (np) in place of the mean of the Poisson distribution ( ), so that the formula becomes P(x) = (np)x. e-np / x! 74

Poisson as an approximation of Binomial Differences decrease as n increases, p decreases

• Mean Poisson Distribution Characteristics • Variance and Standard Deviation where = number of successes in a segment of unit size t = the size of the segment of interest 76

Using Poisson Tables t X 0. 10 0. 20 0. 30 0. 40 0. 50 0. 60 0. 70 0. 80 0. 90 0 1 2 3 4 5 6 7 0. 9048 0. 0905 0. 0045 0. 0002 0. 0000 0. 8187 0. 1637 0. 0164 0. 0011 0. 0000 0. 7408 0. 2222 0. 0333 0. 0003 0. 0000 0. 6703 0. 2681 0. 0536 0. 0072 0. 0007 0. 0001 0. 0000 0. 6065 0. 3033 0. 0758 0. 0126 0. 0016 0. 0002 0. 0000 0. 5488 0. 3293 0. 0988 0. 0198 0. 0030 0. 0004 0. 0000 0. 4966 0. 3476 0. 1217 0. 0284 0. 0050 0. 0007 0. 0001 0. 0000 0. 4493 0. 3595 0. 1438 0. 0383 0. 0077 0. 0012 0. 0000 0. 4066 0. 3659 0. 1647 0. 0494 0. 0111 0. 0020 0. 0003 0. 0000 Example: Find P(x = 2) if =. 005 and t = 100 77

Graph of Poisson Probabilities Graphically: =. 005 and t = 100 X t = 0. 50 0 1 2 3 4 5 6 7 0. 6065 0. 3033 0. 0758 0. 0126 0. 0016 0. 0002 0. 0000 P(x = 2) =. 0758 78

Poisson Distribution Shape • The shape of the Poisson Distribution depends on the parameters and t: t = 0. 50 t = 3. 0 79

Example 7. 12… The number of typographical errors in new editions of textbooks varies considerably from book to book. The number of errors is Poisson distributed with a mean of 1. 5 per 100 pages. The instructor randomly selects 100 pages of a new book. What is the probability that there are no typos? • That is, what is P(X=0) given that = 1. 5? “There is about a 22% chance of finding zero errors” 80

Poisson Distribution… As mentioned on the Poisson experiment slide: The probability of a success is proportional to the size of the interval Thus, knowing an error rate of 1. 5 typos per 100 pages, we can determine a mean value for a 400 page book as: =1. 5(4) = 6 typos / 400 pages. 81

Example 7. 13… For a 400 page book, what is the probability that there are no typos? P(X=0) = “there is a very small chance there are no typos” 82

Example 7. 13… For a 400 page book, what is the probability that there are five or less typos? P(X≤ 5) = P(0) + P(1) + … + P(5) This is rather tedious to solve manually. A better alternative is to refer to Poisson Table… …k=5, =6, and P(X ≤ k) =. 446 “there is about a 45% chance there are 5 or less typos” 83

Case Study 1. 6 On Monday mornings, the First National Bank has only one teller window open for deposits & withdrawals. Experience has shown that the average number of arriving customers in a 4 -minute interval on Monday mornings is 2. 8, and each teller can serve more than that number efficiently. These random arrivals at this bank on Monday mornings are Poisson distributed. 1. 2. 3. What is the probability that on a Monday morning exactly 6 customers will arrive in a 4 -minute interval? What is the probability that no one will arrive at the bank to make a deposit or withdrawal during a 4 -minute interval? What is the probability that exactly 3 people will arrive at the bank during a 2 -minute period on Monday mornings to make a deposit or a withdrawal? What is the probability that 5 or more customers will arrive during an 8 -minute period?

Case Study 1. 7 Following is the count of customers in a restaurant every 5 minutes from 7 p. m. to 8 p. m. every Saturday night for 3 weeks. 1. 2. 3. 4. 5. Week 1 3 6 4 6 2 3 1 5 1 0 3 3 Week 2 1 2 4 0 2 6 5 4 2 5 3 4 Week 3 5 3 5 4 7 3 4 8 1 3 Calculate l What is the probability that no customers arrive during any given 5 minute interval? What is the probability that 6 or more customers arrive during any given 5 -minute interval? What is the probability that during a 10 minute interval fewer than 4 customers arrive? What is the probability that between 3 & 6 (both inclusive) customers arrive in any 10 minute interval?

Case Study 1. 8 A medical researcher estimates that 0. 00004 of the population has a rare blood disorder. If the researcher randomly selects 100, 000 people from the population, what is the probability that 7 or more people will have the rare blood disorder? What is the probability that more than 10 people will have the rare blood disorder? Suppose the researcher gets more than 10 people who have the rare blood disorder in the sample of 100, 000 but that the sample was taken from a particular geographic region. What might the researcher conclude form the results?

Case Study 1. 9 Colour blindness appears in 1% of the people in a certain population. 1. 2. How large must a random sample be if the probability of its containing a colour blind person is to be 0. 95 or more? What is the probability that a sample of 100 will contain a) b) No colour blind person 2 or more colour blind persons?

Probability Density Functions… Unlike a discrete random variable, a continuous random variable is one that can assume an uncountable number of values. We cannot list the possible values because there is an infinite number of them. Because there is an infinite number of values, the probability of each individual value is virtually 0. 88

Point Probabilities are Zero Because there is an infinite number of values, the probability of each individual value is virtually 0. Thus, we can determine the probability of a range of values only. • E. g. with a discrete random variable like tossing a die, it is meaningful to talk about P(X=5), say. • In a continuous setting (e. g. with time as a random variable), the probability the random variable of interest, say task length, takes exactly 5 minutes is infinitesimally small, hence P(X=5) = 0. • It is meaningful to talk about P(X ≤ 5). 89

Probability Density Function… • A function f(x) is called a probability density function (over the range a ≤ x ≤ b if it meets the following requirements: 1) f(x) ≥ 0 for all x between a and b, and f(x) area=1 a b x 2) The total area under the curve between a and b is 1. 0 90

The Uniform Distribution Probability Distributions Continuous Probability Distributions Uniform Normal 91

The Uniform Distribution • The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable 92

The Uniform Distribution The Continuous Uniform Distribution: f(x) = where f(x) = value of the density function at any x value a = lower limit of the interval b = upper limit of the interval 93

Uniform Distribution Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6: 1 f(x) = 6 - 2 =. 25 for 2 ≤ x ≤ 6 f(x). 25 2 6 x 94

Uniform Distribution… • Consider the uniform probability distribution (sometimes called the rectangular probability distribution). • It is described by the function: f(x) a b x area = width x height = (b – a) x =1 95

Example 8. 1(a)… The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2, 000 gallons and a maximum of 5, 000 gallons. f(x) 2, 000 5, 000 x Find the probability that daily sales will fall between 2, 500 and 3, 000 gallons. Algebraically: what is P(2, 500 ≤ X ≤ 3, 000) ? 96

Example 8. 1(a)… • P(2, 500 ≤ X ≤ 3, 000) = (3, 000 – 2, 500) x =. 1667 f(x) 2, 000 5, 000 x • “there is about a 17% chance that between 2, 500 and 3, 000 gallons of gas will be sold on a given day” 97

Example 8. 1(b)… • The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2, 000 gallons and a maximum of 5, 000 gallons. f(x) 2, 000 5, 000 x • What is the probability that the service station will sell at least 4, 000 gallons? • Algebraically: what is P(X ≥ 4, 000) ? 98

Example 8. 1(b)… • P(X ≥ 4, 000) = (5, 000 – 4, 000) x =. 3333 f(x) 2, 000 5, 000 x • “There is a one-in-three chance the gas station will sell more than 4, 000 gallons on any given day” 99

Example 8. 1(c)… • The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2, 000 gallons and a maximum of 5, 000 gallons. f(x) 2, 000 5, 000 x • What is the probability that the station will sell exactly 2, 500 gallons? • Algebraically: what is P(X = 2, 500) ? 100

Example 8. 1(c)… • P(X = 2, 500) = (2, 500 – 2, 500) x =0 f(x) 2, 000 5, 000 x • “The probability that the gas station will sell exactly 2, 500 gallons is zero” 101

Case Study 2. 1 The retail price of a medium-sized box of a well known brand of cornflakes ranges from $2. 80 to $3. 14. 1. Assuming that these prices are uniformly distributed, what are the average price and standard deviation of prices in this distribution? 2. If a price is randomly selected from this list, what is the probability that it will be between $3. 00 & $3. 10 102

Case Study 2. 2 The average fill volume of a nregular can of soft drink is 12 ounces. Suppose the fill volumes of these cans ranges from 11. 97 to 12. 03 ounces and is uniformly distributed. 1. What is the height of this distribution? 2. What is the probability that a randomly selected can contains more than 12. 01 ounces of fluid? 3. What is the probability that the fill volume is between 11. 98 and 12. 01 ounces? 103

The Normal Distribution Probability Distributions Continuous Probability Distributions Uniform Normal 104

The Normal Distribution • ‘Bell Shaped’ • Symmetrical • Mean, Median and Mode are Equal Location is determined by the mean, μ Spread is determined by the standard deviation, σ The random variable has an infinite theoretical range: + to f(x) σ x μ Mean = Median = Mode 105

Many Normal Distributions By varying the parameters μ and σ, we obtain different normal distributions 106

The Normal Distribution Shape f(x) Changing μ shifts the distribution left or right. σ μ Changing σ increases or decreases the spread. x 107

Finding Normal Probabilities Probability is the Probability is measured area under the curve! under the curve f(x) by the area P (a x b) a b x 108

Probability as Area Under the Curve The total area under the curve is 1. 0, and the curve is symmetric, so half is above the mean, half is below f(x) 0. 5 μ x 109

Empirical Rules What can we say about the distribution of values around the mean? There are some general rules: f(x) μ ± 1σ encloses about 68% of x’s σ μ 1σ σ μ μ+1σ x 68. 26% 110

The Empirical Rule • • μ ± 2σ covers about 95% of x’s μ ± 3σ covers about 99. 7% of x’s 2σ 3σ 2σ μ 95. 44% x 3σ μ x 99. 72% 111

Importance of the Rule • If a value is about 2 or more standard deviations away from the mean in a normal distribution, then it is far from the mean • The chance that a value that far or farther away from the mean is highly unlikely, given that particular mean and standard deviation 112

The Standard Normal Distribution • • • Also known as the “z” distribution Mean is defined to be 0 Standard Deviation is 1 f(z) 1 0 z Values above the mean have positive z-values, values below the mean have negative z-values 113

The Standard Normal • Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normal distribution (z) • Need to transform x units into z units 114

Translation to the Standard Normal Distribution • Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and dividing by its standard deviation: 115

The Normal Distribution… • The normal distribution is the most important of all probability distributions. The probability density function of a normal random variable is given by: • It looks like this: • Bell shaped, • Symmetrical around the mean … 116

The Normal Distribution… • Important things to note: The normal distribution is fully defined by two parameters: its standard deviation and mean The normal distribution is bell shaped and symmetrical about the mean Unlike the range of the uniform distribution (a ≤ x ≤ b) Normal distributions range from minus infinity to plus infinity 117

Standard Normal Distribution… • A normal distribution whose mean is zero and standard deviation is one is called the standard normal distribution. 0 1 1 • As we shall see shortly, any normal distribution can be converted to a standard normal distribution with simple algebra. This makes calculations much easier. 118

Normal Distribution… • The normal distribution is described by two parameters: • its mean and its standard deviation. Increasing the mean shifts the curve to the right… 119

Normal Distribution… The normal distribution is described by two parameters: its mean and its standard deviation. Increasing the standard deviation “flattens” the curve… 120

Calculating Normal Probabilities… • We can use the following function to convert any normal random variable to a standard normal random variable… 0 Some advice: always draw a picture! 121

Calculating Normal Probabilities… We can use the following function to convert any normal random variable to a standard normal random variable… This shifts the mean of X to zero… 0 122

Calculating Normal Probabilities… We can use the following function to convert any normal random variable to a standard normal random variable… 0 This changes the shape of the curve… 123

Calculating Normal Probabilities… Example: The time required to build a computer is normally distributed with a mean of 50 minutes and a standard deviation of 10 minutes: 0 What is the probability that a computer is assembled in a time between 45 and 60 minutes? Algebraically speaking, what is P(45 < X < 60) ? 124

Calculating Normal Probabilities… P(45 < X < 60) ? …mean of 50 minutes and a standard deviation of 10 minutes… • P(45 < X < 60) ? 0 125

Calculating Normal Probabilities… OK, we’ve converted P(45 < X < 60) for a normal distribution with mean = 50 and standard deviation = 10 to P(–. 5 < Z < 1) [i. e. the standard normal distribution with mean = 0 and standard deviation = 1] so Where do we go from here? ! 126

Calculating Normal Probabilities… • P(–. 5 < Z < 1) looks like this: • The probability is the area • under the curve… • We will add up the • two sections: • P(–. 5 < Z < 0) and • P(0 < Z < 1) 0 –. 5 … 1 127

Calculating Normal Probabilities… We can use Table 3 in Appendix B to look-up probabilities P(0 < Z < z) We can break up P(–. 5 < Z < 1) into: P(–. 5 < Z < 0) + P(0 < Z < 1) The distribution is symmetric around zero, so (multiplying through by minus one and re-arranging the terms) we have: P(–. 5 < Z < 0) = P(. 5 > Z > 0) = P(0 < Z <. 5) Hence: P(–. 5 < Z < 1) = P(0 < Z <. 5) + P(0 < Z < 1) 128

Calculating Normal Probabilities… • How to use Table 3… This table gives probabilities P(0 < Z < z) First column = integer + first decimal Top row = second decimal place P(0 < Z < 0. 5) P(0 < Z < 1) P(–. 5 < Z < 1) =. 1915 +. 3414 =. 5328 129

Calculating Normal Probabilities… • Recap: The time required to build a computer is normally distributed with a mean of 50 minutes and a standard deviation of 10 minutes What is the probability that a computer is assembled in a time between 45 and 60 minutes? P(45 < X < 60) = P(–. 5 < Z < 1) =. 5328 “Just over half the time, 53% or so, a computer will have an assembly time between 45 minutes and 1 hour” 130

Example • If x is distributed normally with mean of 100 and standard deviation of 50, the z value for x = 250 is • This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100. 131

Comparing x and z units μ = 100 σ = 50 100 0 250 3. 0 x z Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z) 132

The Standard Normal Table • The Standard Normal table gives the probability from the mean (zero) up to a desired value for z . 4772 Example: P(0 < z < 2. 00) =. 4772 0 2. 00 z 133

The Standard Normal Table The column gives the value of z to the second decimal point The row shows the value of z to the first decimal point . . . 2. 0 . 4772 P(0 < z < 2. 00)2. 0 =. 4772 The value within the table gives the probability from z = 0 up to the desired z value 134

General Procedure for Finding Probabilities To find P(a < x < b) when x is distributed normally: • Draw the normal curve for the problem in terms of x • Translate x-values to z-values • Use the Standard Normal Table 135

Z Table example • Suppose x is normal with mean 8. 0 and standard deviation 5. 0. Find P(8 < x < 8. 6) Calculate z-values: 8 8. 6 x 0 0. 12 Z P(8 < x < 8. 6) = P(0 < z < 0. 12) 136

Z Table example • Suppose x is normal with mean 8. 0 and standard deviation 5. 0. Find P(8 < x < 8. 6) m=8 =5 8 8. 6 P(8 < x < 8. 6) m=0 =1 x 0 0. 12 z P(0 < z < 0. 12) 137

Solution: Finding P(0 < z < 0. 12) Standard Normal Probability Table (Portion) z . 00 . 01 P(8 < x < 8. 6) = P(0 < z < 0. 12) . 02 . 0478 0. 0. 0000. 0040. 0080 0. 1. 0398. 0438. 0478 0. 2. 0793. 0832. 0871 0. 3. 1179. 1217. 1255 0. 00 Z 0. 12 138

Finding Normal Probabilities • Suppose x is normal with mean 8. 0 and standard deviation 5. 0. • Now Find P(x < 8. 6) Z 8. 0 8. 6 139

Finding Normal Probabilities • Suppose x is normal with mean 8. 0 and standard deviation 5. 0. • Now Find P(x < 8. 6). 5000. 0478 P(x < 8. 6) = P(z < 0. 12) = P(z < 0) + P(0 < z < 0. 12) =. 5 +. 0478 =. 5478 Z 0. 00 0. 12 140

Upper Tail Probabilities • Suppose x is normal with mean 8. 0 and standard deviation 5. 0. • Now Find P(x > 8. 6) Z 8. 0 8. 6 141

Upper Tail Probabilities • Now Find P(x > 8. 6)… P(x > 8. 6) = P(z > 0. 12) = P(z > 0) - P(0 < z < 0. 12) =. 5 -. 0478 =. 4522. 0478 . 5000 Z 0 0. 12 . 50 -. 0478 =. 4522 Z 0 0. 12 142

Lower Tail Probabilities • Suppose x is normal with mean 8. 0 and standard deviation 5. 0. • Now Find P(7. 4 < x < 8) 7. 4 8. 0 Z 143

Lower Tail Probabilities Now Find P(7. 4 < x < 8)… The Normal distribution is symmetric, so we use the same table even if z-values are negative: . 0478 P(7. 4 < x < 8) = P(-0. 12 < z < 0) =. 0478 7. 4 8. 0 Z 144

le T ab t ion ibu i st r al D o rm rd N nda Sta 0. 00 0. 01 0. 02 0. 03 0. 04 0. 05 0. 06 0. 07 0. 08 0. 09 0. 0000 0. 0040 0. 0080 0. 0120 0. 0160 0. 0199 0. 0239 0. 0279 0. 0319 0. 0359 0. 1 0. 0398 0. 0438 0. 0478 0. 0517 0. 0557 0. 0596 0. 0636 0. 0675 0. 0714 0. 0753 0. 2 0. 0793 0. 0832 0. 0871 0. 0910 0. 0948 0. 0987 0. 1026 0. 1064 0. 1103 0. 1141 0. 3 0. 1179 0. 1217 0. 1255 0. 1293 0. 1331 0. 1368 0. 1406 0. 1443 0. 1480 0. 1517 0. 4 0. 1554 0. 1591 0. 1628 0. 1664 0. 1700 0. 1736 0. 1772 0. 1808 0. 1844 0. 1879 0. 5 0. 1915 0. 1950 0. 1985 0. 2019 0. 2054 0. 2088 0. 2123 0. 2157 0. 2190 0. 2224 0. 6 0. 2257 0. 2291 0. 2324 0. 2357 0. 2389 0. 2422 0. 2454 0. 2486 0. 2517 0. 2549 0. 7 0. 2580 0. 2611 0. 2642 0. 2673 0. 2704 0. 2734 0. 2764 0. 2794 0. 2823 0. 2852 0. 8 0. 2881 0. 2910 0. 2939 0. 2967 0. 2995 0. 3023 0. 3051 0. 3078 0. 3106 0. 3133 0. 9 0. 3159 0. 3186 0. 3212 0. 3238 0. 3264 0. 3289 0. 3315 0. 3340 0. 3365 0. 3389 1. 0 0. 3413 0. 3438 0. 3461 0. 3485 0. 3508 0. 3531 0. 3554 0. 3577 0. 3599 0. 3621 1. 1 0. 3643 0. 3665 0. 3686 0. 3708 0. 3729 0. 3749 0. 3770 0. 3790 0. 3810 0. 3830 1. 2 0. 3849 0. 3869 0. 3888 0. 3907 0. 3925 0. 3944 0. 3962 0. 3980 0. 3997 0. 4015 1. 3 0. 4032 0. 4049 0. 4066 0. 4082 0. 4099 0. 4115 0. 4131 0. 4147 0. 4162 0. 4177 1. 4 0. 4192 0. 4207 0. 4222 0. 4236 0. 4251 0. 4265 0. 4279 0. 4292 0. 4306 0. 4319 1. 5 0. 4332 0. 4345 0. 4357 0. 4370 0. 4382 0. 4394 0. 4406 0. 4418 0. 4429 0. 4441 1. 6 0. 4452 0. 4463 0. 4474 0. 4484 0. 4495 0. 4505 0. 4515 0. 4525 0. 4535 0. 4545 1. 7 0. 4554 0. 4564 0. 4573 0. 4582 0. 4591 0. 4599 0. 4608 0. 4616 0. 4625 0. 4633 1. 8 0. 4641 0. 4649 0. 4656 0. 4664 0. 4671 0. 4678 0. 4686 0. 4693 0. 4699 0. 4706 1. 9 0. 4713 0. 4719 0. 4726 0. 4732 0. 4738 0. 4744 0. 4750 0. 4756 0. 4761 0. 4767 2. 0 0. 4772 0. 4778 0. 4783 0. 4788 0. 4793 0. 4798 0. 4803 0. 4808 0. 4812 0. 4817 2. 1 0. 4826 0. 4830 0. 4834 0. 4838 0. 4842 0. 4846 0. 4850 0. 4854 0. 4857 2. 2 0. 4861 0. 4864 0. 4868 0. 4871 0. 4875 0. 4878 0. 4881 0. 4884 0. 4887 0. 4890 2. 3 0. 4896 0. 4898 0. 4901 0. 4904 0. 4906 0. 4909 0. 4911 0. 4913 0. 4916 2. 4 0. 4918 0. 4920 0. 4922 0. 4925 0. 4927 0. 4929 0. 4931 0. 4932 0. 4934 0. 4936 2. 5 0. 4938 0. 4940 0. 4941 0. 4943 0. 4945 0. 4946 0. 4948 0. 4949 0. 4951 0. 4952 2. 6 0. 4953 0. 4955 0. 4956 0. 4957 0. 4959 0. 4960 0. 4961 0. 4962 0. 4963 0. 4964 2. 7 0. 4965 0. 4966 0. 4967 0. 4968 0. 4969 0. 4970 0. 4971 0. 4972 0. 4973 0. 4974 2. 8 0. 4974 0. 4975 0. 4976 0. 4977 0. 4978 0. 4979 0. 4980 0. 4981 2. 9 0. 4981 0. 4982 0. 4983 0. 4984 0. 4985 0. 4986 3. 0 0. 4987 0. 4988 0. 4989 0. 4990

Using the Normal Table • What is P(Z > 1. 6) ? P(0 < Z < 1. 6) =. 4452 z 0 1. 6 P(Z > 1. 6) =. 5 – P(0 < Z < 1. 6) =. 5 –. 4452 =. 0548 146

Using the Normal Table • What is P(Z < -2. 23) ? P(Z < -2. 23) P(0 < Z < 2. 23) P(Z > 2. 23) z -2. 23 0 2. 23 P(Z < -2. 23) = P(Z > 2. 23) =. 5 – P(0 < Z < 2. 23) =. 0129 147

Using the Normal Table • What is P(Z < 1. 52) ? P(0 < Z < 1. 52) P(Z < 0) =. 5 z 0 1. 52 P(Z < 1. 52) =. 5 + P(0 < Z < 1. 52) =. 5 +. 4357 =. 9357 148

Using the Normal Table • What is P(0. 9 < Z < 1. 9) ? P(0 < Z < 0. 9) P(0. 9 < Z < 1. 9) z 0 0. 9 1. 9 P(0. 9 < Z < 1. 9) = P(0 < Z < 1. 9) – P(0 < Z < 0. 9) =. 4713 –. 3159 =. 1554 149

Example 8. 2 • The return on investment is normally distributed with a mean of 10% and a standard deviation of 5%. What is the probability of losing money? • We want to determine P(X < 0). Thus, 150

Example 8. 2 If the standard deviation is 10 the probability of losing money is P(X < 0) Thus, increasing the standard deviation increases the probability of losing money, which is why the standard deviation (or the variance) is a measure of risk. 151

Finding Values of Z… Often we’re asked to find some value of Z for a given probability, i. e. given an area (A) under the curve, what is the corresponding value of z (z. A) on the horizontal axis that gives us this area? That is: P(Z > z. A) = A 152

Finding Values of Z… What value of z corresponds to an area under the curve of 2. 5%? That is, what is z. 025 ? Area =. 50 Area =. 025 Area =. 50–. 025 =. 4750 If you do a “reverse look-up” for. 4750, you will get the corresponding z. A = 1. 96. Since P(z > 1. 96) =. 025, we say: z. 025 = 1. 96 153

Finding Values of Z… Other Z values are Z. 05 = 1. 645 Z. 01 = 2. 33 154

Using the values of Z Because z. 025 = 1. 96 and - z. 025= -1. 96, it follows that we can state P(-1. 96 < Z < 1. 96) =. 95 Similarly P(-1. 645 < Z < 1. 645) =. 90 155

Normal Distribution as an approximation of the Binomial Distribution

Case Studies 2. 3 – 2. 5 What is the probability of obtaining a score greater than 700 on a GMAT test that has a mean of 494 and a standard deviation of 100? Assume GMAT scores are normally distributed. For the same GMAT examination, what is the probability of randomly drawing a score that is 550 or less? What is the probability of randomly obtaining a score between 350 & 450 on the GMAT exam? 157

Case Study 2. 6 An IQ test was given to students of a high-school & it was found that the IQs were approximately normally distributed with mean 90 and variance 100. Estimate the proportion of students with IQ 1. Greater than 110 2. Between 70 & 110 3. Between 100 & 120 4. Not more than 110 158

Case Studies 2. 7 Tool workers are subject to work-related injuries. One disorder, caused by strains to the hands & wrists is called carpal tunnel syndrome. It strikes as many as 23, 000 workers per year. The US Labour Department estimates that the average cost of this disorder to employers & insurers is approximately $30, 000 per injured worker. Suppose these costs are normally distributed, with a standard deviation of $9, 000. 1. What proportion of these costs are between $15, 000 and $45, 000? 2. What proportion of these costs are greater than $50, 000? 3. Suppose that the standard deviation is unknown, but 90. 82% of the costs are more than $7, 000. What would be the value of the standard deviation? 4. Suppose that the mean is unknown, but the standard deviation is still $9, 000. How much would the average cost be if 79. 95% of the costs were less than $33, 000? 159