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1 Econ 240A Power Three. 2 Summary: Week One Descriptive Statistics –measures of central tendency –measures of dispersion Distributions of observation.

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Presentation on theme: "1 Econ 240A Power Three. 2 Summary: Week One Descriptive Statistics –measures of central tendency –measures of dispersion Distributions of observation."— Presentation transcript:

1 1 Econ 240A Power Three

2 2 Summary: Week One Descriptive Statistics –measures of central tendency –measures of dispersion Distributions of observation values –Histograms: frequency(number) Vs. value Exploratory data Analysis –stem and leaf diagram –box and whiskers diagram

3 3 Probability The Gambler Kenny Rogers 20 Great Years

4 4 Outline Why study probability? Random Experiments and Elementary Outcomes Notion of a fair game Properties of probabilities Combining elementary outcomes into events probability statements probability trees

5 5 Outline continued conditional probability independence of two events

6 6 Why study probability? Understand the concept behind a random sample and why sampling is important –independence of two or more events understand a Bernoulli event –example; flipping a coin understand an experiment or a sequence of independent Bernoulli trials

7 7 Cont. Understand the derivation of the binomial distribution, i.e. the distribution of the number of successes, k, in n Bernoulli trials understand the normal distribution as a continuous approximation to the discrete binomial understand the likelihood function, i.e. the probability of a random sample of observations

8 8 Concepts Random experiments Elementary outcomes example: flipping a coin is a random experiment –the elementary outcomes are heads, tails example: throwing a die is a random experiment –the elementary outcomes are one, two, three, four, five, six

9 9 Concept A fair game example: the probability of heads, p(h), equals the probability of tails, p(t): p(h) = p(t) =1/2 example: the probability of any face of the die is the same, p(one) = p(two) = p(three) = p(four) =p(five) = p(six) = 1/6

10 10 Uncertainty in Life Demography –Death rates –Marriage –divorce

11 11 Uncertainty in Life: US (CDC)

12 12

13 13 Probability of First Marriage by Age, Women: US (CDC)

14 14 Cohabitation: The Path to Marriage?: US(CDC)

15 15 Race/ethnicity Affects Duration of First Marriage

16 Properties of probabilities Nonnegative –example: p(h) probabilities of elementary events sum to one –example p(h) + p(t) = 1

17 17 Flipping a coin twice: 4 elementary outcomes heads tails heads tails heads tails h, h h, t t, h t, t

18 18 Throwing Two Dice, 36 elementary outcomes

19 19 Larry Gonick and Woollcott Smith, The Cartoon Guide to Statistics

20 20 Combining Elementary Outcomes Into Events Example: throw two dice: event is white die equals one example: throw two dice and red die equals one example: throw two dice and the sum is three

21 21 Event: white die equals one is the bottom row Event: red die equals one is the right hand column

22 22 Event: 2 dice sum to three is lower diagonal

23 Operations on events The event A and the event B both occur: Either the event A or the event B occurs or both do: The event A does not occur, i.e.not A:

24 Probability statements Probability of either event A or event B –if the events are mutually exclusive, then probability of event B

25 25 Probability of a white one or a red one: p(W1) + p(R1) double counts

26 Two dice are thrown: probability of the white die showing one and the red die showing one

27 27 Probability 2 dice add to 6 or add to 3 are mutually exclusive events Probability of not rolling snake eyes is easier to calculate as one minus the probability of rolling snake eyes

28 Problem What is the probability of rolling at least one six in two rolls of a single die? –At least one six is one or two sixes –easier to calculate the probability of rolling zero sixes: (5/36 + 5/36 + 5/36 + 5/36 + 5/36) = 25/36 –and then calculate the probability of rolling at least one six: 1- 25/36 = 11/36

29 29 1 2 3 4 5 6 1 2 3 4 5 6 Probability tree 2 rolls of a die: 36 elementary outcomes, of which 11 involve one or more sixes

30 30 Conditional Probability Example: in rolling two dice, what is the probability of getting a red one given that you rolled a white one? –P(R1/W1) ?

31 31 In rolling two dice, what is the probability of getting a red one given that you rolled a white one?

32 Conditional Probability Example: in rolling two dice, what is the probability of getting a red one given that you rolled a white one? –P(R1/W1) ?

33 Independence of two events p(A/B) = p(A) –i.e. if event A is not conditional on event B –then

34 34 Concept Bernoulli Trial –two outcomes, e.g. success or failure –successive independent trials –probability of success is the same in each trial Example: flipping a coin multiple times

35 Problem 6.28 Distribution of a retail store purchases classified by amount and method of payment

36 36 Problem (Cont.) A. What proportion of purchases was paid by debit card? B. Find the probability a credit card purchase was over $100 C. Determine the proportion of purchases made by credit card or debit card

37 Problem 6.28

38 38 Problem (Cont.) A. What proportion of purchases was paid by debit card? 0.36 B. Find the probability a credit card purchase was over $100 C. Determine the proportion of purchases made by credit card or debit card

39 39 Problem (Cont.) A. What proportion of purchases was paid by debit card? B. Find the probability a credit card purchase was over $100 p(>$100/credit card) = 0.23/0.47 = 0.489 C. Determine the proportion of purchases made by credit card or debit card

40 40 Problem (Cont.) A. What proportion of purchases was paid by debit card? B. Find the probability a credit card purchase was over $100 C. Determine the proportion of purchases made by credit card or debit card –note: credit card and debit card purchases are mutually exclusive –p(credit or debit) = p(credit) + p (debit) = 0.47 + 0.36

41 41 Problem 6.61 A survey of middle aged men reveals that 28% of them are balding at the crown of their head. Moreover, it is known that such men have an 18% probability of suffering a heart attack in the next ten years. Men who are not balding in this way have an 11% probability of a heart attack. Find the probability that a middle aged man will suffer a heart attack in the next ten years.

42 42 Middle Aged men Bald P (Bald and MA) = 0.28 Not Bald

43 43 Middle Aged men Bald P (Bald and MA) = 0.28 Not Bald P(HA/Bald and MA) = 0.18 P(HA/Not Bald and MA) = 0.11

44 44 Probability of a heart attack in the next ten years P(HA) = P(HA and Bald and MA) + P(HA and Not Bald and MA) P(HA) = P(HA/Bald and MA)*P(BALD and MA) + P(HA/Not BALD and MA)* P(Not Bald and MA) P(HA) = 0.18*0.28 + 0.11*0.72 = 0.054 +.0792 = 0.1296

45 45 Summary: Probability Rules Addition: P(A or B) = P(A) + P(B) – P(A and B) –If A and B are mutually exclusive, P(A and B) = 0 Subtraction: P(A) = 1 – P( not E) Multiplication: P(A and B) = P(A/B) P(B) –If A and B are independent, then P(A/B) = P(B)


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