Overview Both chapters and 6 deal with a similar concept probability distributions. The difference is that chapter concerns itself with discrete probability distribution while chapter 6 covers continuous probability distributions. Recall the rough difference between discrete and continuous as found when we discussed data. Discrete - countable Continuous - measurable This difference makes it possible to create tables for discrete distribution. Continuous distributions, however, have infinitely many outcomes, and so it is impossible to create tables. Every probability distribution is a relative frequency table (recall the definition of probability), and so we will be using these tables, or the histogram for the table, to find probabilities for certain events. The table (or histogram) is intended to give us the shape of the probability. Chapter All probability distributions have certain characteristics. For eample the entire amount of probability must sum to, and all individual probabilities must be between and. For discrete distributions that means: Consider the following eamples: P( X ) and P( X ) e: let X = tossing a coin three times and counting the number of heads... These values come from a tree diagram or from the sample space: S = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} P(X = ) 8 8 8 8 e: let X = roll a die times and count the number of fives Again these come from a tree diagram. In this case the sample space does not contain equally likely outcomes e: Let X be a random variable whose probability function, P(X = ), is given as follows: P( X ) for =,,,.6..6. P(X = ) ( ) 6 ( ) ( ) 6 6 6 ( 6 ) ( 6 ) ( 6 ) 6 ( 6 ) P(X = )
One can notice that the area of each rectangle is the same as the probability from the table. Therefore, when finding probabilities, the areas of rectangles are used. Consider the following generic eample. P(X = )..7..7. 6. 7.8 8. 9.. For eample, to find the probability, P( X 7), we either need to sum the appropriate values on the table (see in blue) or total the appropriate areas of rectangles (see in blue). Both answers would be the same.... 6 7 8 9 If the question was changed slightly to read P( X 7) then we would just eclude the rectangle labeled 7. That is: P( X 7) P( X or X or X or X 6 or X 7) whereas: P( X 7) P( X or X or X or X 6) In addition to finding probabilities, we also want to do what we did with data in general. Recall that with data we tried to answer three basic questions: Where is the data centered - Centrality (mean, median, mode, midrange, etc.) How spread out is the data - Dispersion (range, standard deviation, variance) What is the shape of the data - Distribution Well, up to now, we have only discussed distribution or shape of the probability. Finding the mean and standard deviation for a discrete distribution can also be found using the table. The formulas are not too difficult, but it is a little tedious to process the table. The formulas are: Mean: P( ) Standard Deviation: P() e: Prev. P( X ) P(X = ) P( ) 9 P( ) 8 7 so: P( ) P( ) () 6 6 An application of discrete random variable probability distributions is Epected Value. Epected Value is a long term average. Thus, the formula is the same as the mean: E P( )
Following are a couple of eamples. e: Suppose a yearly insurance premium for an individual is $6. Suppose also that this individual has a.999 probability of living throughout the coverage period (the year). If the coverage amount is $,, what is the epected value for the insured? Event P() P() Live -6..999 -.9 Die 998.. 9.9-6. P( ) So, the epected value for the estate (on the year) is to lose $6 (of the $6 premium) e: Suppose you buy a $ raffle ticket with the potential of prizes (vacation, laptop, dinner for two, coffee coupon). Find the epected return for each player. Event P() P() Vacation,999..999 Laptop 699..699 Dinner 7..7 Coffee 9..87 Lose -.99698989999 -.9969899 -.997997 So, the epected return for a player is to lose appro. 9 cents for every dollar ticket purchased. One of the premier eamples of a discrete random variable is the binomial random variable. A binomial eperiment must follow four requirements. The eperiment must have a fied number of trials (n) The trials must be independent Each trial must have all outcomes classified into two categories (even though the sample space may have more than two simple events) The probabilities must remain constant for each trial A couple of classic eamples of a binomial eperiment is tossing a coin and rolling a die. Of the aforementioned four conditions, the primary one that helps people recognize that they are working with a binomial eperiment is that the outcomes fall into two categories (hence the bi in binomial). This does not mean that there are only two outcomes (like heads or tails on a coin toss). The outcomes to rolling a die can also be categorized into two outcomes (get a or don t get a ). Consider the following eample: e: Consider rolling a die four times and counting the s. 6 6 6 6 6 6 fives fives fives fives fives fives fives five fives fives fives five fives five five fives # of s P( X ) ( ) ( ).8 # of paths 6 6 ( ) ( ).88 6 6 6 ( ) ( ).7 6 6 ( ) ( ). 6 6 ( ) ( ).776 6 6 /6 for each /6 for each
In binomial eperiments the two categorical outcomes are generically labeled as success and failure. These are not intended to be equivalent to what you would consider to be a success. For eample, if you were studying parachutes not opening, then a parachute not opening would be considered a success for the eperiment. This leads to a generalization for all binomial eperiments. The probability function for a binomial distribution (the function that describes all probabilities) is: n P( X ) C p q for n =,,,..., n- n where: n = fied number of trials p = prob of success per trial q = prob of failure per trial = number of successes in n trials So, for our last eample, the probability function would be: P( X ) C ( ) ( ) 6 6 e: Assume (from a previous study of coffee drinkers) that any individual has a probability of. of drinking coffee (thus a probability of.9 of not drinking coffee). If we randomly select people from the population, find the following a) the probability function for this binomial eperiment For this eperiment n =, p =., and q =.9 P ( X ) C (.) (.9) where =,,,,, b) the probability that all five drink coffee P(all drink coffee) = P ( X ) C (.) (.9). 86 c) the probability that eactly two people drink coffee P( drink coffee) = P X C ( ) (.) (.9) (.68) (79).76 d) the probability that at least two people drink coffee P(at least two drink coffee) = P(less than two drink coffee) = [ P ( X or X ) ] = [ P (X ) + P (X ) ] = C (.) (.9) C (.) (.9) =.7999 89.9898. 686696 Another nice feature of our binomial random variable is that it is easier to find the mean and standard deviation. For any discrete r.v. P( ) OLY for Binomial np P() npq e: The last eample with coffee drinking. Recall n =, p =. and q =.9 So, np (.). npq (.) (.9).99777
Chapter 6 As noted earlier, the difference between the discrete random variables from chapter and the continuous random variables from chapter 6 is that discrete random variables have finite number of outcomes (and thus can have a table made), but continuous random variables have infinitely many outcomes which means no tables. Graphically, the tables produced a relative frequency histogram that we noted would help us understanding that areas of rectangles was equivalent to probabilities. Discrete Rectangles Continuous - Curve... 6 7 8 9 6 7 8 9 The idea that probability equates to area is the same in both discrete and continuous, but the process to get the answer is very different. In general, to find the area under a curve you need calculus. However, even then, the curve needs to be nice. Since there are infinitely many curves that satisfy the conditions of being a probability distribution ( P() and Σ P() = ), we will initially only deal with one the ormal Distribution. μ Unfortunately, there are infinitely many different normal curves, so we will focus on one at the start the Standard ormal Distribution (μ =, σ = ). I will always use the random variable letter Z when I m referring to the standard normal dist. Because μ = and σ =, every position on the number line is at its own z-score (hence why this is sometimes called the z-dist). A table is available to find areas for this distribution (table A- in fold-out). However, the only areas you can find on the table is for regions of the type shown below (see the picture at the top of the table) The shaded area is the probability P(Z z) z
Use table A- to do the following eamples: e: Find P(Z.) =.89 e: Find P( Z.) = P(Z.). =.89. =.9 e: Find P(-. Z.66) = P(Z.66) - P(Z -.) =.7.79 =.77 e: (Alternate method for symmetry) Find P(-. Z.) = P( Z.) = (.9) e: Find P(Z.) = - P(Z.) =.89 =.7 One should now be asking how to do non-standard normal distributions. Well, the answer is to make them into standard normal distributions by z-scoring everything. a X b P ( a X b) P P( c Z d) e: Given a random variable that is nonstandard normal with mean μ = and σ = 6, find P( X 69.8) P( X 69.8) P( Z.).89..9 e: Given X is normal with μ = 7 and σ =, find PX ( 8.67) P( X 8.67) P( X 8.67) P( Z.) =.976 =.6 We need to interpolate..97..976..98. e: Given X is normal with μ = and σ =, find P(8. X ) P(8. X ) P(. Z.68) = P( Z.68) P( Z.).68.77 =.766 -.69.68.766 =.676.69.79. Two eamples of using a normal distribution are the Central Limit Theorem and Approimating a Binomial with a ormal. (sections 6. and 6.6) Whereas the generic non-standard normal problem is asking to find the probability for one representative from the population, the Central Limit Theorem (CLT) is what is used when we are being asked questions about a group, or sample, from the population. The central limit theorem states that the population of all possible sample means (-bar s) from a population will be approimately normal with a mean and standard deviation as follows: and n otice that this means that the larger the sample size, the smaller the standard deviation for the population of all possible sample means would be. This will affect the probability.
e: Suppose men have head breadths that are normally distributed with μ = 6 and σ = a) If one man is chosen from the population, find P(X < 6.) Since this is a non-standard normal, we need to change to standard (μ = 6, σ = ) X 6 6. 6 P( < 6.) P P Z.79 b) If men are selected, find P( < 6.). ote that is normally distributed with mean 6 and standard deviation. n 6 6. 6 Thus, P( < 6.) P P Z.977.. The last application of normal distributions forces us back to chapter. Often the number of trials in a binomial problem is very large. As such there would be a tremendous number of rectangles to find the area of more than most people want to find. It turns out that as the number of trials increase, the histogram for a binomial distribution begins to look normalish. Provide we have the condition that np and nq, then the histogram is sufficiently normal shaped that we can use a normal curve to approimate the area of rectangles in a binomial histogram. e: Let n =, where p = and q =.76 Find P(X 6) ormal curve must be in the same central location as the binomial histogram and of the same breadth. That is, the same mean and standard dev. 6 6 For bino: np npq 6 6 We would need to shade a section under the normal curve beginning at 69. to have approimately the same area as the blue shaded rectangles of the binomial that begin with rectangle 6. So, P( X 6) P( Y 69.) Binomial ormal
Continuing, we get 69. 6 P( X 6) P( Y 69.) P Z P( Z.8) P( Z.8) 6.9898...9896..8.9898.8..9898.6 rounds to.