Two tyes of Random Variables: ) Discrete random variable has a finite number of distinct outcomes Examle: Number of books this term. ) Continuous random variable can take on any numerical value within a range of values. Examle: Cost of books this term. Probability Model consists of all values, x, of a random variable, X, along with the robabilities for each value denoted P(X = x). Examle: An insurance comany offers a death and disability olicy that ays $0,000 when you die or $5,000 if you are disabled. It charges a remium of only $50 a year for this benefit. Suose the death rate is one out of every eole, and that another out of suffer from some kind of disability. Is the comany likely to make a rofit selling such a lan? Create a robability model Find the exected value The robability model for this insurance olicy can be dislayed as: Policyholder Outcome Death Disability Neither Payout, x 0,000 5,000 0 997 Probability, P(X = x) The Exected Value of a random variable is the value we exect a random variable to take on or the theoretical long-run average value. It is denoted µ for oulation mean or E ( x) for exected value. The exected value for a discrete random variable is found by summing each value by its robability. E( X ) x P( X x) Note: Be sure that every ossible outcome is included when finding E (X) and verify that you have a valid robability model to start with. Examle: What is the exected value of the ayout from the revious examle? E( X ) x P( X x) So our total ayout comes to $0,000, or $0 er olicy. 997 $0, 000 $5, 000 $0 $0. The Variance of a random variable is the exected value of the squared deviation from the mean. The variance of a discrete random variable is found by calculating. Var( X ) x P( X x) The Standard Deviation of a random variable is the square root of the variance. SD( X ) Var( X ).
Examle: The exected value (or mean) is not what actually haens to any articular olicyholder. No individual olicy actually costs the comany $0. Since we are dealing with random events, some olicyholders receive big ayouts, others nothing. Because the insurance comany must anticiate this variability, it needs to know the standard deviation of the random variable. Policyholder Outcome Payout x Death 0,000 Disability 5,000 Neither 0 Probability P(X = x) 997 Var( X ) x P( X x) Deviation ( x - µ) (0,000 0) = 9,980 (5,000 0) = 4,980 (0 0) = 0 997 9980 4980 0 000 000 000 49, 600 then comute for the standard deviation, SD( X ) Var( X ) 49, 600 $386.78 The insurance comany can exect an average ayout of $0 er olicy, with a standard deviation of $386.78. More about Means and Variances: Adding or subtracting a constant to each value of a random variable shifts the mean but does not change the variance or standard deviation: E( X c) E( X) c Var( X c) Var( X) Multilying each value of a random variable by a constant multilies the mean by that constant and the variance by the square of the constant: E( ax ) ae( X) Var( ax) a Var( X) The mean of the sum of two random variables is the sum of the means: E( X Y ) E( X) E( Y ) The mean of the difference of two random variables is the difference of the means: E( X Y ) E( X) E( Y ) If two random variables are indeendent, then variance of their sum or difference is always the sum of the variances. Var( X Y ) Var( X) Var( Y )
Continuous Variables Random variables that can take on any value in a range of values are called continuous random variables. Continuous random variables have means (exected values) and variances. When two indeendent continuous random variables have Normal models, so does their sum or difference. STEP-BY-STEP Examle,. 378: Packaging Stereos A comany manufactures small stereo systems. At the end of the roduction line the stereos are ackaged and reared for shiing. In stage, called Packing, workers collect all the system comonents, ut each in lastic bags and then lace everything inside a rotective Styrofoam form. The acked form then moved onto stage, called Boxing. There, workers lace the form and acking instructions in a cardboard box, close, seal and label for shiing. The comany says that times required for the acking stage can be described by a Normal model with a mean of 9 minutes and a standard deviation of.5 minutes. The times for the boxing stage can also be modeled as Normal, with a mean of 6 minutes and a standard deviation of minute. a) What is the robability that acking two consecutive systems takes over 0 minutes? Let P = time for acking the first system P = time for acking the second T = total time to ack two systems = P + P Normal Model Assumtion: We are told that both random variable follow Normal models. Indeendence Assumtion: We can reasonably assume that the two acking times are indeendent. Find the exected value: E( T ) E P P E P E P 9 9 8 minutes Since the times are indeendent, Var( T ) Var P P Var P Var P.5.5 4.50 SD( T ) 4.50. minutes Use a model T with N ( 8,.). 0 8 z 0.94. P T 0 0.94 0.736 P z 8 0 OR normalcdf 0,000,8,. 0.736 There s a little more than 7% chance that it will take a total of over 0 minutes to ack two consecutive stereo systems. b) What ercentage of the stereo systems takes longer to ack than to box? (Hint: Let D = the difference in times to ack and box a system) P = time for acking a system B = time for boxing a system 3
D = difference in times to ack and box a system = P B The robability that it takes longer to ack than to box a system is the robability that the difference P B is greater than zero. Normal Model Assumtion: We are told that both random variable follow Normal models. Indeendence Assumtion: We can reasonably assume that the two acking times are indeendent. Find the exected value: E( D) E( P B) E( P) E( B) 9 6 3 minutes Since the times are indeendent, Var( D) Var( P B) Var( P) Var( B).5 3.5 SD( D) 3.5.80 minutes Use a model D with N ( 3,.80). 0 3 z.67.80 P D 0.67 0.955 P z OR normalcdf 0,000,3,.80 0.955 0 3 About 95% of all stereo systems will require more time for acking than for boxing. 4
) There are only two ossible outcomes (called success or failure) on each trial. ) The robability of success, denoted, is the same on every trial. 3) The trials are indeendent. NOTE: The indeendence assumtion is violated whenever the oulation is finite and we samle without relacement. When we don t have an infinite oulation, the trials are not indeendent. If that assumtion is violated, it is still okay to roceed as long as the samle is smaller than 0% of the oulation. A Binomial model tells us the robability for a random variable that counts the number of successes in a fixed number of Bernoulli trials. Suose the random variable X = the number of successes in n observations. Then X is a binomial random variable if: ) There are only two outcomes: success or failure. ) The robability of success,, is the same for each observation. 3) The n observations are indeendent. 4) There is a fixed number of n observations. Binomial models require two arameters, n, the number of trials; and, the robability of success. We denote this by binom (n, ). n! In n trials, there are nck ways to have k! n k! k successes (i.e., total number of ways to arrange k out of n objects) Read nc k as n choose k. Note: n! = n (n-), and n! is read as n factorial. Binomial robability model for Bernoulli trials: n = number of trials = robability of success q = = robability of failure X = # of successes in n trials x n x P X x ncx q n nq A Geometric robability model tells us the robability for a random variable that counts the number of Bernoulli trials until the first success. Suose the random variable X = the number of trials required to obtain the first success. Then X is a geometric random variable if: ) There are only two outcomes: success or failure. ) The robability of success,, is the same for each observation. 3) The n observations are indeendent. 4) The variable of interest is the number of trials required until the first success. Because n is not fixed, there could be an infinite number of X values. However, the robability that X is a very large number is more and more unlikely. Therefore the robability histogram for a geometric distribution is always right skewed. Geometric models are comletely secified by one arameter,, the robability of success, and are denoted by Geom(). Geometric robability model for Bernoulli trials: = robability of success q = = robability of failure X = # of trials until the first success occurs x P X x q q q 5
Finding Geometric and Binomial Probabilities using the TI-Calc Go to nd VARS The robability density function (or df) assigns a robability to each value of X; individual outcome The cumulative density function (or cdf) calculates the sum of the robabilities u to X. geometdf( used to find the robability that it takes a certain number of trials to get a success. When using, you ut in geometdf(robability, # of trials) geometcdf( finds the sum of the robabilities of several ossible outcomes. It calculates the robability of finding the first success on or before the x th trial. geometdf (robability of occurrence, number of trials) Examle ) What is the robability that the first son is the fourth child born? 4 geometdf ( 0. 5, ) 0. P X 4 065 Examle ) What is the robability that the first son born is at most the fourth child? 4 geometcdf ( 0. 5, ) 0. P X 4 9375 Examle 3) What is the robability that the first son born is at least the third child? geometcdf (., ). P X 3 0 5 0 5 binomdf( allows us to find the robability of an individual outcome. Known as binomdf(n,, X), where n stands for number of trials, stands for robability of success, and X is the desired number of successes. binomcdf( allows us to find the total robability of getting X or fewer successes among n trails. Known as binomcdf(n,, X), n stands for number of trials, stands for robability, and X stands for number of successes from x number to 0. binomdf (number of trials, robability of occurrence, number of secific events or successes) To find P(exactly x successes in n trials) use binomdf(n,, x) To find P(at most x successes in n trials) use binomcdf(n,, x) To find P(at least x successes in n trials) use - binomcdf(n,, x -) Examle 4) A genetic trait of one family manifests itself in 5% of the offsring. If eight offsring are randomly selected, find the robability that the trait will aear in a) exactly three of them. P X 3 binomdf 8 0 5 3) 0 076 (,.,. b) at least 5. P X 5 binomcdf 8, 0 5 4 0 073 (., ). Examle 5) A new sales gimmick has 30% of the M&M s covered with seckles. These groovy candies are mixed randomly with the normal candies as they are ut into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the seckles. a) Is this situation a Bernoulli trial? Exlain. "at least" X is the comlement of "at most" X-. To find P(the first is exactly x) use geometdf(, x) To find P(the first is at most x) use geometcdf(, x) To find P(the first is at least x) use - geometcdf(, x -) Yes. There are two ossible outcomes: success = seckled, failure = ordinary (not seckled). The robability of success = 0.30 is constant. 6
Trials can be assumed indeendent there is no reason to believe that finding a seckled candy reveals anything about whether the next one out of the bag will have seckles. We can also assume that the samle is less than 0% of all M&M's so even if the trials are not indeendent, the 0% condition allows us to roceed. b) What s the robability that the first seckled candy is the fourth one we draw from the bag? 3 P st seckled is 4th ( 0. 70)( 0. 70)( 0. 70)( 0. 30 ) OR 0. 70 0. 30 0. 09 c) What s the robability that the first seckled candy is the tenth one? 9 P st seckled is 0th 0. 70 0. 30 0. 0 d) What s the robability we find the first seckled one among the first three we look at? P st seckled P nd is first one seckled P 3 rd is fi rst one seckled 0. 30 0. 70 0. 30 0. 70 0. 30 0. 657 e) How many do we exect to have to check, on average, to find a seckled one? E( X) 0. 30 3. 33 Examle 6) Suose each child born to Jay and Kay has robability 0.5 of having blood tye O. If Jay and Kay have 5 children, what is the robability that exactly of them have tye O blood? Let X = number of children with tye O blood in the 5 children. ) There are only two outcomes: success (has tye O) or failure (not tye O). ) The robability of success, tye of blood,, is 0.5 for each of the 5 observations. 3) Each of the 5 observations is indeendent, since one child s blood tye will not influence the next child s blood tye. 4) There is a fixed number of observations, n = 5. So X is a binomial random variable. The following table shows the robability distribution function, df, for the binomial random variable, X. x 0 3 4 5 P X x 0.373 0.3955 0.637 0.0879 0.046 0.000 5 P X 0 P none have blood tye O 0. 75 binomdf 5, 0. 5, 0 0. 373,.,.,.,. 3 P exactly 3 has blood tye binomdf,.,.,.,.,.,. P X P exactly has blood tye O binomdf 5 0 5 0 3955 P X P exactly has blood tye O binomdf 5 0 5 0 637 P X O 5 0 5 3 0 0879 P X 4 P exactly 4 has blood tye O binomdf 5 0 5 4 0 046 P X 5 P exactly 5 has blood tye O binomdf 5 0 5 5 0 000 7
The following table shows the cumulative distribution function, cdf, for the binomial random variable, X. x 0 3 4 5 P X x 0.373 0.3955 0.637 0.0879 0.046 0.000 P F X X 0 0.373 P X 0.638 P X 0.8965 P X 3 0.9844 P X 4 0.9990 P X 5.0000 Binomcdf ( 5, 0.5, 0) = 0.373 Binomcdf ( 5, 0.5, ) = 0.638 Binomcdf ( 5, 0.5, ) = 0. 8965 Binomcdf ( 5, 0.5, 3) = 0.9844 Binomcdf ( 5, 0.5, 4) = 0. 9990 Binomcdf ( 5, 0.5, 5) =.0000 Examle 7) Suose that 0% of the light bulbs in a strand of Christmas tree lights are defective. a) What is the robability of getting the first defective light bulb on the third try? P X 3 0. 80 0. 0 0. 8 b) What s the robability that it will take at least 5 trials to get the first defective bulb? 4 P X 5 ( 0. 80)( 0. 80)( 0. 80)( 0. 80 ) OR 0. 80 0. 4096 c) What s the robability that it will take at most 3 tries to get the first defective light bulb? P X 3 P X P X P X 3 0. 0 0. 80 0. 0 0. 80 0. 0 0. 488 d) What s the robability that there will be no defective light bulbs among the first five light bulbs selected? 6 0. 80 P X 5 0. 377 Examle 8) Sam and the Geometric Model Postini is a global comany secializing in communication security. The comany monitors over billion Internet messages er day and recently reorted that 9% of e-mails are sam! Let s assume that your e-mail is tyical - 9% sam. We ll also assume you aren t using a sam filter, so every message gets dumed in your inbox. Since sam comes in many different sources, we ll consider tour messages to be indeendent. Overnight your inbox collects e-mail. When you first check your e-mail in the morning, about how many sam e-mails should you exect to have to wade through and discard before you find a real message? What s the robability that the 4 th message in your inbox is the first one that isn t sam? There are two outcomes: a real message (success) and sam (failure) The robability of success, = 0.9 = 0.09 Assume that the messages arrive indeendently and in a random order, so I can use the model Geom(0.09). Let X = number of e-mails EX ( ).. 0 09 P( X 4) P sam sam sam real 3 0. 9 0. 09 0. 0678 On average, I exect to check just over e-mails before I find a real message. There s slightly less that a 7% chance that my first real message will be the 4 th one I check. 8
Examle 9) Sam and the Binomial Model Using the information from the revious examle, suose your inbox contains 5 messages. What are the mean and standard deviation of the number of real messages you should exect to find in your inbox? What is the robability that you will find only or real messages? Assume that messages arrive indeendently and at random, with the robability of success (real message) = 0.9 = 0.09. Let X = number of real messages among 5, so I can use the model Binom(5, 0.09). E( X ) n 5( 0. 09). 5 SD( X) nq 5 0. 09 0. 9. 43 or P X P X P X 4 3 5C 0. 09 0. 9 5C 0. 09 0. 9 0. 340 0. 777 0. 57 Among 5 e-mail messages, I exect to find an average of.5 that aren t sam, with a standard deviation of.43 messages. There s just over 50% chance that or e-mails will be real messages. When dealing with a large number of trials in a Binomial situation, making direct calculations of the robabilities becomes tedious (or outright imossible). Fortunately, the Normal model comes to the rescue As long as the Success/Failure Condition holds, we can use the Normal model to aroximate Binomial robabilities. Success/failure condition: A Binomial model is aroximately Normal if we exect at least 0 successes and 0 failures: n 0 and nq 0 When we use the Normal model to aroximate the Binomial model, we are using a continuous random variable to aroximate a discrete random variable. So, when we use the Normal model, we no longer calculate the robability that the random variable equals a articular value, but only that it lies between two values. Examle 0) Sam and the Normal aroximation to the Binomial From the Postini examle, we know that 9% of e-mail messages are sam. Recently you installed a sam filter. You observe that over the ast week it okayed only 5 of 4 e-mails you received, classifying the rest as junk. Should you worry that the filtering is too aggressive? What is the robability that no more than 5 of 4 e-mails is a real message? Assume that messages arrive randomly and indeendently with the robability of success (real message) = 0.9 = 0.09. The model Binom(5, 0.09) alies but will be hard to work with. Checking conditions for the Normal aroximation: These messages reresent less than 0% of all e-mail traffic. n = (4)(0.09) = 7.98 real messages and nq = (4)(0.9) = 94.08 sam messages, both far greater than 0. It is okay to aroximate this binomial robability by using a Normal model. u n 4( 0. 90) 7. 98 nq 4 0. 09 0. 9 0. 79 9
5 7. 98 P X 5 P z. 0 79 P z. 3 0. 9834 Among my,4 e-mails, there s over a 98% chance that no more than 5 of them were real messages, so the filter may be working roerly. JUST CHECKING: From a revious chater, we noted that the Pew Research Center reorted that they are able to contact only 76% of the randomly selected households drawn for a telehone survey. ) Exlain why these hone calls can be considered Bernoulli trials. There are two outcomes: contact and no contact The robability of contact is constant, = 0.76 The random calls are indeendent ) Which of the models (Geometric, Binomial, Normal) would you use to model the number of successful contacts from a list of 000 samled households? Exlain. Binomial, with n = 000, = 0.76. u n nq 000( 0. 76) 760 and 000 0. 76 0. 4 3. 5 3) Pew further reorts that even after they have contacted a household, only 38% agree to be interviewed, so the robability of getting a comleted interview for a randomly selected household is only 0.9. Which of the models of this chater would you use to model the number of households Pew has to call before they get the first comleted interview? Geometric, with = 0.9. MORE EXERCISES ) Suose a comuter chi manufacturer rejects 3% of the chis roduced because they fail resale testing. What is the robability that the seventh chi you test is the first bad one you find? 7 PX ( 7) 0. 97 0. 03 0. 05 ) The drilling record for an oil comany suggests that the robability the comany will hit oil in roductive quantities at a certain offshore location is 0.0. Suose the comany lans to drill a series of wells. a) What is the robability that the 7 th well drilled is the first roductive well? P( 7 th is good) P( 6 bad, then good) 6 0. 80 0. 0 0. 054 b) What is the robability that the first roductive well was one of the first 3 wells drilled? P( st or nd or 3 rd is good) ( 0. 0) 0. 80 0. 0 0. 80 0. 0 0. 4880 c) How many wells would you exect to drill before you hit oil? Exected Value. 5 wells 0 0 d) If the comany drills 0 wells, what is the robability that 4 of them will be roductive? 0
4 6 P( 4 out of 0) 0C4 0. 0 0. 80 0. 088 e) If the comany drills 0 wells, how many wells will the comany exect to be roductive? With what standard deviation? Exected Value n 0 0. 0 wells Standard Deviation nq 0 0. 0 0. 80. 65 wells 3) Assume that 3% of eole are left-handed. Find the robability of each outcome described below. a) The first lefty is the fifth erson chosen. 4 P( 4 rights then a left) 0. 87 0. 3 0. 0745 b) The first lefty is the second or third erson. P( nd erson or 3 rd erson) ( 0. 87)( 0. 3) 0. 87 0. 3 0. c) If you were on a committee of 5 eole, what is the robability that there are exactly 3 lefties in the grou? 3 P( 3 out of 5) 5C3( 0. 3) 0. 87 0. 07 d) What is the robability that there is at least one lefty among the 5 eole? P( or more out of 5) P( no lefties) 0 5 5C00. 3 0. 87 0. 50 Assume that 3% of eole are left-handed. If we select 5 eole at random, find the robability of each outcome described below. e) There are at least 3 lefties in the grou. P( 3 or 4 or 5 out of 5) 0. 07 0. 00 0. 00004 0. 08 f) There are no more than 3 lefties in the grou. P( 3 or fewer out of 5) P( 0) P( ) P( ) P( 3) P( 4 or 5 out of 5) 0. 00 0. 00004 0. 999 4) Out of the 5 eole, how many lefties do you exect? With what standard deviation? 5( 0. 3) 0. 65 lefties in a grou of 5 eole