Statistics For Business And Economics 8th Edition Answers

Statistics For Business And Economics 8th Edition Answers – Explain the basic concepts and definitions of probability. Use a Venn diagram or tree diagram to represent simple probabilities. Calculate the general rules of probability. Calculate the conditional probabilities. Determine whether the events are statistically independent. Use the theorem Copyright © 2013 Pearson Education, Inc. Post as P Ch. 3-2

3 Important terms 3.1 Random experiment – a process that leads to an uncertain outcome Main outcome – the possible outcome of a random experiment Sample space (S) – the sum of all possible outcomes of a random experiment (E) – any subset of the sample space Main results Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-3

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4 Important Terms (cont.) Intersection of Events – If A and B are two events in a sample space S, then the intersection, A ∩ B, is the set of all outcomes in S that A and B S AB B B Copyright © 2013 Pearson Education, Inc . Publishing as Prentice Hall Ch. 3-4

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5 Important Conditions (cont.) A and B are mutually exclusive if they have no common initial result, that is, the set A ∩ B is empty S A B Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-5

6 Important Terms (cont.) Union of Events – If A and B are two events in a sample space S, then the union, A U B, is the sum of all outcomes in S that are associated with A or B S. The region represents A U B A B . Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-6

7 Important conditions (continued) Events E1, E2, …, Ek are collectively complete events if E1 U E2 U U Ek = S, i.e. events completely cover the sample space. The completion of an event A is the sum of all initial outcomes in the sample space that are not associated with A. S Complement is specified. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-7

Example 8 Let the sample space be the set of all possible outcomes of a dice roll: S = [1, 2, 3, 4, 5, 6] Let A be the event “a number is rolled”, let B be the event “The number is rolled” is rolled at least 4 inches, then A = [2, 4, 6] and B = [4, 5, 6] Copyright © 2013 Pearson Education, Inc. Prentice Hall Publishing Ch. Like 3-8

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(continued) S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6] Complements: Intersections: Unions: Copyright © 2013 Pearson Education, Inc . Published as Prentice Hall Ch. 3-9

(continued) S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6] Mutually exclusive: A and B are not mutually exclusive. Results 4 and 6 are collectively complete: A and B are not collectively complete A U B does not include 1 or 3 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-10

3.2 Probability – The probability of an unknown event occurring (always between 0 and 1) 1 1 for each event 0 ≤ P(A) ≤ 1 A.5 Improbability Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-11

There are three ways to estimate the probability of an unknown event: 1. Classical probability 2. Relative frequency probability 3. Subjective probability Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-12

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All results in the sample space are assumed to appear equally. Classical probability of an event A: requires the number of outcomes in the sample space Copyright © 2013 Pearson Education, Inc. 3-13

The number of possible orders is the sum of the possible ways to arrange x objects in x order! Read as “x factor” Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-14

(Continuous) interaction: The number of possible arrangements when x elements are to be selected from a set of n elements and arranged in order [with (n – x) elements remaining]. Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-15

Use the combination formula to determine the number of combinations of n cases where k is taken when n! = n(n-1)(n-2)…(1) 0! = 1 by definition Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-16

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(continued) Combinations: The number of combinations of x objects chosen from n is the number of possible choices that can Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-17

Suppose two letters are to be selected from A, B, C, D and arranged in order. How many variations are possible? The solution is the number of sequences with n = 4 and x = 2. The permutations are AB AC AD BA BC BD CA CB CD DA DB DC Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-18

(Continued) Suppose two letters are to be chosen from A, B, C, D. How many combinations are possible (ie, the order does not matter)? Solution The number of combinations is AB (same as BA) BC (same as CB) AC (same as CA) BD (same as DB) AD (same as DA) CD (same as DC) Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-19

Three Approaches (Continued) 2. Probability Relative Frequency Limit The proportion of times an event A occurs over a large number of trials, n 3. Subjective Probability An individual’s opinion or belief about the likelihood of an event Copyright © 2013 Pearson Education, Inc. . . Publishing as Prentice Hall Ch. 3-20

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1. If A is every event in the sample space, then 2. Let A be an event in S and Oi denote the original outcome. Then (the notation means the sum of all principal scores in A) 3. P(S) = 1 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-21

3.3 Law of Addition: Law of Addition: Probability of two events coming together Published as Prentice Hall Ch. 3-22

23 Probability Table The probabilities and joint probabilities of two events A and B are summarized in this table: B A Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-23

Consider a standard deck of 52 cards of four suits: ♥ ♣ ♦ ♠ Let event A = card is ace Let event B = card of red suit Copyright © 2013 Pearson Education, Inc. 3-24

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(continued) P(Red U Ace) = P(Red) + P(Ace) – P(Red ∩ Ace) = 26/52 + 4//52 = 28/52 Don’t count two together! Suit Type All Red Black Ace 2 2 4 No Ace 24 24 48 All 26 26 52 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-25

Conditional probability is the probability of an event given that another event has occurred: the conditional probability of A that B has occurred the conditional probability of B given that A has occurred Copyright © 2013 Pearson Education, Inc . Publishing as Prentice Hall Ch. 3-26

Among used cars, 70% have air conditioning and 40% have CD players. 20% of vehicles have both. Given that the car has air conditioning, what is the probability that the car has a CD player? That is, we want to find P(CD | AC). Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-27

(Continued) Among used cars, 70% have air conditioning (AC) and 40% have CD players (CD). 20% of vehicles have both. CD No CD Total AC 0.2.5.7 No AC.2.1.3 Total.4.6 1.0 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-28

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(Continued) Regarding air conditioning, we are only looking at the top end (70% of cars). Of these, 20% have a CD player. 20% 70% is 28.57%. CD No CD Total AC 0.2.5.7 No AC.2.1.3 Total.4.6 1.0 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-29

P(red ∩ ace) = P(red) Publishing as Prentice Hall ch. 3-31

Two events are statistically independent if and only if: Events A and B are independent when the probability of one event is not affected by the other if A and B are independent if P(B) > 0 if P(A) > 0 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-32

(continued) For multiple events: E1, E2, , Ek are statistically independent if and only if: Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Ch. 3-33

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Among the available used cars, 70% have air conditioning (air conditioning) and 40% are used.

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