The Bernoulli distribution has only two outcomes, Y, success or failure, with values one or zero. The probability of success is p.

Size: px

Start display at page:

Download "The Bernoulli distribution has only two outcomes, Y, success or failure, with values one or zero. The probability of success is p."

William Chapman
5 years ago
Views:

1 The Bernoulli distribution has onl two outcomes, Y, success or failure, with values one or zero. The probabilit of success is p. The probabilit distribution f() is denoted as B (p), the formula is: f () p ( p) E ( Y ) p ( ) p( p) VAR Y Let Y be or depending on the outcome of tail or head, respectivel. Since the coin is balanced, p is.5. f() A binomial probabilit distribution determines the probabilities for the number of wins (successes) Y in n identical wagers (trials) where these conditions are met: n identical trials outcomes are either a success or a failure the probabilit of success is p

2 all trial outcomes are independent events The probabilit distribution f() is denoted as BIN (n, p), the formula is: f n () p ( ) n p E ( Y ) np ( ) np( p) VAR Y Toss a balanced coin twice, Y is the number of heads. A carton of parts, half are defective, half are good. Sample two parts replacing them after each draw. Y is the number of defective parts.!"#$ %&!"#$ (.5) (.5) (.5) f() f()

3 Toss a balanced coin three times, Y is the number of heads. A carton of parts, half are defective, half are good. Sample three parts replacing them after each draw. Y is the number of defective parts. A manufacturer makes a particular to in either red or ellow. After making equall size batches of tos of either color, the mix them for special packaging in sets of three. Collectors value sets of all one color. What proportion of packages have all one color? '!"#$ (.5) (.5) (.5) (.5) f() ( ) * '!"#$ f() Elementar Events / HHH / HHT HTH THH / HTT THT TTH / TTT %& '!"#$ f()....

4 ! "# $% Suppose the probabilit of manufacturing a defective auto part is %. What is the probabilit of shipping at least one defective part in a shipment of parts? In a shipment of parts, Y is the number of defective parts shipped. Y is BIN (,.) f x x x () x..9 +!!"+$..9 f() ETC. & * The event shipping zero defective parts is the complement of shipping at least one defective part. Therefore, we ma use the basic rule of probabilit that states for an event A, P(A ) P(A). Let A {shipping at least one defective part} and A {shipping zero defective parts}. From the above calculation, P(A ).9. So the probabilit of shipping al least one defective part is P(A)

5 ,) & - Excel Formulas Excel Values f() f() BINOMDIST(,,.,).6 BINOMDIST(,,.,).7 BINOMDIST(,,.,).97 BINOMDIST(,,.,).57 BINOMDIST(,,.,).6 5 BINOMDIST(5,,.,).9 6 BINOMDIST(6,,.,). 7 BINOMDIST(7,,.,). BINOMDIST(,,.,). 9 BINOMDIST(9,,.,). BINOMDIST(,,.,). %& +!!"+$ f()

6 &%' ( The ABC Media Compan claims that its advertisements reach percent of the population in its marketing area. After advertising with ABC, the Widget Compan did not enjo increased sales; therefore, it wants to test ABC s per cent claim. Widget Compan sampled people and asked them if the had seen the advertisement. Beforehand, Widget decided to reject the percent claim if two or fewer people (% or less) in the surve saw the advertisement. If ABC s percent claim is true, what is the probabilit that it is rejected in a given sample? Y is the number in the surve that responds positivel. Y is distributed as BIN (,.) if the claim is true.. /* +!!"!$ f() F() ABC s claim will be rejected is if the event {Y } occurs. The probabilit is P(Y ) f() + f() + f() Thus, Widget will be wrongl accusing ABC with probabilit approximatel 6.7 percent, or in other words, about 7 to odds. This represents too high a risk of wrongl accusing someone. Tpicall, we like this risk to be a low value somewhere around five to 95 odds, i.e. with probabilit.5. Therefore, what value of would let P(Y ).5? From the table we see that P(Y ).66. This is close enough to.5, so the answer is. That is, with probabilit of.5, we take the risk of wrongl confronting ABC when there is one or fewer people who see the advertisement in a sample of people. Note that the cumulative distribution function was indirectl invoked in this example. Indeed, P(Y ) F(). In man instances, the cumulative distribution function is more useful than the probabilit mass function.

7 ) *#'+"+,#' What is the probabilit of winning at least one number with the single one bet? Use the binomial distribution BIN (, /6). Y is the number one s. The event {Y } indicates there are no wins and is complementar to {Y }, the event of winning at least one number. Therefore, P(Y ) - P(Y ) X (/6) X (5/6) - 5/6 9/6. Let Y denote the plaer s winnings (not net) for a one-dollar bet. What is the expected monetar value? * & No. "ONE's" f() f() E(Y).99696

8 We see that the expected monetar value is approximatel 9. cents. Therefore, each plaer looses 7.7 cents to the banker. We found this same monetar value earlier when we calculated the probabilities in a much different and more difficult wa.

Expected Value 7/7/2006

Expected Value 7/7/2006 Definition Let X be a numerically-valued discrete random variable with sample space Ω and distribution function m(x). The expected value E(X) is defined by E(X) = x Ω x m(x), provided