Probability Review. Pages Lecture 1: Probability Theory 1 Main Ideas 2 Trees 4 Additional Examples 5

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1 Probability Review Pages Lecture 1: Probability Theory 1 Main Ideas 2 Trees 4 Additional Examples 5 Lecture 2: Random Variables (RVs) & Distributions 8 Definitions, Notation, & Ideas 8 Additional Examples 11 Summary & Addendum to Lectures Lecture 3: Binomial Distribution 16 Bernoulli Distribution: Mean and Variance 16 Binomial Mean, Variance and Probabilities 17 Combinations and Trees 18 More Examples 20 Binomial Tables 21 Lecture 4: Normal Distribution 24 The Normal Table 25 Finding Probabilities for Standard Normal RVs 27 Finding Probabilities for General Normal RVs 28 Expectations and Variances of Sums of RVs Variables 29 These notes summarize the ideas and provide examples. A change in font color (usually to the red Comic Sans font illustrated in this footnote) indicates exercise solutions that the reader is encouraged to work through. In some cases, these solutions are in text boxes. Occasional references are made to the following textbook: Bruce Bowerman, Richard O Connell, Emily Murphree, & J.B. Orris (2015), Essentials of Business Statistics, 5 th Edition, McGraw-Hill/Irwin. Bruce Cooil, 2015

2 Lecture 1: Probability Concepts and Notation Reference: Bowerman, O Connell, Orris & Murphree (5 th Ed). Ch. 4: , especially: Outline: A. Definitions and Ideas 1. Sample Space, Events and Complements 2. Probability of Intersection 3. Probability of Union B. Conditional Probability 1. Definition 2. Multiplication Rule 3. TREES 4. Independence (vs. Mutually Exclusive Events) Imagine an industry where there have been 200 new products in the last few years. We categorize them two ways: according to how successful they eventually were, and according to whether or not a preliminary evaluation indicated success. Define the outcomes or events as follows: G = event that my preliminary evaluation is "good"; S = event that product is huge success; M = event that product is medium success (will not be continued); F = event that product is a failure (specific examples: software or hardware products, books, movies). S M F TOTAL G GG TOTAL These 200 products are a sample space. Sample space refers to all possible outcomes of a statistical experiment. Here, the statistical experiment is the choice of one product. Events Any subset of the sample space is an event: S, M, F, G, and G are all events. Each occurs with a certain probability, e.g., P[G] = 150/200 = GG is the complement of G,i.e., the event G does not occur, P[ G ] = 1 - P[G] =.25 (or 50/200). Page 1 of 29

3 Intersections The intersection of two events refers to the event where they both occur. For example, G1S refers to the event that the preliminary evaluation is good and the product went on to be a huge success: note that both G and S occur for 60 products, so P[G 1 S] / P[ G and S ]=60/200 = Similarly, P[Ḡ 1 M] = 20/200 = 0.1; P[S 1 F] = 0/200 = 0. Unions The union of two events refers to the event of either or both occurring. So for example, P[GcS] / P[ G or S ]= 160/200 = 0.80, P[McG ] = 110/200 = One can use Poincaré's formula (also known as the Addition Rule, p.166) to calculate the probability of a union: P(ScG) = P(S) + P(G) - P(S1G) = 70/ /200-60/200 (because 60/200 is included in each of the first two probs.) = 160/200 = Also for mutually exclusive events P[ScM] = P[S] + P[M] because S and M are mutually exclusive so that P[S1M]=0. Note: P[McG] = ( )/200 = 0.85 or 85%; P[Fc M] = ( )/200= 0.65 or 65%. General Comments McF is the same as FcM. M1F is the same as F1M (this is the null or empty set, sometimes written `). Page 2 of 29

4 S M F TOTAL G PP[ GG MM] = 2222/ GG TOTAL S M F TOTAL G PP [SS FF] = 00/ GG TOTAL S M F TOTAL G PP [MM GG] = ( )/ GG TOTAL S M F TOTAL G PP [FF MM] = ( )/ GG TOTAL Page 3 of 29

5 Probabilities Corresponding to Table on Page 1 S M F TOTAL G GG TOTAL (I've simply divided all of numbers in the original table by the sample size of 200.) The Same Information Organized in a Tree Step 1 Step 2 OUTCOME PROBABILITY S (60/150=.4) G 1 S.3 G(.75) M (.4) G 1 M.3 F (.2) G 1 F.15 S (.05/.25=0.2 ) G G 1 S.25(.2)=.05 GG(.25 ) M (.1/.25 = 0.4 ) G G 1 M.25(0.4)=.1 F ( 0.4 ) G 1 F.25(0.4)=.1 Conditional Probability The parenthetical probabilities next to S, M, and F in the diagram above are conditional probabilities. The conditional probability of S given G is written as P[S G] and defined as: P[S G] = P[S1G]/P[G]. Thus, P[S G] =.3/.75 = 0.4. Note that this is different from P[G S], since P[G S] = P[G1S]/P[S] =.3/.35 = Page 4 of 29

6 Multiplication Rule The multiplication rule allows us to find the probability of an intersection from a conditional probability and the unconditional probability of the event on which we are conditioning. The rule is simply a consequence of the definition of conditional probability. This rule is used to derive the event probabilities in the last column of a probability tree. In general for any two events A and B: P[A1B]= P[A B] P[B]. (Also: P[A1B] = P[B A] P[A].) This works because we're recovering the numerator of the conditional probability by multiplying by its denominator! Independence Note that the probability of M is the same whether or not G occurs since, P[M G] = 0.4, and P[M] = 0.4. When this happens, we say that M and G are independent. If M and G are independent, it s also true that P[G M] = P[G] (=.75). In general, two events A and B are independent if P[B A] = P[B] (or if P[A B]=P[A]). An equivalent definition: independence of A and B means P(A1B)=P(A)P(B). Main Ideas Notation: definitions of A1B=B1A, AUB=BUA (these operations are commutative!) Rule of Unions (Poincaré's formula): P(AUB) = P(A) + P(B) - P(A1B) Definition of Conditional Probability: (P(A#B) / [P(A1B)/P(B)] Multiplication Rule: P(A1B) = P(A#B) P(B) Independence: A & B are independent if P(A1B) = P(A)P(B) Or, equivalently, if P(A B) = P(A) Or, equivalently, if P(B A) = P(B) Page 5 of 29

7 More Examples (Lecture 1) 1. A manufacturing process falls out of calibration 10% of the time. When it is calibrated, defective nails are produced with a probability of 1%. When it is not calibrated, the probability of producing a defect is 0.1. Let C=event process is calibrated; D=event of producing defect. Outcome Probability D(.01) C 1 D.009 C (0.9) D _ (0.99) C 1 D _.891 C _ (0.1) D(.1) C _ 1 D 0.01 D _ (0.90) C _ 1 D _ 0.09 a. What is the overall probability of getting a defective? P(D) = P (C 1 D) + P(C _ 1 D) = (Using Tree above) =.019 b. What is the probability that I will sample a defective and the process is out of calibration (i.e., what is the probability of the intersection of these two events)? P(C _ 1 D) = 0.01 (based on tree above) c. I randomly select 1 nail that has just been produced. It's defective! What's the probability the process needs calibration? P(C _ D) = P(C _ 1 D) /P(D) = 0.01/0.019 = I have two career opportunities. Opportunity A will occur with probability 0.4. Opportunity B will occur with probability 0.3. Assume the opportunities are independent. a. What is the probability I will have both opportunities? P( A 1 B) = P(A) * P(B) = 0.4(0.3) = 0.12 b. What is the probability that I will have at least 1 opportunity? P( A c B) = P(A) + P(B) - P(A 1 B) = = 0.58 Page 6 of 29

8 3. An individual has deadly virus XX with probability of A patient who has the virus will always test positive on an inexpensive screening test. A patient who does not have the virus still has a 0.05 probability of testing positive on the screen. If I test positive, what is the probability that I have the virus? Let V= event that I have the virus; T= event that I test positive. V(.001) _ V(.999) Outcome Probability T (1) V 1 T.001(1) =.001 T(0) V 1 T.001(0) = 0 _ T (.05) V T(.95) V 1 T.999(.05) = _ 1 T.999(.95) = The question asks for P(V T) = P (V 1 T)/P(T) By definition of conditional probability = 0.001/( ) Using the Tree = or about 2% ************************************************************** For diagnostic tests, like the one above, the probabilities in the first and last branches of the tree are typically used to evaluate the diagnostic efficacy of the test: C 1 st branch: P[T V] (which is 1" in this illustration) is referred to as the sensitivity of the test; C last branch: P[T V](which is 0.95" in this illustration) is referred to as the specificity of the test. Notice that in the product example at the beginning of lecture 1, the preliminary evaluation of the product could be thought of as a diagnostic test. How effective is it for predicting S (event that product will be a huge success)? C Sensitivity: P[G S] = (see bottom of p.4); C Specificity: P[G S]= 0.2/0.65 = 0.31 (using tree on p. 4). Page 7 of 29

9 Lecture 2: Random Variables and Probability Distributions Reference: Ch. 5: , App. 5.3; Outline C Random Variable(RV): What Is It and Why Use It? C Probability Distributions of RV C Expected Value (or Mean) of RV C Variance (and Standard Deviation) of RV C Mean and Variance of Linear Function of RV (If Y = a + bx, then μ Y = a + bμ X and σ Y = bσ X ) Random Variables A random variable (rv) is simply a way of representing numerical outcomes that occur with different probabilities. A discrete rv can assume only a countable number of values. A continuous rv assumes an uncountably infinite number of values corresponding to the points on an interval (or more than one interval). EXAMPLE: I'm a consultant whose income depends on the number of projects that I finish. Let X represent the number of projects that I complete in a month: X=# of projects completed in month. X is an example of a discrete rv. Assume that I undertake 2 new projects in a month, there is a probability of 0.6 that I will complete each one, and the projects are independent. A probability tree provides one way to analyze this situation. Finish this one! Let S successful completion of project i i Step 1 Step 2 (1 st Project) (2 nd Project) Outcomes Probability (.6) 2 successes.36 S 2 S 1 (.6) (.4) success, failure.24 S 2 S 2 (.6) failure, success.24 S 1 (.4) S 2 (.4) 2 failures.16 Page 8 of 29

10 The PROBABILITY DISTRIBUTION of X can be summarized with the function P(x). Lecture 2 Pg. 2 x P(x) xp(x) (x-μ) 2 (x-μ) 2 P(x) x 2 P(x) (0-1.2) (0.16) 0 = (1-1.2) (0.48) 0.48 = (2-1.2) (0.36) 1.44 =0.64 TOTAL: This is the mean of X; common notation: μ X Or E(X) This is the variance of X; 2 common notation: X. F X = (0.48) ½ = 0.69 This is the mean of the random variable X 2 or E(X 2 ). Page 9 of 29

11 Lecture 2 Pg. 3 Expected Value (Or Mean) of X: μ / E(X) / 3 xp(x) The expected value of a random variable is simply the weighted average of its possible values, where each value is weighted by its probability. In English it is often referred to as: "the population mean (or average)," "the distribution mean," or "the expected value." Variance of X: σ 2 / E[(X-μ) 2 ] / 3 (x-μ) 2 P(x) The variance may also be calculated as σ 2 = 3 x 2 P(x) - μ 2. Note the first term in this last expression is just the expectation or average value of X 2. Because the variance is in squared units, people often use its square root σ, the standard deviation, as an alternative measure of how "spread-out" the distribution is. Standard Deviation of X: σ = [Variance of X] ½ Another Example: Suppose Y is my total income in thousands per month: Y = X where X is the number of projects completed. Use E(X) (μ X ), and Var(X) (σ 2 X ) to find: E(Y), Var(Y) and σ Y. μ Y = μ X = 5+(10*1.2)= 17 thousand dollars σ 2 Y = 102 σ 2 X = 100(0.48) =48 (thousand dollars)2 σ Y = 10 σ X = 10 (0.69) = 6.9 thousand dollars Page 10 of 29

12 Lecture 2 Pg. 4 Suppose Y = a + bx, where "a" and "b" are just constants (not random variables). We can find the mean of Y as the same linear function of the mean of X, [Mean of Y] = a + b[mean of X]. What about the variance? The variance of Y is [Variance of Y] = b 2 [Variance of X], and if we take square roots of both sides of last equation: [Standard Deviation of Y] = b[standard Deviation of X]. Example: The distribution of the number of UHD TVs sold at an electronics store is shown below. 0.5 Probability Distribution of UHD TVs Sold Per Week X= UHD TVs Sold Per Week Probability Probability UHD TVs Sold Per Week Find the mean and standard deviation of this distribution. E(X) = 0(.03) + 1(0.2) + 2(.5) + 3(.2) + 4(.05) + 5(.02) = 2.1; Var(X) = (0-2.1) 2 (.03) + (1-2.1) 2 (.2) + (2-2.1) 2 (.5) + (3-2.1) 2 (.2) + (4-2.1) 2 (.05) + (5-2.1) 2 (.02) = 0.89 ==> σ X = Page 11 of 29

13 Lecture 2 Pg. 5 Consider the distribution of Y = 5 + 2X. Probability Scatterplot of Probability vs C3 X Y Probability Y = 5 + 2X Note that we can simply find the mean and standard of Y directly, using the mean and standard deviation of X: E(Y) = 5 + 2*E(X) = 5 + 2(2.1) = 9.2 ; σ Y = 2* σ X = 2(0.94) = 1.9. Another Example Find the mean, variance and standard deviation of the rv X = profit on 10 thousand dollar investment (after 1 year) in thousands of dollars. Possible Values of X: Probability: E(X) = -10(0.3) + 0(0.4) + 20(0.3) = 3 thousand dollars; Var(X) = E(X 2 )-(E(X)) 2 = [100(0.3) + 0(0.4) +400(0.3)] -9 =141. Thus: σ X = %141 = thousand dollars. Also find the mean and standard deviation of the broker s income, I, where I (in thousands of dollars) is defined as: I = X. E(I) = (3) = 3.3 thousand dollars ; σ I = 0.1(σ X ) = 0.1(11.87)= thousand dollars. Page 12 of 29

14 Summary & Addendum to Lectures 1 & 2 Big Picture So Far Lecture 1 Elements of Probability Theory: Definitions & Notation A Central Idea: Conditional Probability (It s an important concept by itself, but it also provides a way of explaining independence) Probability Trees Provide a way of summarizing (and reviewing) all of the major ideas and notation Lecture 2 Great way of solving problems that require the calculation of a conditional probability Definitions of RVs Probability Distributions Means and Variances Page 13 of 29

15 At the End of Lecture 2 We Looked at Applications Where It s Useful to Remember That: If Y = a + bx Then: E(Y) = a + be(x) (OR: µ = a + bµ ) Y X σ = bσ Y X Remembering this can save a lot of labor. What Happens With Sums Of RVs? The Mean (Or Expectation) Of A Sum Of RVs Is The Sum Of The Means Of Those RVs If The RVs Are Independent, Then The Variance Of A Sum Of RVs Is The Sum Of The Variances Of Those RVs Page 14 of 29

16 Example: Define independent RVs: N 1 = profit from investment 1 N 2 = profit from invesment 2 T= total profit = N 1 + N 2 Assume: E(N 1 ) = 2, E(N 2 ) = 3 (millions $) σ 2 N 1.5, = σ N 2 = Find the Expectation and Standard Deviation of Total Profit: E(T) = = 5 million Var(T) = , So σ = (4) 1/2 = 2 (This Idea Applies to Any Type of RV, So That Any Specific Application Understates Its Importance.) Imagine How Much Work This Would Be If I Had to Go Back and Calculate the Mean and Variance Directly from the Distribution of T! Page 15 of 29

17 Lecture 3: The Binomial Distribution Reference: Ch. 4: 4.6; Ch. 5: 5.3 Outline:! Bernoulli Distribution μ = p, σ 2 = p(1-p)! Binomial Distribution μ = np, σ 2 = np(1-p) We Consider Means, Variances and Calculating Probabilities (Directly or with Tables) Bernoulli Distribution Examples: (1) X = # of times I get a 6" on one roll of a die. P[X=1] = P[getting a 6" on a roll of a die] = 1/6 P[X=0] = 5/6. E[X] = 0(5/6) + 1(1/6) = 1/6. Var(X) = E(X 2 )- μ 2 = 'x 2 P(x) - μ 2 =[0 2 (5/6) (1/6)] -(1/6) 2 = 1/6-(1/6) 2 OR 1/6(1-1/6). (2) 20% of all customers prefer brand A to brand B. I interview just 1 customer. Consider the distribution of rv X, where: X = # (of 1) who prefer A to B. P[X=1] = 0.2, P[X=0]= 0.8, E(X) = 0.2. Var(X)= 0.2(0.8) = Examples Above Are Illustrations of Bernoulli Random Variables A Bernoulli random variable is simply a variable, X, that indicates whether or not something happens. In general, X might be thought of as X = # of successes in 1 trial (or attempt). Suppose success occurs with probability p, then: P[X=1]= prob. of 1 success = p; P[X=0] = 1-p AND: E(X)= p (the prob. of success), Var(X) = p - p 2 = p(1-p) = [prob. of success]*[prob. of failure]. Bottom Line: The mean of a Bernoulli random variable is the probability of success; its variance is the product of the probability of success and the probability of failure. Page 16 of 29

18 Binomial Distribution Examples of Binomial Random Variables (rvs): (1) Y = Number of times "6" comes up in 3 rolls of die. Possible Values: 0, 1, 2, 3. (2) Suppose there are in 4 independent business deals, & each will result in a profit with probability of 0.7. Consider the distribution of Y = # of profitable deals (of 4). Possible Values: 0, 1, 2, 3, 4. (3) Assume 20% of all customers (an infinite population) prefer brand A to brand B. Only 10 customers are interviewed. Consider the distribution of Y = # of customers of 10 who prefer brand A to brand B. Possible Values: 0, 1, 2, 3, 4,..., 10. A General Binomial Random Variable As illustrated above, many applications require that we work with a rv that represents a slight generalization the Bernoulli rv, Y = # of successes in n trials, where the trials are assumed to be independent & each results in a success with the same probability p. Y is referred to as a binomial rv (& the sequence of trials is sometimes referred to as a Bernoulli process). Note that a binomial rv is really just the sum of n Bernoulli rvs (where n = number of trials). Consequently, the mean and variance of Y are just n times the mean and variance of the corresponding Bernoulli rv: E[Y] = n C p ; Var[Y] = n C p(1-p) (because p and p(1-p) are the Bernoullli mean and variance). Probabilities can be found in Table A.1, App. A (pp ) or probabilities can be calculated directly using the following formula: P[Y=y] = prob. of y successes in n trials = [# of sequences of n trials that have exactly y successes] * [probability of each such sequence] n! y!( n p y n y y n y = = (also written ). y)! ( 1 p) n y p ( 1 p ) Page 17 of 29

19 Regarding Notation n y n! is referred to as n choose y and it represents y!( n y)! the number of ways of selecting y items from a larger group of n items (or as the number of ways y successes can occur in n trials). Each term in this formula is a factorial and represents the product of all positive integers up to and including that number, e.g., n! = n(n-1)(n-2)...(3)(2)(1). Examples: 5 factorial = 5! = 5(4)(3)(2)(1) = ! 3 choose 2 = 2!( 3 2)! = [3(2)(1)]/[1 (2)(1)] = 3. Use a tree to verify that this really counts the number of ways that exactly 2 successes can occur in 3 trials. Trial 1 Trial 2 Trial 3 Outcomes S SSS S S F SSF S SFS F F SFF S FSS S F F FSF S FFS F F FFF Also, 3 choose 3"= 3! 3!( 3 3)! =3!/[0!3!] = 1 (Since 0! = 1 by definition). Page 18 of 29

20 4! 4 choose 2"= = 4!/[2!2!] = 6. 2!( 4 2)! Use a tree to verify this is number of ways that exactly 2 successes can occur in 4 trials. Number of ways of having exactly 2 successes in 4 trials Trial 1 Trial 2 Trail 3 Trial 4 Outcomes S SSSS S F SSSF S S SSFS F F SSFF S S SFSS S F F SFSF S SFFS F F SFFF S FSSS S F FSSF S S FSFS F F FSFF F S FFSS S F F FFSF S FFFS F F FFFF Page 19 of 29

21 Sometimes the trees are much too complicated. For example, suppose I want to count the number of ways 3 successes can occur in 10 trials. By far the easiest way is to mechanically use the formula: 10! 3!(10 3)! 10*9*8*7! 3! 7! 10*9* ! 10 choose 3"= =. Notice that in general: n choose y is the same as n choose (n-y). For example, 10 choose 3" is equal to 10 choose 7." Arithmetically, this is obviously true, but logically it also must be true: when I choose 3 from 10, I am simultaneously not choosing the other 7, so the number of subsets of size 3 must be equal to the number of subsets of size 7. Applications to the Binomial Example 2 on Page 2: Example 2: Write down the probability distribution of Y = # of profitable deals (of 4), p = 0.7. This has a Binomial distribution (n=4, p=0.7). E(Y)= 4(0.7) = 2.8 ; Var(Y) = 4(0.7)(0.3) = 0.84; σ Y = (0.84) ½ = Probability of exactly 2 profitable deals: P[Y = 2] =! 2! ( 0. 7 ) ( 0. 3 ) Probability of at least 2 profitable deals: P[Y > 2] = P(2)+P(3)+P(4)= 1-[P(0)+P(1)] ;! ( 0. 7) 0! 4! ( 0. 3) ( 0. 3) P(0)= ; (Also see Table.) P(1)= 4! 1 ( 0. 7 ) ( 0. 3 ) ! 3! So, P[Y > 2] = 1-( ) = (Or UseTable.) Probability of at most 2 profitable deals: P[Y < 2] = P(0)+ P(1) + P(2) = = Binomial Tables are provided on the next pages. These are in Bowerman, et al., but are originally from Hildebrand, D. & Ott, L.(1991) Statistical Thinking for Managers, Boston: PWS-Kent. Page 20 of 29

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24 Last Example! Define r.v. W as: W = # of business failures (of 8 new ventures). W is binomial with n=8, p= 0.4. Find the following probabilities: 8! (0.4) 6!(8 6)! P[W=6] = = P[W >6] = P(7) + P(8) 6 (0.6) 2 = = P[ W # 6] = = What is the most probable outcome? Answer: W=3. 8! ( 0. 4) 3!( 8 3)! P[W=3] = = Find the mean and variance of W: 3 ( 0. 6) 5 E[W] = np = 8*0.4 = 3.2; Var[W] = np(1-p)= 3.2*0.6 = Page 23 of 29

25 Lecture 4: Normal Distribution Reference: Ch. 6: (especially 6.3); Appendix 6.3 Outline:! Description of Normal & Standard Normal Distributions! Standardizing to Find Probabilities! Sums of Normal RVs Description The normal distribution is a continuous, bell-shaped distribution (e.g., see the figures in section 6.3 or at By standardizing any normal rv, it is possible to calculate the probabilities with respect to a standard normal distribution (i.e., a normal distribution with mean of 0 and standard deviation of 1). The tables on the next two pages provides a tabulation of standard normal cumulative probabilities. This table is also in section 6.3 of the text, inside the back cover, and in Appendix A of the text (Table A.3). What is the probability that a standard normal rv, Z, will lie within 1 standard deviation of 0 (its mean)? P[-1 #Z # 1] = [Table Value at 1]- [Table value at -1] = = Note that there is no probability content at individual points so that, P[-1 #Z # 1] = P[-1 < Z < 1] (i.e., whether or not we include the values at the ends of this interval, the probability remains unchanged). Based on the normal distribution, here are general (and somewhat dangerous) empirical guides that are often used when people think of any symmetric and continuous distribution:! approximately 68% of observations fall within 1 standard deviation of the mean! approximately 95% fall within 2 standard deviations! nearly all fall within 3 standard deviations. Page 24 of 29

26 Lecture 4 Pg. 2 Standard Normal Cumulative Probability Table 0 Cumulative probabilities for NEGATIVE z-values are shown in the following table: z Page 25 of 29

27 Lecture 4 Pg. 3 Cumulative probabilities for POSITIVE z-values are in the following table: z From Page 26 of 29

28 What is the probability that a standard normal rv. will fall between and 2.22? P[-1.25 #Z #2.22]= Lecture 4 Pg. 4 [Table Value at 2.22] - [Table value at -1.25] = = What is the probability it will be greater than -2.51? P[Z > -2.51] = 1 - [Table Value at -2.51] = = Less than 1.55? P[Z < 1.55]= [Table value at 1.55] = Page 27 of 29

29 Lecture 4 Pg. 5 EXAMPLE Suppose X = profit (or loss) from an investment (in thousands of dollars). Assume that we know from previous experience with this type of investment that X is normal (μ = 5,σ = 2). What is P[X > 10.5]? =P[ Z > (10.5-5)/2 ] = P[Z > 2.75) = 1 - [Table Value at 2.75] = = What is P[X >2.44]? = P[ Z > (2.44-5)/2 ] = P[Z > -1.28] =1- [Table Value at -1.28] = = What is the probability that I will lose money? (i.e., P[X<0]?) = P[ Z < (0-5)/2 ] = P[Z < -2.5] =[Table Value at -2.5] = What is P[3 < X < 7]? = P[ {(3-5)/2} < Z < {(7-5)/2} ] = P[-1<Z<1] = [Table Value at 1]- [Table value at -1] = = Page 28 of 29

30 Lecture 4 Pg. 6 Recall Two Useful Principles from Lecture 2 1) The Expectation of a Sum of RVs = Sum of the Expectations. 2) If the RVs are also independent, The Variance of the Sum of RVs = Sum of the Variances. Normal Rvs Also Have the Following Remarkable Property 3) The RV Created By Summing Independent Normal RVs Also Has a Normal Distribution!!!! Application: Assume the following RVs are in units of $1,000: X 1 = profit (or loss) from investment 1, X 1 is Normal(μ = 5, σ 2 = 4); X 2 = profit (or loss) from investment 2, X 2 is Normal(μ = 10, σ 2 = 5). (Also assume these RVs are independent.) Consider: T = total profit from both investments = X 1 + X 2 What is the distribution of T? (Be sure to specify its mean and variance.) T is normal with μ = 5+10 = 15 & σ 2 = = 9 ( and σ = 3). Find P[T>10]. = P[Z > (10-15)/3] = P[Z>-1.67] = 1 - [Table Value at -1.67] = Finally, what is the distribution of I = income to a broker = X X 2 (in $1000)? I has a normal distribution with μ = 5 + (0.1)E(X 1 ) + (0.2)E(X 2 ) = 5+ (0.1)5 + (0.2)10 = 7.5; σ 2 = (0.1) 2 Var(X 1 ) + (0.2) 2 Var(X 2 ) = (0.1) 2 4+ (0.2) 2 5 = 0.24 ; Thus: σ = (0.24) ½ = Page 29 of 29