Answer Key: Quiz2-Chapter5: Discrete Probability Distribution

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1 Economics 70: Applied Business Statistics For Economics & Business (Summer 01) Answer Key: Quiz-Chapter5: Discrete Probability Distribution The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages (x) in the city has the following probability distribution. x f (x ) ) The mean and the standard deviation for the number of electrical outages (respectively) are a. 0 and 0.8 b..6 and 5.77 c. 3 and 0.01 d. 0.6 and x f (x) ( x u) *0.8=0 (0-0.6)^= *.15=0.15 (1-0.6)^= *.04=0.08 (-0.04)^= *.01=0.03 (3-0.01)^= Totals Mean = = 0. 6 Variance = ( x µ ) = Standard deviation= ( x u) = = ) Four percent of the customers of a mortgage company default on their payments. A sample of five customers is selected. What is the probability that exactly two customers in the sample will default on their payments? a b c d Answer: C n n! x n x 5! = = p (1 p) = (0.04) (0.96) = 10*0.04 *0.96 = x x!( n x)!!3! 3) Twenty percent of the students in a class of 100 are planning to go to graduate school. The mean and standard deviation of this binomial distribution is a. 4 and 0 b. and 16 c. 16 and d. 0 and 4 e) 4 and 17 E( x) = np = 0. *100 = 0 σ = np(1 p) = (0.)(100)(0.8) = 4 4) A production process produces % defective parts. A sample of five parts from the production process is selected. What is the probability that the sample contains exactly two defective parts? a. 0.0 b

2 Economics 70: Applied Business Statistics For Economics & Business (Summer 01) Answer Key: Quiz-Chapter5: Discrete Probability Distribution c d e) none of the above Answer: C n n! = = x x!( n p x)! x (1 p) n x = 5! (0.0)!3! (0.98) 5) X is a random variable with the probability function: f(x) = X/6 for X = 1, or 3 The expected value of X is a) b).333 c) d).000 Answer: B x f (x) ( x u) 1 1/6= * = (1-0.6)^= /6=0.333 *0.3333= (-0.04)^= /6=0.50 3*0.150 =0.150 (3-0.01)^= Totals Mean = E( x) = x = = = 10*0.004* = Number of fatal accidents Number of days ) The table above shows the police record of a metropolitan area kept over the past 300 days show the number of fatal accidents. What is the probability that in a given day there will less than 3 accidents? a) 0.5 b) 10 c) 0. d) 0.8 x # of fatal accidents f (x) /300 = /300 = /300 = /300 = /300 =0.05 Using the computed relative frequencies, the cumulative relative frequencies for less than 3 accidents is p ( x < 3) = p( x = 0) + p( x = 1) + p( x = ) = 0.15 = = Number of fatal accidents Number of days

3 Economics 70: Applied Business Statistics For Economics & Business (Summer 01) Answer Key: Quiz-Chapter5: Discrete Probability Distribution 7) The table above shows the police record of a metropolitan area kept over the past 300 days show the number of fatal accidents. What is the probability that in a given day there will be no accidents? a) 0.00 b) 0.15 c) 1.00 d) 0.85 Answer: B x # f (x) /300 = /300 = /300 = /300 = /300 =0.05 N = 300 p ( x = 0) = 0.15 Number of new clients Probability f (x) ) The table above is for Roth is a computer-consulting firm. The number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that is shown above. The expected number of new clients per month is: a) 1 b) 6 c) 0 d) 3.05 Number of clients Probability f (x) *0.05= *0.10= *0.15= *0.30= *0.0= *0.10= *0.05=0.30 E( x) = x = = 3.05 Number of new clients Probability f (x)

4 Economics 70: Applied Business Statistics For Economics & Business (Summer 01) Answer Key: Quiz-Chapter5: Discrete Probability Distribution 9) The table above is for Roth is a computer-consulting firm. The number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that is shown above. The standard deviation is: a..047 b.3.05 c. 1 d Number of Probability ( x u) * clients f (x) *0.05=0 (0-3.05)^*.05= *0.10=0.10 (1-3.05)^*.10= *0.15=0.30 (-3.05)^*.15= *0.30=1.05 (3-3.05)^*.30= *0.0=1.80 (4-3.05)^*.0= *0.10=0.50 (5-3.05)^*.10= *0.05=0.30 (6-3.05)^*.05=0.435 ( ) ( ) 3.05 ( x µ ) = Number of goals Probability f (x) St. Dev.= ( x u) =.0475 = ) The table above shows the probability distribution for the number of gals the lions soccer team makes per game. What is the expected number of goals per game? a).35 b) 0 c) 1 d), since it has the highest probability Number of goals Probability f (x) *0.05= *0.15= *0.35= *0.30= *0.15=0.60 E ( x) = x = =.35 Daily Sales (In $1,000) Probability f (x)

5 Economics 70: Applied Business Statistics For Economics & Business (Summer 01) Answer Key: Quiz-Chapter5: Discrete Probability Distribution 11) The table above shows the probability distribution for the daily sales at Michael s Co. The probability of having sales of at least $50,000 is: a) 0.90 b) 0.5 c) 0.30 d) 0.10 Using the relative frequencies given on the table, we need to calculate the cumulative relative frequencies for sales greater than or equal to $50,000. p ( x $50,000) = p( x = $50,000) + p( x = $60,000) + p( x = $70,000) = = ) Bob would rather risk getting a parking ticket than pay the parking fee at a campus parking facility. Bob parks at the facility Monday through Friday. The probability of getting a ticket on any given weekday is 15%. Answer the questions based on a five-day period. What is the probability that Bob will get a ticket during the week (Monday-Friday)? a) b) c) d) n = 5, p = 0.15 p n p = n p(1 p) 1 4 The probability that bob will get late during the week is: f ( x = 1) = 5(0.15) (0.85) = ) Suppose you are receiving a shipment of a product in lots of 5,000. You randomly select 5 units from each lot and accept the shipment if the number of defective units is less than 3. Find the probability of accepting a lot, if the actual fraction of defectives in the lot is 0.01 (1%). a) b) c) d) n = 5, p = 0.01 x n x = n p(1 p) 0 5 f ( x = 0) = 5(0.01) (0.99) = f ( x = 1) = 5(0.01) (0.99) = f ( x = ) = 5(0.01) (0.99) = The probability that less than 3 items will be defective is p ( x < 3) = p( x = 0) + p( x = 1) + p( x = ) = = ) Bob would rather risk getting a parking ticket than pay the parking fee at a campus parking facility. Bob parks at the facility Monday through Friday. The probability of getting a ticket on any given weekday is 15%. Answer the questions based on a five-day period. 16) What is the probability that Bob will get at least one ticket during the week? a) b)

6 Economics 70: Applied Business Statistics For Economics & Business (Summer 01) c) d) n = 5, p = 0.15 Answer Key: Quiz-Chapter5: Discrete Probability Distribution x n x = n p(1 p) 1 4 (x = 1) = 5(0.15) (0.85) = 3 f f (x = ) = 5(0.15) (0.85) = f (x = 3) = 5(0.15) (0.85) = f (x = 4) = 5(0.15) (0.85) = f (x = 5) = 5(0.15) (0.85) = The probability that Bob will get at least 1 ticket during the week is: f ( x 1) = p( x = 1) + p( x = ) + p( x = 3) + p( x = 4) + p( x = 5) = = The easier way to solve this particular problem is to define the problem f ( x 1) = 1 p( x = 0) 0 5 Where p ( x = 0) = 5(0.15) (0.85) = and therefore f ( x 1) = = ) The E70 common departmental final has 30 multiple questions. Each question has 5 choices. If you guessed the answers to all the questions, how many questions should you expect to answer correctly? a) 5 b) 6 c) 7 d) 14 Answer: B The number of correct guesses, x, has a binomial distribution with n = 30, p = 0. That is because there are 5 answers in each question, the probability to guess each question correctly is 1/5=0%. Therefore the expected number of questions that can be guessed correctly is: E ( x) = 0.*30 = 6 6

MATH 264 Problem Homework I

MATH 264 Problem Homework I MATH Problem Homework I Due to December 9, 00@:0 PROBLEMS & SOLUTIONS. A student answers a multiple-choice examination question that offers four possible answers. Suppose that the probability that the