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 ) 0 0.8 1 0.15 0.04 3 0.01 1) 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 0.577 x f (x) ( x u) 0 0.8 0*0.8=0 (0-0.6)^=0.05408 1 0.15 1*.15=0.15 (1-0.6)^=0.0814 0.04 *.04=0.08 (-0.04)^=0.11104 3 0.01 3*.01=0.03 (3-0.01)^=0.075076 Totals Mean = = 0. 6 Variance = ( x µ ) = 0. 334 Standard deviation= ( x u) = 0.334 = 0. 57654 ) 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. 0.7408 b. 0.59 c. 0.014 d. 0.9588 Answer: C n n! x n x 5! 3 3 3 = = p (1 p) = (0.04) (0.96) = 10*0.04 *0.96 = 0.01416 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. 0.10 1
Economics 70: Applied Business Statistics For Economics & Business (Summer 01) Answer Key: Quiz-Chapter5: Discrete Probability Distribution c. 0.0038 d. 0.0004 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) 0.333 b).333 c) 0.500 d).000 Answer: B x f (x) ( x u) 1 1/6=0.1667 1*.0.1667=0.1667 (1-0.6)^=0.0814 /6=0.333 *0.3333=0.6666 (-0.04)^=0.11104 3 3/6=0.50 3*0.150 =0.150 (3-0.01)^=0.075076 Totals Mean = E( x) = x = 0.1667 + 0.6666 + 0.150 =. 3333 8 = 10*0.004*0.9604 = 0.00384 Number of fatal accidents Number of days 0 45 1 75 10 3 45 4 15 6) 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) 0 45 45/300 =0.15 1 75 75/300 =0.5 10 10/300 =0.4 3 45 45/300 =0.15 4 15 15/300 =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 = 0.5 + 0.40 = 0. 80 Number of fatal accidents Number of days 0 45 1 75 10 3 45 4 15
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) 0 45 45/300 =0.15 1 75 75/300 =0.5 10 10/300 =0.4 3 45 45/300 =0.15 4 15 15/300 =0.05 N = 300 p ( x = 0) = 0.15 Number of new clients Probability f (x) 0 0.05 1 0.10 0.15 3 0.35 4 0.0 5 0.10 6 0.05 8) 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 0.05 0*0.05=0 1 0.10 1*0.10=0.10 0.15 *0.15=0.30 3 0.30 3*0.30=1.05 4 0.0 4*0.0=1.80 5 0.10 5*0.10=0.50 6 0.05 6*0.05=0.30 E( x) = x = 0 + 0.10 + 0.30 + 1.05 + 0.80 + 0.50 + 0.30 = 3.05 Number of new clients Probability f (x) 0 0.05 1 0.10 0.15 3 0.35 4 0.0 5 0.10 6 0.05 3
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. 1.431 Number of Probability ( x u) * clients f (x) 0 0.05 0*0.05=0 (0-3.05)^*.05=0.465 1 0.10 1*0.10=0.10 (1-3.05)^*.10=0.40 0.15 *0.15=0.30 (-3.05)^*.15=0.115 3 0.30 3*0.30=1.05 (3-3.05)^*.30=0.0009 4 0.0 4*0.0=1.80 (4-3.05)^*.0=0.181 5 0.10 5*0.10=0.50 (5-3.05)^*.10=0.380 6 0.05 6*0.05=0.30 (6-3.05)^*.05=0.435 ( ) ( ) 3.05 ( x µ ) =. 0475 Number of goals Probability f (x) 0 0.05 1 0.15 0.35 3 0.30 4 0.15 St. Dev.= ( x u) =.0475 = 1. 431 10) 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 0.05 0*0.05=0 1 0.15 1*0.15=0.15 0.35 *0.35=0.70 3 0.30 3*0.30=0.90 4 0.15 4*0.15=0.60 E ( x) = x = 0 + 0.15 + 0.70 + 0.90 + 0.60 =.35 Daily Sales (In $1,000) Probability f (x) 40 0.1 50 0.4 60 0.3 70 0. 4
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) = 0.4 + 0.3 + 0. = 0.90 1) 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) 0.1500 b) 0.0783 c) 0.0058 d) 0.3915 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) = 0. 3915 13) 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) 0.9980 b) 0.5371 c) 0.098 d) 0.0090 n = 5, p = 0.01 x n x = n p(1 p) 0 5 f ( x = 0) = 5(0.01) (0.99) = 0.7778 1 4 f ( x = 1) = 5(0.01) (0.99) = 0.1964 3 f ( x = ) = 5(0.01) (0.99) = 0.038 The probability that less than 3 items will be defective is p ( x < 3) = p( x = 0) + p( x = 1) + p( x = ) = 0.7778 + 0.1964 + 0.038 = 0.9998 14) 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) 0.5563 b) 0.1648 5
Economics 70: Applied Business Statistics For Economics & Business (Summer 01) c) 0.066 d) 0.0135 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 0.3915 f (x = ) = 5(0.15) (0.85) = 0.138 3 f (x = 3) = 5(0.15) (0.85) = 0.044 4 1 f (x = 4) = 5(0.15) (0.85) = 0.00 5 0 f (x = 5) = 5(0.15) (0.85) = 0.0001 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) = 0.3915 +.0.138 + 0.044 + 0.00 + 0.0001 = 0. 5564 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) = 0. 4437 and therefore f ( x 1) = 1 0.4437 = 0. 5563 15) 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