ࡱ>  K|[ }bjbj 12ΐΐu++888888888;8N;KJKJKJKJ $P h8 ƖD 88KJKJ{L 8KJ8KJ u\KJw80NQ8Z .Z Z 80  N Z  + 7: November 20, 2011: Basic Probability Rule (Material for final exam Fall 2011) Given  EMBED Equation.3  and EMBED Equation.3 . Find EMBED Equation.3 . Answer:  EMBED Equation.3  From the given table find  EMBED Equation.3  BCDKTotalE737522F8611429Total15918951 Answer:  EMBED Equation.3 160/493=0.3245436105 A family has two children. Using b to stand for boy and g for girl in ordered pairs write the sample space then find the event E that the family has exactly one daughter and the event F the family has at least one daughter. Now identify which one is true E and F are independent (B) E and F are dependent E and F are mutually exclusive that means  EMBED Equation.3  (D) None of the given Answer: (B) Dependent Roll a die twice. Write the sample space. Find the following probabilities:  EMBED Equation.3  Answer: 5/6 Suppose events A and B are independent and that p(A) = 0.4 and p(B) = 0.6. Find EMBED Equation.3 . Answer: 0.24 Suppose events A and B are independent and that p(A) = 0.3 and p(B) = 0.6. Find EMBED Equation.3  Answer: 0.72 Given that p(A) = 0.4, p(B) = 0.5 and  EMBED Equation.3 . Find  EMBED Equation.3 . Answer: 0.70 Events  EMBED Equation.3 ,  EMBED Equation.3  and  EMBED Equation.3  form a partition of the sample space S with probabilities  EMBED Equation.3 ,  EMBED Equation.3 and  EMBED Equation.3  If E is an event in S with  EMBED Equation.3 ,  EMBED Equation.3  and  EMBED Equation.3  then find the probability of  EMBED Equation.3  Answer: 0.47 A package, say A, of 24 crocus bulbs contains eight yellow, eight white, and eight purple crocus bulbs. A package, say B, of 24 crocus bulbs contains six yellow, six white and 12 purple crocus bulbs. One of the two packages is selected at random. If three bulbs from this package were planted and all three yielded white flowers, compute the conditional probability that the package A was selected. If the three bulbs yielded one yellow, one white, and one purple flower, compute the conditional probability that the package B was selected. Solution: Suppose WWW means three white bulbs, then  EMBED Equation.3 ,  EMBED Equation.3 , also  EMBED Equation.3  a)  EMBED Equation.3  b)  EMBED Equation.3  How many different license plates can be made using 2 letters followed by 4 digits, if Letters and digits may be repeated. Answer: 6760000 Letters may be repeated, but digits are not repeated. Answer: 3407040 Neither letters nor digits are repeated. Answer: 3276000 License plates in a small state have three letters followed by four numbers. How many possible license plates are possible if the letters have to be different? Repeated numbers are also not allowed. Answer: 78624000 A group of musicians consists of 3 percussionists, 4 cellists and 4 violinists. A subgroup needs to be picked consisting of 1 percussionists, 1 cellist and up to 2 violinist. How many such sub groupings are possible? Answer: 204 Write the universal set for a two children family: use b for boys and g for girls. Also determine if the set that the first child a girl and the set that the second child is a boy are mutually exclusive. Write all information in the spaces provided below: Write universal set U = { Second child is a boy A = { First child is a girl B = { The intersection:  EMBED Equation.3 = { Conclusion: Not mutually exclusive Probability Distribution Random variables of the discrete type In  HYPERLINK "http://en.wikipedia.org/wiki/Probability_theory" \o "Probability theory" probability theory, a  HYPERLINK "http://en.wikipedia.org/wiki/Probability_distribution" \o "Probability distribution" probability distribution is called discrete if it is characterized by a  HYPERLINK "http://en.wikipedia.org/wiki/Probability_mass_function" \o "Probability mass function" probability mass function. Thus, the distribution of a  HYPERLINK "http://en.wikipedia.org/wiki/Random_variable" \o "Random variable" random variable X is discrete, and X is then called a discrete random variable, if  EMBED Equation.3  If a random variable is discrete, then the  HYPERLINK "http://en.wikipedia.org/wiki/Set_(mathematics)" \o "Set (mathematics)" set of all values that it can assume with non-zero probability is  HYPERLINK "http://en.wikipedia.org/wiki/Finite_set" \o "Finite set" finite or  HYPERLINK "http://en.wikipedia.org/wiki/Countably_infinite" \o "Countably infinite" countably infinite, because the sum of uncountably many positive  HYPERLINK "http://en.wikipedia.org/wiki/Real_number" \o "Real number" real numbers (which is the  HYPERLINK "http://en.wikipedia.org/wiki/Least_upper_bound" \o "Least upper bound" least upper bound of the set of all finite partial sums) always diverges to infinity. Given a random experiment with an outcome space S, a function X that assigns to each element s in S one and only one real number X(s) = x is called a random variable, like a function of s. The space of X is the set of real numbers  EMBED Equation.3 , where  EMBED Equation.3  means the element s belongs to S. The probability mass function (pmf)  EMBED Equation.3  of a discrete random variable X is a function that satisfies the following properties:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Example 1. Suppose that X has a discrete uniform distribution on  EMBED Equation.3 and its pmf is  EMBED Equation.3 . As a general case we may write pmf as  EMBED Equation.3  Example 2. Roll a 4 sided die twice and let X equal the larger of the two outcomes if they are different and common value if they are the same. The outcome space for this experiment is  EMBED Equation.3 , where we assume that each of these 16 points has probability  EMBED Equation.3 . Then  EMBED Equation.3  and  EMBED Equation.3 . Looking at the pattern one can easily find the pmf  EMBED Equation.3  Exercise Let the pmf of X be defined by  EMBED Equation.3 , a) draw a bar graph and a b) probability histogram For each of the following, determine the constant c, so that  EMBED Equation.3 satisfies the conditions of being a pmf for a random variable X,  EMBED Equation.3  Answer: c = 10  EMBED Equation.3  Answer: c = 55  EMBED Equation.3  Answer: c = 3  EMBED Equation.3  Answer: c = 1/30  EMBED Equation.3  Answer: c = n(n+1)/2 Mathematical expectation In  HYPERLINK "http://en.wikipedia.org/wiki/Probability_theory" \o "Probability theory" probability theory and  HYPERLINK "http://en.wikipedia.org/wiki/Statistics" \o "Statistics" statistics, the expected value (or expectation value, or mathematical expectation, or mean, or first moment) of a  HYPERLINK "http://en.wikipedia.org/wiki/Random_variable" \o "Random variable" random variable is the  HYPERLINK "http://en.wikipedia.org/wiki/Lebesgue_integral" \o "Lebesgue integral" integral of the random variable with respect to its  HYPERLINK "http://en.wikipedia.org/wiki/Probability_measure" \o "Probability measure" probability measure. For discrete random variables this is equivalent to the probability- HYPERLINK "http://en.wikipedia.org/wiki/Weighted_sum" \o "Weighted sum" weighted sum of the possible values. The term "expected value" can be misleading. It must not be confused with the "most probable value." The expected value is in general not a typical value that the random variable can take on. It is often helpful to interpret the expected value of a random variable as the long-run average value of the variable over many independent repetitions of an experiment. When it exists, mathematical expectation E satisfies the following properties: If c is a constant,  EMBED Equation.3  If c is a constant and u is a function,  EMBED Equation.3  If  EMBED Equation.3  and  EMBED Equation.3  are constants and  EMBED Equation.3  and  EMBED Equation.3  are functions, than  EMBED Equation.3  Example 1. Let X have the pmf  EMBED Equation.3 . Find  EMBED Equation.3  Solution:  EMBED Equation.3 , verify. Example 2. Let X have the pmf  EMBED Equation.3 . Find mean =  EMBED Equation.3  and also  EMBED Equation.3  and variance  EMBED Equation.3  and also standard deviation EMBED Equation.3 . Solution: Mean =  EMBED Equation.3   EMBED Equation.3  Variance  EMBED Equation.3  and  EMBED Equation.3  Example 3. A politician can emphasize jobs or the environment in her election campaign. The voters can be concerned about jobs or the environment. A payoff matrix showing the utility of each possible outcome is shown.  EMBED Equation.3  The political analysts feel there is a 0.39 chance that the voters will emphasize jobs. Which strategy should the candidate adopt to gain the highest utility Environment b) Jobs Explain mathematically. Solution: For the environment the expected value is  EMBED Equation.3  On the other hand for jobs the expected value is  EMBED Equation.3 . So the preference will go for a) Environment (because of higher expected value). Exercise Find mean and standard deviation of the following:  EMBED Equation.3  Answer: Mean 15, SD = 7.07  EMBED Equation.3  Answer: Mean 5, SD = 0  EMBED Equation.3  Answer: Mean 1.67, SD = 0.75 Given  EMBED Equation.3 , determine  EMBED Equation.3 Var EMBED Equation.3 and mean= EMBED Equation.3  Answer:  EMBED Equation.3   EMBED Equation.3  Bernoulli trials and the Binomial distribution A Bernoulli experiment is a random experiment, the outcome of which can be classified in but one of two mutually exclusive and exhaustive ways, say, success or failure (life or death, head or tail, 3 or not 3 etc. The pmf of a Bernoulli trail is  EMBED Equation.3  and we say that the random variable x has Bernoulli distribution. The mean of Bernoulli trial is given as  EMBED Equation.3 , verify. The variance of Bernoulli trial is  EMBED Equation.3  Example 1. In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased, the probability of purchasing two winning tickets is  EMBED Equation.3  One may use calculator as follows (TI) 2nd DISTR 0 (binompdf) (8, 0.20, 2) will display 0.29360128 Example 2. In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased, the probability of purchasing at best 6 winning tickets is  EMBED Equation.3  One may use calculator as follows (TI) 2nd DISTR A (binomcdf) (8, 0.20, 6) will display 0.99991552 Example 3. In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased. Find the probability of purchasing at least 6 winning tickets. Hint. Find  EMBED Equation.3  or  EMBED Equation.3  Example 4. A quiz consists of 24 multiple choice questions. Each question has 5 possible answers, only one of which is correct. If you answer the questions completely based on guessing, what is the probability that You will answer exactly 4 wrong? You will answer exactly 4 correctly? You will answer at least 20 correctly? You will answer at most 3 wrong? You will answer at most 3 correctly? Solution: The probability that you will answer one question wrong is  EMBED Equation.3 . The probability of answering exactly 4 wrong is a binomial probability of B(24, 0.8, 4), which is  EMBED Equation.3 , which is almost zero. If you use TI calculator use binompdf (24, 0.8, 4). Check your calculator using the following code:  2nd DISTR 0 binompdf (24, .8, 4) The probability that you will answer exactly 4 correct is B(24, 0.2, 4) = 0.196 At least 20 correct  EMBED Equation.3  = 20 correct + 21 correct + 22 correct + 23 correct + 24 correct =  EMBED Equation.3 . It is easy to use calculator with binomcdf as follows:  EMBED Equation.3  At most three wrong:  EMBED Equation.3  At most three correct:  EMBED Equation.3  Example 5. A computer manufacturer tests a random sample of 28 computers. The probability that a computer is non defective is 91.3%. What is the probability that: Exactly 7 computers are defective? Answer: 0.006605 At least two computers are defective? Answer: 0.7131689 At most two computers are defective? Answer: 0.555224 Example 6. A quiz consists of 10 multiple choice questions, each with 4 possible choices. For someone who makes random guesses for all of the questions, find the probability of passing if the minimum passing grade is 90%. Solution:  EMBED Equation.3  Example 8. A student claims that he has extrasensory perception (ESP). A coin is flipped 25 times, and a student is asked to predict the outcome in advance. He gets 20 out of 25 correct. What was the probability that he would have done at least this well if he had no EPS? Solution:  EMBED Equation.3  Exercise 1. Toss a fair coin 12 times. How many possible outcomes do you have? What is the probability of getting a) exactly 7 heads, b) at least 7 heads, c) at most 7 heads? Answer: 4056, a) 19.34% b) 38.72% c) 80.62% Exercise 2. A student claims that he has extrasensory perception (ESP). A coin is flipped 30 times, and a student is asked to predict the outcome in advance. He gets 25 out of 30 correct. What was the probability that he would have done at least this well if he had no EPS? Answer: 0.0001625 Exercise 3. A quiz consists of 20 multiple choice questions, each with 5 possible choices. For someone who makes random guesses for all of the questions, find the probability of passing if the minimum passing grade is 80%. Answer: 0.0000000138 Example 4. A computer manufacturer tests a random sample of 30 computers. The probability that a computer is defective is  EMBED Equation.3 %. What is the probability that: Exactly 7 computers are defective? At least two computers are defective? At most two computers are defective? Exercise 5. In the instant lottery with 10% winning tickets, if X is equal to the number of winning tickets among 20 tickets that are purchased, find the probability of purchasing at best 7 winning tickets, at least 7 winning tickets, no more than 6 winning tickets, no less than 6 winning tickets Exercise 6. The rates of on-time flights for commercial jets are continuously tracked by the U.S Department of transportation. Recently, Southwest Air had the best rate with 80% of its flights arriving on time. A test is conducted by randomly selected 16 Southwest flights and observing whether they arrive on time. Find the probability that exactly 4 flights arrive on time The probability that at least 4 flights arrive on time At best 4 flights arrive on time Random variable of the continuous type: A random variable is a function X that assigns to each element s in the outcome space S one and only one corresponding real number X(s) = x. The space of X is the set of real numbers  EMBED Equation.3 is an interval. In discrete case the S is the set of discrete points. In the continuous case we call the integrable function EMBED Equation.3 , a probability density function (pdf) which satisfies the following:  EMBED Equation.3  b)  EMBED Equation.3  c) The probability of the event  EMBED Equation.3  is  EMBED Equation.3  Probability Distribution Function: A function F is a distribution function of the random variable X iff the following conditions are satisfied: F is non decreasing i.e.,  EMBED Equation.3 , or  EMBED Equation.3  for all  EMBED Equation.3  F is continuous c) F is normalized i.e.,  EMBED Equation.3  Example 1. Evaluate the integral  EMBED Equation.3  Solution:  EMBED Equation.3  Example 2. Show that  EMBED Equation.3  is a probability density function. Solution (Hint): Show that  EMBED Equation.3  and  EMBED Equation.3  Example 3. Let Y be a continuous random variable with pdf  EMBED Equation.3  and the distribution function is defined by  EMBED Equation.3  Find mean  EMBED Equation.3  (check the integral) and Variance  EMBED Equation.3  (check the integral). Find also the standard deviation  EMBED Equation.3 . Example 4. The probability density function of a continuous random variable x is given by the function  EMBED Equation.3 . Find its corresponding distribution function, mean, variance and standard deviation, interval of one standard deviation of mean, two standard deviation of mean and three standard deviation of mean. Solution: The distribution function is the integral of the 1 pdf function over the real line. Draw the graph of the pdf and notice that F(0) = 0, 0 1 2 F(1) = 1/2 and F(2) = 1. We also notice that distribution function is zero, i.e.,  EMBED Equation.3  when x < 0. The distribution function over the interval  EMBED Equation.3  is  EMBED Equation.3  The distribution function over the interval  EMBED Equation.3  is  EMBED Equation.3  The distribution function over the interval  EMBED Equation.3  is  EMBED Equation.3  Thus we have the probability distribution function defined as follows:  EMBED Equation.3  You can calculate mean, standard deviation and variance. Look at example 3. Example 5. Show the following function is a probability distribution function  EMBED Equation.3  Solution: We need to check the following properties: Check that F is non-decreasing.  EMBED Equation.3  shows that F is not decreasing. For continuity check that  EMBED Equation.3  and  EMBED Equation.3  F is normalized:  EMBED Equation.3  The function F(x) defined above is probability distribution function. Exercise: For each of the following functions, i) find the constant c so that f (x) is a pdf of the random variable X, ii) find the distribution function F(x)  EMBED Equation.3 and iii) sketch f (x) and F(x), iv) find also  EMBED Equation.3  .  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Sketch the graph of the following pdf f (x), then find and sketch the probability distribution function F(x) on the real line. Review example 4.  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Section 9.4 The Normal Distribution  INCLUDEPICTURE "http://mathworld.wolfram.com/images/eps-gif/NormalDistribution_651.gif" \* MERGEFORMATINET  A normal distribution of a random variable X with  HYPERLINK "http://mathworld.wolfram.com/Mean.html" mean  EMBED Equation.3  and  HYPERLINK "http://mathworld.wolfram.com/Variance.html" variance  EMBED Equation.3 is a statistic distribution with  HYPERLINK "http://mathworld.wolfram.com/ProbabilityDensityFunction.html" probability density function (pdf)  INCLUDEPICTURE "http://mathworld.wolfram.com/images/equations/NormalDistribution/NumberedEquation1.gif" \* MERGEFORMATINET (1) on the domain  INCLUDEPICTURE "http://mathworld.wolfram.com/images/equations/NormalDistribution/Inline4.gif" \* MERGEFORMATINET . While statisticians and mathematicians uniformly use the term "normal distribution" for this distribution, physicists sometimes call it a Gaussian distribution and, because of its curved flaring shape, social scientists refer to it as the "bell curve." De Moivre developed the normal distribution as an approximation to the  HYPERLINK "http://mathworld.wolfram.com/BinomialDistribution.html" binomial distribution, and it was subsequently used by Laplace in 1783 to study measurement errors and by Gauss in 1809 in the analysis of astronomical data (Havil 2003, p.157). The normal distribution is an extremely important  HYPERLINK "http://psychology.wikia.com/wiki/Probability_distribution" \o "Probability distribution" probability distribution in many fields. It is a family of distributions of the same general form, differing in their location and scale  HYPERLINK "http://psychology.wikia.com/wiki/Parameter" \o "Parameter" parameters: the  HYPERLINK "http://psychology.wikia.com/wiki/Mean" \o "Mean" mean ("average") and  HYPERLINK "http://psychology.wikia.com/wiki/Standard_deviation" \o "Standard deviation" standard deviation ("variability"), respectively. The standard normal distribution is the normal distribution with a mean of zero and a standard deviation of one (the green curve in the plots below). It is often called the bell curve because the graph of its  HYPERLINK "http://psychology.wikia.com/wiki/Probability_density_function" \o "Probability density function" probability density resembles a bell.  HYPERLINK "http://images2.wikia.nocookie.net/psychology/images/1/1b/Normal_distribution_pdf.png"  INCLUDEPICTURE "http://images3.wikia.nocookie.net/psychology/images/thumb/1/1b/Normal_distribution_pdf.png/800px-Normal_distribution_pdf.png" \* MERGEFORMATINET  If a  HYPERLINK "http://psychology.wikia.com/wiki/Random_variable" \o "Random variable" random variable X  HYPERLINK "http://images.wikia.com/wikitex/images/2/2b/2b8/113286d41777791b5a45b5d20fbd71.png"  has this distribution, we write  HYPERLINK "http://images.wikia.com/wikitex/images/2/2b/2b8/113286d41777791b5a45b5d20fbd71.png"  INCLUDEPICTURE "http://images.wikia.com/wikitex/images/2/2b/2b8/113286d41777791b5a45b5d20fbd71.png" \* MERGEFORMATINET ~  HYPERLINK "http://images.wikia.com/wikitex/images/c/c8/c83/72d9bf0dd9d2fb1294c9935aa7bc81.png"  INCLUDEPICTURE "http://images.wikia.com/wikitex/images/c/c8/c83/72d9bf0dd9d2fb1294c9935aa7bc81.png" \* MERGEFORMATINET . If  HYPERLINK "http://images.wikia.com/wikitex/images/d/d6/d6a/ddd9ac06ff294dbf700fbcba74ad01.png"  INCLUDEPICTURE "http://images.wikia.com/wikitex/images/d/d6/d6a/ddd9ac06ff294dbf700fbcba74ad01.png" \* MERGEFORMATINET and  HYPERLINK "http://images.wikia.com/wikitex/images/8/8a/8ab/12465b7784c5f9f1b6a9b1cbf74981.png"  INCLUDEPICTURE "http://images.wikia.com/wikitex/images/8/8a/8ab/12465b7784c5f9f1b6a9b1cbf74981.png" \* MERGEFORMATINET , the distribution is called the standard normal distribution and the probability density function reduces to  HYPERLINK "http://images.wikia.com/wikitex/images/8/8d/8de/943f01099203d1915931eb386fed31.png"  INCLUDEPICTURE "http://images.wikia.com/wikitex/images/8/8d/8de/943f01099203d1915931eb386fed31.png" \* MERGEFORMATINET  Area under a normal curve: For a standard normal variate z, the normal distribution has mean zero and standard deviation one with pdf  EMBED Equation.3  The area under the standard normal distribution curve for  EMBED Equation.3 . We have now the difficulty to evaluate the integral without having the knowledge of multivariable calculus and polar coordinate form. But this difficulty we can manage using standard values from the table 5 of Normal distribution at page # 423 or using our calculator. Look at example 3. Important Information: All normal density curves satisfy the following property which is often referred to as empirical rule: 68.26% of the observations fall within 1 standard deviation of mean. 95.44% of the observations fall within 2 standard deviation of mean 99.74% of the observations fall within 3 standard deviation of mean Note: Within 5 standard deviation of mean we assume 100% data points. Example 1. Find the mean and standard deviation of the normal distribution whose pdf is given as  EMBED Equation.3  Solution: Compare with the standard formula of pdf for the normal distribution and find that  EMBED Equation.3 . Example 2. Write the pdf of a normal distribution with mean 3 and variance 16. Solution: We have  EMBED Equation.3 , the pdf of the normal distribution is given as  EMBED Equation.3  Example 3. Find the area under the normal curve with mean zero and standard deviation one for the standard variate  EMBED Equation.3 .  Solution: From table 5a:   EMBED Equation.3  For this value choose row with 1.2 and column 0.04.  0 1.24 Using calculator:  EMBED Equation.3  The calculator code:  2nd DISTR 2 normalcdf (-5, 1.24) =0.8925120 Example 4. Find the area under the normal curve with mean zero and standard deviation one for the standard variate  EMBED Equation.3 .  Solution: From table 5a:   EMBED Equation.3  For this value choose row with 1.2 and column 0.04.  0 1.24 Using calculator:  EMBED Equation.3  The calculator code: 2nd DISTR 2 normalcdf ( 1.24, 5) =0.1074875 Example 5. Find the area under the normal curve with mean zero and standard deviation one for  EMBED Equation.3 .  Solution: From table 5a:   EMBED Equation.3    -0.12 0 1.24 Using calculator:  EMBED Equation.3  The calculator code:  2nd DISTR 2 normalcdf (-0.12, 1.24) =0.4402707 Example 6. Suppose x is a normally distributed random variable with mean 10.2 and standard deviation 1.5. Find each of the following probabilities.  EMBED Equation.3 .  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Draw normal curve and show the region bounded by the normal curve and the x values. Solution: from calculator a)  EMBED Equation.3  try similar way for b), and c),   EMBED Equation.3  Try for e). 6.1 10.2 13.3 Exercise Set The physical fitness of an athlete is often measured by how much oxygen the athlete takes in (which is recorded in millimeters per kilogram, ml/kg). The maximum oxygen uptake for elite athletes has been found to be 80 with a standard deviation 9.2. Assume that distribution is approximately normal. What is the probability that an elite athlete has a maximum oxygen uptake of at least 75 ml/kg? Answer: 70.66% What is the probability that an elite athlete has a maximum oxygen uptake of 65 ml/kg or lower? Answer: 5.15% Consider someone with a maximum oxygen uptake of 26 ml/kg. Is it likely that this person is an elite athlete? Answer: No The combined score of SAT 1 test are normally distributed with mean of 998 and a standard deviation of 202. If a college includes a minimum score of 800 among its requirements, what percentage of students do not satisfy that requirement? Answer: 16.35% IQ score are normally distributed with mean of 100 and a standard deviation 15. Mensa is an international society that has one and only one qualification for membership, a score in the top 2 on an IQ test. What IQ score should one have in order to be eligible for Mensa? Answer: hint: (x-100)/15 = invnorm(0.98), x = 130.81 In a typical region of 90,000 people, how many are eligible for Mensa? Answer: 90,000 (0.02) = 1800 Using diaries for many weeks, a study on the lifestyle of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.6 hours of sleep, with a standard deviation of 2.56 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed. What is the probability that a visually impaired student gets less than 6.1 hours of sleep? Answer: 8.58% What is the probability that a visually impaired student gets between 6.3 and 10.35 hours of sleep? Answer: 51.65% Forty percent of students get less than how many hours of sleep on a typical day? Answer: 8.95 hours Healthy people have body temperatures that are normally distributed with a mean of 98.20 degree Fahrenheit and a standard deviation of 0.62 degree Fahrenheit. If a healthy person is randomly selected, what is the probability that he or she has a body temperature above 98.9 degree Fahrenheit? Answer: 12.94% A hospital wants to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 1% of healthy people to exceed it? Answer: hint: (x-98.2)/.62 = invnorm(0.99), 99.64 The heights of a large group of people are assumed to be normally distributed. Their mean height is 68 inches, and the standard deviation is 4 inches. What percent of these people are taller than 73 inches? Answer: 10.56% Suppose a population is normally distributed with a mean of 24.6 and a standard deviation of 1.3. What percent of the data will lie between 25.3 and 26.8? 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L^`LhH. ^`hH.MsoDataStore-!wHGLJ54JMNRA==2.!wItem /DPropertiesHU      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Oh+'0_  4 @ L Xdlt|MAT 211 - Final Mathematics Normal.dotmFiroz19Microsoft Office Word@e'@6G=@rh%@ rVdG4^VT$m   /&" WMFCN hhlVT$m EMFhd \KhC   Rp@Times New Roman <RQ`2<4  $Q`2<4 Id/14< <:d/1 <XG* Times ew Roman4h8'1x%1<:dv% % %   TX@E@?LtNovember 20, 2011: ?T;;4`A44:;:;:;&T9 X@E@?LxBasic Probability RuleN;-!4G4:A;@ !!&;SA!4TT: V X@E@: ?LP TW X@E@W ?L|(Material for final exam'n:'43!:!':4'!@;!3:;`TT$X@E@?LP |T%X@E@%? L`Fall 2011G;!!::;;TTX@E@?LP)&TTTX@E@?LP ? Rp@Times New Roman <RQ`2<4  $Q`2<4 Id/14< L;d/1 <XG* Times ew Roman4h8'1x%1L;dv% % %  TTV@E@LP |-Rp@Times New Roman xRQ`2xp \$Q`2xp Id/1px \<d/1 <XG* Times ew Roman19118'1x%1\<dv% % % Rp@"Arial xRQ`2xp \$Q`2xp Id/1px l=d/1 <X3.* Arial0l28'1x%1l=dv% % % % % %  % % % TX"8@E@#LP1.2% % % TT#m8@E@##LP K% % % Tpnz:@E@n#LXGiven H2,2!F(GDICzv N!b K!b     { !% %     Rp@Times New Roman !%w.&1@\'X'%&1\'ܿX'%\'0X'}ĭ X'V&1!OEd&1СaJ'|$,'1}0'10}+'1t^Zwtp$w" !dv%       `RpTimes New Roman1w[&w !%w@: @s wwdb ָwwV !b @Ties New Romandv% T 1 ;3??#L{0,1,2,3,4,5,6,7,8,9},{1,2,3,5,6,7}`RpTimes New Roman1w[&w !%w@(5( !{0,1,2,3,4,5,6,7,8,9,,| A| ]|5((@Ties New `wdv% ( TX= ;3??LPUA`RpSymbol|!A|H]| T! . 5 . !UA( !{0,1,2,3,4,56,|. A|]|5 . . ]|5(̫Ti |`|]|dv% ( TX 83??LP==Rp"System|o"A|H]|c Ao"* . 5 . !== . !UA( !{0,12,|. A|]|5 . . ]|5 . |5(̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TTu  :@E@u #LP T`  :@E@ #LTand,23!F(GDIC N!b K!b        !% %     Rp@Times New Roman !%w !%w:2w !%w` !  2^Zwؚ|| |w!V!V @w !!Vo" !H]|c Bo"o"!!5 U?w Ut^Zwtp$wt" !dv%        `RpTimes New Roman1w[&w !%wA|H]| U!!]| wwd8 qwwV !8 q@Ties New Romandv% T 9;,-?? Ld{3,4,6,7,9}F`RpTimes New Roman1w[&w !%wA|5 !{3,4,6,7,9}8 qw,&" WMFC hh|A|]|5@Ties New `wdv% ( TT0  ;,-??LPB`RpSymbol|o"A|H]|c Co" . d5 . !B5 !{3,4,6,7,9},|. A|]|f5 . . ]|5̫Ti |`|]|dv% ( TT N 8,-??LP=Rp"System|!A|H]| V!* . 5 . !=5 . !B5 !{3,46,|. A|]|5 . . ]|f5 . |5̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % Tx:@E@#L\. Find722!F(GDIC]N!b K!b      !% %     Rp@Times New Roman !%w !%w:2w !%w` !  8^Zwؚ|| |wo"V!Vc D !!V! !H]| W!!!!>5 D?wc Dt^Zwtp$w4'" !dv%       6`RpTimes New Roman1w[&w !%wA|H]|c Do"o"]| wwdb ظwwV !b @Ties New Romandv% TX;d??LP()zj`RpTimes New Roman1w[&w !%wA| . 5 . !()db ظw,|. A|]|5 . . @Ties New `wdv% ( T`;d??LTpABtVJRpSymbol|!A|H]| X!!H5! !pAB !()d,|!A|]|L5!!]|5 . ̫Ti |`|]|dv% ( TT2d??LP`TTq98d??LPRp"System|o"A|H]|c Eo"* . 5 . !5! !pAB !(),|. A|]|5 . . ]|L5!|5 . ̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TT\t:@E@\#LP.TTu:@E@u#LP -!"  T|C@E@L\Answer: H2'H,!!F(GDICCMQ!b K!b     DN P!% %     Rp@Times New Roman !%w !%w:2w !%w` !  >^Zwؚ|| |w!V!V @w !!Vo" !H]|c Fo"o"!!&5 Y?w Yt^Zwtp$w4" !dv%      &% 6 nRpTimes New Roman1w[&w !%wA|H]| Y!!]| wwd8 swwV !8 s@Ties New Romandv% TTNTA?E?LP46`RpTimes New Roman1w[&w !%wA| . 5 . !4wd8 sw,|. A|]|5 . . @Ties New Romandv% ( TXA?E?LP()zRpTimes New Roman|WA|x]| ZW . r5 . !() . !4wd,|. A|]|t5 . . ]|5 . ̫Ti |`|]|dv% ( TTMRA?E?LP5j`RpTimes New Roman|!A|H]| ! . 5 . !55 . !() . !4w,|. A|]|5 . . ]|t5 . |5 . ̫ |`|]|dv% ( T`U1A?E?LTpABtVJRpSymbol|!A|H]| [!!<5! !pAB !55 . !() . ,|!A|]|@5!!]|5 . |t5 . ̫ .  |`|]|dv% ( TTpA?E?LP`TX A?E?LP=&% Rp &" WMFC hmh"System|!A|H]| !* . 5 . !=! !pAB !55 . ,|. A|]|5 . . ]|@5!|5 . ̫ .  |`|]|dv% ( % % 0 % K( ( ( % % " % ( " " FGDIC" % % % TT@E@LP -Rp@Times New Roman <RQ`2<4  $Q`2<4 Id/14< |>d/1 <XG* Times ew Roman4h8'1x%1|>dv% % %  TTN@E@LP % % % % TX"@E@LP2.2% % % TT#m@E@#LP K% % % Tn@E@nLFrom the given table find 7!2N2,12,2,2,!23!F(GDIC 2!b K!b       !% %     Rp @Times New Roman !%w !%w:2w !%w` ! %w^Zwؚ|| |w!V!V @w !!V! !H]| !!7P5 \?w \t^Zwtp$w" !dv%        6`RpTimes New Roman1w[&w !%wA|H]| \!!]| wwdx! wwV ! @Ties New Romandv% Tlb ,-??LX(|)()Pj`RpTimes New Roman1w[&w !%wA|!5! !(|)()x! w,|!A|H]|5!!@Ties New `wdv% ( Tp ,-??LXpCFpCFJJ`RpSymbol|!A|H]|c H!!R5! !pCFpCF !(|)()x!,|!A|H]|X5!!]|5!̫Ti |`|]|dv% ( T` ,-??LT+=ZRp"System|!A|H]| ]!*!5! !+=! !pCFpCF !(|)(,|!A|]|5!!]|X5!|5!̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TT I @E@ LP -!" !6 % % %  TTq6@E@qL6 P 2" ! 6 % % %  TT6@E@L 6PBCTT6@E@L 6P 2" !6 % % %  TT,6s@E@,L6PCHTTt6@E@tL6P 2" !!6 % % %  TT6@E@L!6PDHTT6A@E@L!6P 2" !60  % % %  TTO 6 @E@O L60 PKNTT 6 @E@ L60 P 2" !4 6 % % %   Tlq 6C @E@q L4 6 XTotal=2,TTD 6u @E@D L4 6 P 2" ' % Ld252!??%  % Ld252!??% ( ' % Ld2520!??% ( ' % Ld 2 5 2!??% ( ' % Ld 25 2s!??% ( ' % Ld252!??% ( ' % Ld252!??% ( ' % Ld252!??% ( ' % Ld 25 2!??% ( ' % Ld252!??% &" WMFC hMh( ' % Ld2/ 52s!??% ( ' % Ld0 23 50 2!??% ( ' % Ld4 2 54 2M!??% ( ' % Ld 2 5 2!??%  % Ld 2 5 2!??% ( ' % Ld66o!??% ( ' % Ld 6  6o!??% ( ' % Ld66o!??% ( ' % Ld66o!??% ( ' % Ld66o!??% ( ' % Ld0 63 0 6o!??% ( ' % Ld 6  6o!??% ( !  % % %  TTO@E@OL PECTT@E@L P 2" !  % % %  TT@E@L P72TT@E@L P 2" ! % % %  TT7h@E@7LP32TTi@E@iLP 2" !! % % %  TT@E@L!P72TT6@E@L!P 2" !0  % % %  TT]  @E@] L0 P52TT  @E@ L0 P 2" !4   % % %  TX  @E@ L4  P2222TT = @E@ L4  P 2" ' % Ld!??% ( ' % Ld0!??% ( ' % Ld   !??% ( ' % Ld  s!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ld  !??% ( ' % Ld!??% ( ' % Ld/ s!??% ( ' % Ld0 3 0 !??% ( ' % Ld4  4 M!??% ( ' % Ld   !??% ( ' % Ldo!??% ( ' % Ld   o!??% ( ' % Ldo!??% ( ' % Ldo!??% ( ' % Ldo!??% ( ' % Ld0 3 0 o!??% ( ' % Ld   o!??% ( !  % % %  TTR@E@RtL PF=TT@E@tL P 2" !  % % %  TT@E@tL &" WMFC h-hP82TT@E@tL P 2" ! % % %  TT7h@E@7tLP62TTi@E@itLP 2" !! % % %  TX@E@tL!P1122TTO@E@tL!P 2" !0  % % %  TT]  @E@] tL0 P42TT  @E@ tL0 P 2" !4   % % %  TX  @E@ rL4  P2922TT = @E@ rL4  P 2" ' % Ld!??% ( ' % Ld0!??% ( ' % Ld   !??% ( ' % Ld  s!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ld  !??% ( ' % Ld!??% ( ' % Ld/ s!??% ( ' % Ld0 3 0 !??% ( ' % Ld4  4 M!??% ( ' % Ld   !??% ( ' % Ldn!??% ( ' % Ld   n!??% ( ' % Ldn!??% ( ' % Ldn!??% ( ' % Ldn!??% ( ' % Ld0 3 0 n!??% ( ' % Ld   n!??% ( !  % % %  Tl@E@L XTotal=2,TT@E@L P 2" !  % % %  TX@E@L P1522TT)@E@L P 2" ! % % %  TT7h@E@7LP92TTi@E@iLP 2" !! % % %  TX@E@L!P1822TTO@E@L!P 2" !0  % % %  TT]  @E@] L0 P92TT  @E@ L0 P 2" !4   % % %  TX  @E@ L4  P5122TT = @E@ L4  P 2" ' % Ld!??% ( ' % Ld0!??% ( ' % Ld   !??% ( ' % Ld  s!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ld  !??&" WMFC h h% ( ' % Ld!??% ( ' % Ld/ s!??% ( ' % Ld0 3 0 !??% ( ' % Ld4  4 M!??% ( ' % Ld   !??% ( ' % Ldp!??% ( ' % Ld!??%  % Ld!??% ( ' % Ld0!??% ( ' % Ld   p!??% ( ' % Ld   !??% ( ' % Ld  s!??% ( ' % Ldp!??% ( ' % Ld!??% ( ' % Ld!??% ( ' % Ldp!??% ( ' % Ld!??% ( ' % Ld  !??% ( ' % Ldp!??% ( ' % Ld!??% ( ' % Ld/ s!??% ( ' % Ld0 3 0 p!??% ( ' % Ld0 3 0 !??% ( ' % Ld4  4 M!??% ( ' % Ld   p!??% ( ' % Ld   !??%  % Ld   !??% (  % % %  TT]@E@JLP % % % % T|_C@E@L\Answer: H2'H,!!F(GDICC]!b K!b      D^ !% %     Rp @Times New Roman !%w !%w:2w !%w` !  L^Zwؚ|| |w!V!Vc I !!V! !H]| ^!!!!H5 I?wc It^Zwtp$w@&# !dv%        6`RpTimes New Roman1w[&w !%wA|H]|c I!!]| wwdx! _wwV ! _ @Ties New Romandv% Tl_%,-??LX(|)()Pj`RpTimes New Roman1w[&w !%wA|!5! !(|)()x! _w,|!A|H]|5!!@Ties New `wdv% ( TpU_,-??LXpCFpCFJJ`RpSymbol|!A|H]|c J!!R5! !pCFpCF !(|)()x!,|!A|H]|X5!!]|5!̫Ti |`|]|dv% ( T`T],-??LT+=ZRp"System|!A|H]|N k!*!5! !+=! !pCFpCF !(|)(,|!A|]|5!!]|X5!|5!̫ |`|]|dv% ( % % 0 % &" WMFC hhK( % % " % ( " " FGDIC" % % % T_ @E@Lt160/493=0.3245436105222222822222222222TT _ @E@ LP - % % % TT>@E@+LP % % % % TX9"@E@LP3.2% % % TT#8m@E@#LP K% % % Tn;@E@neLA family has two children. Using b to stand for boy and g for girl in ordered pairs write the sample H!,N03,'H2,22!-2H'2122',22!2!230,221!2!1!22!2-!-22,!'H!,2,',N2,!"  THn @E@n *Lspace then find the event E that the fami'2,,,2,2!222,,2,2=2,2,!,NT @E@ ;Lly has exactly one daughter and the event F the family has 03,',3,,022,3,312,!,222-,2,272,!,N03,'!"  Tn @E@ns 5Lat least one daughter. Now identify which one is true,,,'22,2,312-!H3H2,2"0H2,222,'!2-TT  @E@ s LP -!"  TT  @E@ LP -Rp @"Arial xRQ`2xp \$Q`2xp Id/1px ?d/1 <X3.* Arial0l28'1x%1?dv% % %  % % % T`n g @E@nP LT(A)!H"% % % TT e @E@P LP % % % T g @E@P L|E and F are independent=,227-!,22,3,22,2T g @E@P L` Td fg @E@P LT(B) !B!Tg g @E@gP LxE and F are dependent=,227,",2,3,22,2TT  g @E@ P LP - % % % TXnj  @E@n LPC)C!% % % TTj  @E@ LP 2% % % THj @E@ *LE and F are mutually exclusive that means =,227-!,N22,0,3-2'2,2,N,,2'!F(GDIC h p !b K!b      i !% %     Rp@Times New Roman !%w !%w:2w !%w` !  p"^Zwؚ|| |w!V!Vc K !!V! !H]|N l!!!!H5 Kc Kt^Zwtp$wD# !dv%       F`RpTimes New Roman1w[&w !%wA|p"@ wwdp!N mwwV !N m@Ties New Romandv% TX j I `??LPEFb`RpSymbol1w[&w !%wA . 5 . !EFdp!N mw,|. A|]|5 . . @Ties New `wdv% ( TT& ` `??LPf~`RpSymbol|!A|H]|c L!* . H5 . !f5 . !EFdp!,|. A|]|J5 . . ]|5 . ̫Ti |`|]|dv% ( TX" ` `??LP=Rp"System|!A|H]| ˸!* . 5 . != . !f5 . !EF,|. A|]|5 . . ]|J5 . |5 . ̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % Txo j  @E@o L\ Td j @E@ LT(D) !I!Tx&" WMFC hh j  @E@ L\None ofH22,2!TTj  @E@ LP Tj  @E@ L`the given2,13,3TTj  @E@ LP - TT [ @E@D LP - TY  @E@ LxAnswer: (B) DependentH2'H,!"B!I,2,23,2TTY  @E@ LP - TT 9 @E@" LP - % % % TX5 " @E@ LP4.2% % % TT#4 m @E@# LP K% % % Tn7 @E@n KLRoll a die twice. Write the sample space. Find the following probabilities:C2,2,H,,_!,2,',N2,'2,-,7222,!22H212!23,2,'TT 7 ( @E@ LP -!"  % % % !F(GDIC @ !b K!b    `@  !% %     Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w!V!Vc M !!V! !H]| ̸!!!!5 M?wc Mt^Zwtp$w@X# !dv%      `@ fRp@Times New Roman1w[&w !%wA|H]|c M!!]| wwdp!b ݸwwV !b @Ties New Romandv% Tb P ( L4??L()(rolling a sum 5 or more){ww~:pWpppWGRp@Times New Roman1w[&w !%wA|!5! !()(rolling a sum 5 o ,|!A|X ]|5!!@Ties Ne`w`wdv% ( T` e( L4??LTPEP8Rp@Symbol|!A|H]| ͸!!5! !PEP! !()(rolling as,|!A|]|5!!]|5!̫Ti |`|]|dv% ( TT & L4??LP=Rp"System|!A|H]|c N!* . 5 . !=5! !PEP! !()(rl,|. A|]|5 . . ]|5!|5!̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TT @E@ LP % % % % TT;  @E@ LP - T @E@ LdAnswer: 5/6H2'H,!23TT @E@LP - TT@E@sLP -Rp@Times New Roman HRQ`2H@ ,$Q`2H@ Id/1@H @d/1 <XG* Times ew Romant8'1x%1@dv% % %  % % % TX"@E@LP5.2% % % TT#m@E@#LP K% % % Tn@E@nLlSuppose events 82222',,2,2'% % % TT.@E@LPA=% % % TT/G@E@/LP TdH@E@HLTand ,22% % % TT-@E@LPB=% % % TT.G@E@.LP TH4 @E@HLare independent and that ,!,22,3,22,2,222,% % % TT* f @E@5 LPp2% % % TTg  @E@g LP(!% % % TT  @E@ LPA=% % % Tx  @E@ L\) = 0.4!822TT  @E@ LP Td  @E@ LTand ,22% % % TT  @E@ LPp2% % % TT  @E@ LP(!% % % TT  @E@ LPB=% &" WMFC hh% % T +@E@ Lh) = 0.6. Find!823722!F(GDIC+!b K!b      - !% %     Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w!V!V !!V! !H]|c O!!!!5 ?w Θt^Zwtp$wq# !dv%        6`RpTimes New Roman1w[&w !%wA|H]| θ!!]| wwdh!c PwwV !c P@Ties New Romandv% TX~Z??LP()&j`Rp Times New Roman1w[&w !%wA| . 5 . !()dh!c Pw,|. A|]|5 . . @Ties New `wdv% ( T`>??LTpABt`RpSymbol|!A|H]| ϸ!!H5! !pAB !()dh!,|!A|]|L5!!]|5 . ̫Ti |`|]|dv% ( TT??LPRp "System|!A|H]|F !* . 5 . !5! !pAB !(),|. A|]|5 . . ]|L5!|5 . ̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TT@E@LP.TT @E@LP -!"  TTn ~@E@ngLP - TTn|@E@nLPAHT|@E@ Ldnswer: 0.242'H,!223TT|@E@LP - TT[@E@DLP - % % % TX\"@E@LP6.2% % % TT#[m@E@#LP K% % % Tn^@E@nLlSuppose events 82222',,2,2'% % % TT^.@E@LPA=% % % TT/^G@E@/LP TdH^@E@HLTand ,22% % % TT^-@E@LPB=% % % TT.^G@E@.LP TH^4 @E@HLare independent and that ,!,22,3,22,2,222,% % % TT* ^f @E@5 LPp2% % % TTg ^ @E@g LP(!% % % TT ^ @E@ LPA=% % % T ^ @E@  Ld) = 0.3 and !822,22% % % TT ^ @E@ LPp2% % % TT ^ @E@ LP(!% % % TT ^ @E@ LPB=% % % T ^+@E@ Lh) = 0.6. Find!823722!F(GDIC+\!b K!b      -^ !% %     Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w!V!V !!V! !H]|F !!!!5 ?w Иt^Zwtp$w# !dv%        6`Rp Times New Roman1w[&w !%wA|H]| и!!]| wwdh!N pwwV !N p@Ties New Romandv% TX~_Z??LP()&j`RpTimes New Roman1w[&w !%wA| . 5 . !()dh!N pw,|. A|]|5 . . @Ties New `wdv% ( T`>_??LTpABt`Rp Symbol|!A|H]|F !!H5! !pAB !()dh!,|!A|]|L5!!]|5 . &" WMFC hh̫Ti |`|]|dv% ( TTU??LPRp"System|!A|H]| Ѹ!* . 5 . !5! !pAB !(),|. A|]|5 . . ]|L5!|5 . ̫ |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TT^@E@LP -!"  TTnP@E@n9LP - TnM@E@n LdAnswer: 0.72H2'H,!223TTM@E@LP - TT-@E@LP - % % % TX."@E@LP7.2% % % TT#-m@E@#LP K% % % Tn0B@E@n LdGiven that H2,22,% % % TT80t@E@CLPp2% % % TTu0@E@uLP(!% % % TT0@E@LPA=% % % T0%@E@ L`) = 0.4, !822% % % TT0X@E@&LPp3% % % TTY0y@E@YLP(!% % % TTy0@E@zLPB=% % % T0@E@ Lh) = 0.5 and !822,22!F(GDIC. !b K!b      / !% %     Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w! V!VF !!V! !H]| Ҹ!!!!5 ?wF ݘt^Zwtp$w# !dv%        RpTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!c RwwV !c R@Ties New Romandv% TT 0 ??`LP3TT 0 ??H`LP.`TT 0 ??`LP0TT 0( ??s`LP)TT0 ??6`LP(Rp Symbol1w[&w !%wA|5 !(wdh!c RwwV !c R@Ties New Romandv% ( TTC &y ??W`LP=TTf & ??`LP$RpTimes New Roman|!A|H]| Ӹ!*~5 !5 !(wdh!c R wV !̫Ti |`|]||dv% ( TT 0 ??}`LPBTT 0U ??`LPATT0??j`LPpRp "System|!A|H]|F !5 !p5 !5 !(wh! c R wV !̫ |`|]||dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % Tx 0 @E@ L\. Find 722!F(GDIC . !b K!b     ! / !% %     Rp@Times New Roman !%w !%w:2w !%w` ! %w^Zwؚ|| |w!V!V !!V! !H]|F !!!!L5 ?w Ԙt^Zwtp$w# !dv%       Rp Times New Roman1w[&w !%wA|H]| Ը!!]| wwdh!N rwwV !N r@Ties New Romandv% TT 0 ??4`LP)TTr 0 ??6`LP(RpTimes New Roman1w[&w !%wA|5 !(wdh!N rwwV !N r@Ties New Romandv% ( TTd 0 ??`LPBTT 0 ??`L&" WMFC hmhPATT2 0n ??j`LPpRp Symbol|!A|H]|F !^5 !p5 !(wdh!N r wV !̫Ti |`|]||dv% ( TT ! ??JLP`TT &N ??d`LP$TT ! ??JLP`Rp"System|!A|H]| ո!*5 !5 !p5 !(wh! N r wV !̫ |`|]||dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TX 0$@E@ LP. TT%0Q@E@%LP -!"  TTn!@E@n LP - T@E@y LdAnswer: 0.70H2'H,!223TT@E@yLP - TT@E@LP - % % % TX"m@E@XLP8.2% % % TT#mm@E@#XLP K% % % Txno@E@nXL\Events =2,2'!F(GDIC!b K!b      d}!% %     Rp@Times New Roman !%w !%w:2w !%w` ! %w^Zwؚ|| |w! V!VF !!V! !H]| ָ!!!!,5 ?wF t^Zwtp$w4Z6 !dv%       RpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!c TwwV !c T@Ties New Romandv% TTC5y,-??LP1rT@Rp Times New Roman1w[&w !%wA| . 5 . !1wdh!c Tw,|. A|]|5 . . @Ties New `wdv% ( TTTk,-??LPARp"System|!A|H]| ׸! . F5 . !A5 . !1wdh!,|. A|]|H5 . . ]|5 . ̫Ti |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TX,o@E@XLP, !F(GDIC,!b K!b     . l}!% %     Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w! V!VF !!V! !H]| ظ!!!!5 ?wF t^Zwtp$wc6 !dv%       RpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!N twwV !N t@Ties New Romandv% TTpCy??LP2rT@Rp Times New Roman1w[&w !%wA| . 5 . !2wdh!N tw,|. A|]|5 . . @Ties New `wdv% ( TT@k??LPARp"System|!A|H]| ٸ! . F5 . !A5 . !2wdh!,|. A|]|H5 . . ]|5 . ̫Ti |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TTo@E@XLP Td[o@E@XLTand ,22!F(GDIC[!b K!b     ] k}!% %     Rp@Times New Roman !%w !%w:2w !%w` !  Ę^Zwؚ|| |w! V!VF !!V! !H]| ڸ!!!!5 ?wF t^Zwtp$wm6 !dv%       RpGTimes New Ro&" WMFC hMhman1w[&w !%wA|H]|F !!]| wwdh!c VwwV !c V@Ties New Romandv% TTCy\p??LP3rT@Rp Times New Roman1w[&w !%wA| . 5 . !3wdh!c Vw,|. A|]|5 . . @Ties New `wdv% ( TTok\p??LPARp"System|!A|H]| ۸! . F5 . !A5 . !3wdh!,|. A|]|H5 . . ]|5 . ̫Ti |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TTo@E@XLP To@E@X:Lform a partition of the sample space S with probabilities !2!N,2,!222!2,',N2,'2,,,8H22!22,2,'!F(GDICa!b K!b    @  }!% %     Rp@Times New Roman !%w !%w:2w !%w` !  ʘ^Zwؚ|| |w! V!VF !!V! !H]| ܸ!!!!5 ?wF t^Zwtp$wx6 !dv%      @ 3RpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!N vwwV !N v@Ties New Romandv% TTSCyJ??LP1r@Rp Times New Roman1w[&w !%wA| . 5 . !1wdh!N vw,|. A|]|5 . . @Ties New `wdv% ( TlkJ??LX()0.4PE@RpTimes New Roman|!A|H]| ݸ!!F5! !()0.4 !1wdh!,|!A|H]|L5!!]|5 . ̫Ti |`|]|dv% ( TXkJ??LPPAYe@Rp Symbol|!A|H]|F ! . 5 . !PA! !()0.4 !1w,|. A|]|5 . . ]|L5!|5 . ̫ |`|]|dv% ( TTIiJ??LP=Rp"System|!A|H]| ޸!* . 5 . !=5 . !PA! !()0.,|. A|]|5 . . ]|5 . |L5!̫ .  |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TX`o@E@`XLP, !"  !F(GDICm!b K!b    @ o }!% %     Rp@Times New Roman !%w !%w:2w !%w` !  И^Zwؚ|| |w! V!VF !!V! !H]| ߸!!!!5 ?wF t^Zwtp$wD6 !dv%      @ HRpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!c XwwV !c X@Ties New Romandv% TTf??LP2r@Rp Times New Roman1w[&w !%wA| . 5 . !2wdh!c Xw,|. A|]|5 . . @Ties Ne`w`wdv% ( Tl~??LX()0.3PE@RpTimes New Roman|!A|H]| !!F5! !()0.3 !2wdh!,|!A|H]|L5!!]|5 . ̫Ti |`|]|dv% ( TX??LPPAY@Rp Symbol|!A|H]|F ! . 5 . !PA! !()0.3 !2w,|. A|]|5 . . ]|L5!|5 . ̫ |`|]|dv% ( TTZ??&" WMFC h-hLP=Rp"System|!A|H]| !* . 5 . !=5 . !PA! !()0.,|. A|]|5 . . ]|5 . |L5!̫ .  |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % Td@E@LTand ,22!F(GDICX!b K!b    @  }!% %     Rp@Times New Roman !%w !%w:2w !%w` !  Ę^Zwؚ|| |w! V!VF !!V! !H]| !!!!5 ?wF t^Zwtp$w6 !dv%      @ BRpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!N xwwV !N x@Ties New Romandv% TTN??LP3r@Rp Times New Roman1w[&w !%wA| . 5 . !3wdh!N xw,|. A|]|5 . . @Ties New `wdv% ( Tl??LX()0.3PE@RpTimes New Roman|!A|H]| !!F5! !()0.3 !3wdh!,|!A|H]|L5!!]|5 . ̫Ti |`|]|dv% ( TX??LPPAY}@Rp Symbol|!A|H]|F ! . 5 . !PA! !()0.3 !3w,|. A|]|5 . . ]|L5!|5 . ̫ |`|]|dv% ( TTE??LP=Rp"System|!A|H]| !* . 5 . !=5 . !PA! !()0.,|. A|]|5 . . ]|5 . |L5!̫ .  |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TTW@E@WLP -!"  T@E@jLIf E is an event in S with "=',2,3,228H2!F(GDIC!b K!b      }!% %     Rp@Times New Roman !%w !%w:2w !%w` !  ݘ^Zwؚ|| |w! V!VF !!V! !H]| !!!!5 ?wF t^Zwtp$wL6 !dv%       RpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!c ZwwV !c Z@Ties New Romandv% TTV0??LP1r@Rp Times New Roman1w[&w !%wA| . 5 . !1wdh!c Zw,|. A|]|5 . . @Ties New `wdv% ( Tp!E~??LX(|)0.5PE@RpTimes New Roman|!A|H]| !!F5! !(|)0.5 !1wdh!,|!A|H]|L5!!]|5 . ̫Ti |`|]|dv% ( T`!O~??LTPEAJ@Rp Symbol|!A|H]|F !!5! !PEA! !(|)0.5 !1w,|!A|]|5!!]|L5!|5 . ̫ |`|]|dv% ( TT|??LP=Rp"System|!A|H]| !* . "5 . !=5! !PEA! !(|)05,|. A|]|$5 . . ]|5!|L5!̫ .  |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TX@E@jLP, !F(GDIC !b K!b       }!% %  &" WMFC h h   Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w! V!VF !!V! !H]| !!!!5 ?wF t^Zwtp$wį6 !dv%       RpGTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!N zwwV !N z@Ties New Romandv% TT Vk ,-??LP2r@Rp Times New Roman1w[&w !%wA| . 5 . !2wdh!N zw,|. A|]|5 . . @Ties New `wdv% ( TpO! ~,-??LX(|)0.4PE@RpTimes New Roman|!A|H]| !!F5! !(|)0.4 !2wdh!,|!A|H]|L5!!]|5 . ̫Ti |`|]|dv% ( T`! ~,-??LTPEAJ$@Rp Symbol|!A|H]|F !!5! !PEA! !(|)0.4 !2w,|!A|]|5!!]|L5!|5 . ̫ |`|]|dv% ( TT_  |,-??LP=Rp"System|!A|H]| !* . "5 . !=5! !PEA! !(|)04,|. A|]|$5 . . ]|5!|L5!̫ .  |`|]|dv% ( % % 0 % K( % % " % ( " " FGDIC" % % % TT - @E@ jLP Td.  @E@. jLTand ,22!F(GDIC  !b K!b         !% %     Rp@Times New Roman !%w !%w:2w !%w` !  ^Zwؚ|| |w! V!VF !!V! !H]| !!!!5 ?wF t^Zwtp$w6 !dv%       &% S6WS Rp GTimes New Roman1w[&w !%wA|H]|F !!]| wwdh!c \wwV !c \@Ties New Romandv% TT V9 ,-?=?LP3rRp Times New Roman1w[&w !%wA| . 5 . !3wdh!c \w,|. A|]|5 . . @Ties New Romandv% ( Tp P ~,-?=?LX(|)0.3PERp Times New Roman|{A|H]|F {{r5{ !(|)0.3 !3wdh!,|{A|H]|x5{{]|5 . ̫Ti |`|]|dv% ( T` T ~,-?=?LTPEAJRp Symbol|{A|H]| |!5! !PEA !(|)0.3 !3w,|!A|]|5!!]|x5{|5 . ̫ |`|]|dv% ( TT+  |,-?=?LP=& % Rp "System|{A|H]|F {* . N5 . !=5! !PEA !(|)03,|. A|]|P5 . . ]|5!|x5{̫ .  |`|]|dv% ( % % 0 % K( ( ( % % " % ( " " FGDIC" % % % TT  @E@ jLP Tl  @E@ jLXthen 2,2Tl {@E@ jLXfind !22T|K@E@|jLtthe probability of 2,2!23,202!!F(GDICK|!b K!b    ` L \u!% %     Rp@Times New Roman !%w !%w:2w !%w` ! %w^Zwؚ|| |w| V!V @w !!Vp! !H]|F x!x!!!5 ?w ɘt^Zwtp$w06 !dv%      `& % @S6 S ERp Times New Roman& WMFCh h1w[&w !%wA|H]| ||]| wwdh!N |wwV !N |@Ties New Romandv% TT^ ~?O?LPE&% Rp "System1w[&w !%wA| . 5 . !Ewdh!N |w,|. A|]|5 . . @Ties New Romandv% ( % % 0 % K( ( ( % % " % ( " " FGDIC" % % % TT@E@jLP TT@E@jLP - TT@E@LP - To@E@X LdAnswer: 0.47H2'H,!223TTo@E@XLP - TTm@E@LP - TTM@E@6LP - TTK@E@LP -% % 666666666666666666666666666666666666 6 66 6  6 66 6  6 66 6  6 66 6  6 66 6 66666666666666666666  ^`.@Times New Roman--- (2 aP`\November 20, 2011:     ,2 a`\Basic Probability Rule       2 a`\ /2 a`\(Material for final exam       2 a`\ 2 a `\Fall 2011  2 a`\) 2 a`\  @Times New Roman--- 2 tP`\ @Times New Roman---@"Arial---------2 P`\1.--- 2 ]`\ ---2 i`\Given   ,\`,--.  @Times New Roman-. Times New Roman-<2 #{0,1,2,3,4,5,6,7,8,9},{1,2,3,5,6,7}        "   "Times New Roman- 2 UA"Symbol- 2 ==""System------'"System-'''--- 2 `\ 2 `\and ,\`,--.  @Times New Roman-. +Times New Roman-2 + {3,4,6,7,9} "Times New Roman-  2 B"Symbol- 2 =""System- --- --'-'''---2 |`\. Find   ,\`,--.  @Times New Roman-. Times New Roman- 2 ()3"Times New Roman-  2 pAB"Symbol- 2 " 2 ""System- --- --'-'''--- 2 `\. 2 `\ ,\`'2 P`\Answer:  ,\`,--.  @Times New Roman-.- Times New Roman- 2 4"Times New Roman-  2 ()3"Times New Roman-  2 5"Times New Roman-  2 pAB"Symbol-  2 " 2 =("- "System-  ---  --'-'''--- 2 `\ @Times New Roman--- 2 P`\ ---2 P`\2.--- 2 ]`\ ---22 i`\From the given table find       ,\`,--.  @Times New Roman-. (Times New Roman- 2 ((|)()r 0"Times New Roman-  2 pCFpCF#"XSymbol-  2 X+=,&""System-  --- --'-'''--- 2 `\ ,\`',|--- 2 | ',---  2 B 2  ',4---  2 4C 2 4 ',z5---  2 Q5zD 2 ^5z ',{---  2 {K 2 { ',--- 2 TotalF  2  '- @ !{-- @ !{-- @ !3|-- @ !-- @ !?-- @ !-- @ !E-- @ !4-- @ !E5-- @ !z-- @ !?{-- @ !-- @ !8-- @ !-- @ !-- @ !{-- @ !-- @ !-- @ !4-- @ !z-- @ !-- @ !-,|---  2 |E 2 | ',---  2 7 2  ',4---  2 43 2 4 ',z5---  2 S5z7 2 \5z ',{---  2 {5 2 { ',--- 2 22  2  '- @ !{-- @ !3|-- @ !-- @ !?-- @ !-- @ !E-- @ !4-- @ !E5-- @ !z-- @ !?{-- @ !-- @ !8-- @ !-- @ !{-- @ !-- @ !-- @ !4-- @ !z-- @ !-- @ !-,|---  2 |F 2 | ',---  2 8 2  ',4---  2 46 2 4 ',z5--- 2 O5z11  2 `5z ',{---  2 {4 2 { ',--- 2 29  2  '- @ !{-- @ !3|-- @ !-- @ !?-- @ !-- @ !E-- @ !4-- @ !E5-- @ !z-- @ !?{-- @ !-- @ !8-- @ !-- @ !{-- @ !-- @ !-- @ !4-- @ !z-- @ !-- @ !-,/|--- 2 *|/TotalF  2 *|/ ',/--- 2 */15 2 */ ',/4---  2 *4/9 2 *4/ ',/z5--- 2 *O5z/18  2 *`5z/ ',/{---  2 *{/9 2 *{/ ',/--- 2 */51  2 */ '- @ !{-- @ !3|-- @ !-- @ !?-- @ !-- @ !E-- @ !4-- @ !E5-- @ !z-- @ !?{-- @ !-- @ !8-- @ !-- @ !{-- @ !/{-- @ !/{-- @ !3/|-- @ !-- @ !/-- @ !?/-- @ !-- @ !/-- @ !E/-- @ !4-- @ !/4-- @ !E/5-- @ !z-- @ !/z-- @ !?/{-- @ !-- @ !/-- @ !8/-- @ !-- @ !/-- @ !/----  2 <P`\ ---2 NP`\Answer:  ,\`,--.  @Times New Roman-. NTimes New Roman- 2 N(|)()r 1"NTimes New Roman-  2 NpCFpCF#"NSymbol-  2 N+=,&!"System-  --- --'-'''---)2 N*`\160/493=0.3245436105         2 N`\ --- 2 bP`\ ---2 sP`\3.--- 2 s]`\ ---2 sie`\A family has two children. Using b to stand for boy and g for girl in ordered pairs write the sample                   ,\`'J2 i*`\space then find the event E that the fami    d2 ~;`\ly has exactly one daughter and the event F the family has            ,\`'[2 i5`\at least one daughter. Now identify which one is true          2 `\ ,\`' 2 P`\ @"Arial------2 i`\(A) --- 2 `\ ---.2 `\E and F are independent      2 ' `\ 2 R`\(B)  +2 l`\E and F are dependentn       2 `\ ---2 i`\C) --- 2 z`\ ---J2 *`\E and F are mutually exclusive that means         ,\`,--.  @Times New Roman- . Times New Roman- 2 EF"Symbol-  2 f"Symbol-  2 = !"System-  --- --'- '''---2 `\ 2 `\(D)  2 )`\None of   2 ``\ 2 d `\the given   2 `\  2 P`\ +2 P`\Answer: (B) Dependent       2 `\  2 P`\ ---2 P`\4.--- 2 ]`\ ---|2 iK`\Roll a die twice. Write the sample space. Find the following probabilities:           2 ^`\ ,\`'---,\`,--.  @Times New Roman- . 5Times New Roman- 02 5()(rolling a sum 5 or more)'        '5Times New Roman-  2 5PEP((5Symbol-  2 5='"System-  --- --'- '''--- 2 5`\ --- 2 LP`\ 2 _P `\Answer: 5/6    2 _`\  2 qP`\ @Times New Roman- - - ---2 P`\5.--- 2 ]`\ ---"2 i`\Suppose events     - - - 2 `\A --- 2 `\ 2 `\and  - - - 2 `\B --- 2  `\ 12 `\are independent and that     - - - 2 `\p--- 2 `\(- - - 2 `\A ---2 `\) = 0.4   2 `\ 2 `\and  - - - 2 !`\p--- 2 )`\(- - - 2 /`\B --- 2 9`\) = 0.6. Find   ,\`,--.  @Times New Roman- . Times New Roman- 2 ()/!Times New Roman-  2 pAB"Symbol-  2 ""System-  --- --'- '''--- 2 `\. 2 `\ ,\`' 2 i`\  2 i`\A 2 v `\nswer: 0.24  2 `\  2 P`\ ---2 P`\6.--- 2 ]`\ ---"2 i`\Suppose events     - - - 2 `\A --- 2 `\ 2 `\and  - - - 2 `\B --- 2  `\ 12 `\are independent and that     - - - 2 `\p--- 2 `\(- - - 2 `\A ---2  `\) = 0.3 and    - - - 2 !`\p--- 2 )`\(- - - 2 /`\B --- 2 9`\) = 0.6. Find   ,\`,--.  @Times New Roman- . Times New Roman- 2 ()/!Times New Roman-  2 pAB"Symbol-  2 ""System-  --- --'- '''--- 2 `\ ,\`' 2 i`\ 2 i `\Answer: 0.72   2 `\  2 P`\ ---2 #P`\7.--- 2 #]`\ ---2 #i `\Given that   - - - 2 #`\p--- 2 #`\(- - - 2 #`\A ---2 # `\) = 0.4,   - - - 2 # `\p--- 2 #`\(- - - 2 #`\B ---2 ## `\) = 0.5 and    ,\`,--.  @Times New Roman-  .Times New Roman- 2 #3 2 #. 2 #0 2 #) 2 #(Symbol-  2 #= 2 # Times New Roman-  2 #B 2 #A 2 #zp "System-  --- --'- '''---2 #`\. Find   ,\`,--.  @Times New Roman-  .Times New Roman- 2 #T) 2 #(Times New Roman-  2 #EB 2 ##A 2 #pSymbol-  2 "P 2 #4 2 "-"System-  --- --'- '''---2 #]`\.  2 #e`\ ,\`' 2 8i`\ 2 KP `\Answer: 0.70     2 K`\  2 ^P`\ ---2 qP`\8.--- 2 q]`\ ---2 qi`\Events ,\`,--.  @Times New Roman- . tTimes New Roman- 2 t1qTimes New Roman-  2 qA"System-  --- --'- '''---2 q`\, ,\`,--.  @Times New Roman- . tTimes New Roman- 2 t2qTimes New Roman-  2 qA"System-  --- --'- '''--- 2 q`\ 2 q`\and  ,\`,--.  @Times New Roman- . tTimes New Roman- 2 t3qTimes New Roman-  2 qA"System-  --- --'- '''--- 2 q`\ b2 q:`\form a partition of the sample space S with probabilities         ,\`,--.  @Times New Roman- . tTimes New Roman- 2 t1qTimes New Roman-  2 q()0.4sqTimes New Roman-  2 qPAqSymbol-  2 q="System-  --- --'- '''---2 q`\, ,\`',\`,--.  @Times New Roman- . Times New Roman- 2 2vTimes New Roman-  2 v()0.3smTimes New Roman-  2 mPASymbol-  2 ="System-  --- --'- '''---2 `\and  ,\`,--.  @Times New Roman- . Times New Roman- 2 3Times New Roman-  2 ()0.3sTimes New Roman-  2 PASymbol-  2 ="System-  --- --'- '''--- 2 `\ ,\`'42 P`\If E is an event in S with       ,\`,--.  @Times New Roman- . 'Times New Roman- 2 '1Times New Roman-  2 (|)0.5Times New Roman-  2 PEA5Symbol-  2 5="System-  --- --'- '''---2 T`\, ,\`,--.  @Times New Roman- . Times New Roman- 2 2hTimes New Roman-  2 h(|)0.4_Times New Roman-  2 _PEASymbol-  2 ="System-  --- --'- '''--- 2 `\ 2 `\and  ,\`,--.  @Times New Roman- .-  Times New Roman- 2 3Times New Roman-  2 (|)0.3Times New Roman-  2 PEASymbol-  2 =- "System- ---  --'- '''--- 2 /`\ 2 3`\then 3 2 U`\find 3 (2 t`\the probability of   ,\`,--.  @Times New Roman- .-  Times New Roman- 2 E- "System- ---  --'- '''--- 2 `\  2 `\  2 P`\ 2 P `\Answer: 0.47     2 `\  2 P`\  2 P`\  2 P`\ "System- - \\``\\``\\``[[__[[__[[__[[__[[__[[__ZZ^^ZZ^^ZZ^^ZZ^^ZZ^^ZZ^^YY]]YY]]YY]]YY]] ages.wikia.com/wikite  F'Microsoft Office Word 97-2003 Document MSWordDocWord.Document.89qjz ^`hH. PL^P`LhH.Q)Ic>?VL#Wc hu}?<w J20sb A4[67sFkA4Vr[&4\K s{58B/r3,s3Im4-L?j,>o{Sz ,!WW8Num2WW8Num3WW8Num4WW8Num7!4        L#؅                                                                                                                    T        ̸,                 G                          xA\                                           Hw;        (|E        >        FEfHDj  yd"n ~/ f(/,N29HCnCJ1D}GNMIN!O4"T?PTUT1Y:xgeo{q*sWJyh|^p}hR0cTP u?Ohs*{ :Ch% (F%LSagdhttp://psychology.wikia.com/wiki/Probability_density_function%M4http://psychology.wikia.com/wiki/Standard_deviationE &http://psychology.wikia.com/wiki/Mean h+http://psychology.wikia.com/wiki/Parameter}:http://psychology.wikia.com/wiki/Probability_distribution+s7http://mathworld.wolfram.com/BinomialDistribution.htmlO=http://mathworld.wolfram.com/ProbabilityDensityFunction.html-i+http://mathworld.wolfram.com/Variance.html%c'http://mathworld.wolfram.com/Mean.htmlT&*http://en.wikipedia.org/wiki/Weighted_sumd1http://en.wikipedia.org/wiki/Probability_measure,]/http://en.wikipedia.org/wiki/Lebesgue_integralT<-http://en.wikipedia.org/wiki/Random_variable/s(http://en.wikipedia.org/wiki/Statistics{0http://en.wikipedia.org/wiki/Probability_theory7bo/http://en.wikipedia.org/wiki/Least_upper_boundZ6l)http://en.wikipedia.org/wiki/Real_numbersi0http://en.wikipedia.org/wiki/Countably_infinite#Yf(http://en.wikipedia.org/wiki/Finite_setjc/http://en.wikipedia.org/wiki/Set_(mathematics)T<]-http://en.wikipedia.org/wiki/Random_variableHZ7http://en.wikipedia.org/wiki/Probability_mass_functionhW6http://en.wikipedia.org/wiki/Probability_distribution{T0http://en.wikipedia.org/wiki/Probability_theoryode MSA BCambria MathBAh˕F:f*Vd<Vd<!20uu 2QHP $P"! xxMAT 211 - Final MathematicsFiroz!                           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