5.3 The manager of a large computer network has developed the following probability distribution of the number of interruptions per day: a. Compute the expected number of interruptions per day. x p(x) x * p(x) x^2 * p(x) 0 0.32 0 0 1 0.35 0.35 0.35 2 0.18 0.36 0.72 3 0.08 0.24 0.72 4 0.04 0.16 0.64 5 0.02 0.1 0.5 6 0.01 0.06 0.36 SUM 1 1.27 3.29 Mean (µ) of a probability distribution is [x*p(x)] = 1.27 ∑ b. Compute the standard deviation. Variance (s^2) = (x^2 * p[x]) - µ^2 ∑ s^2 = 3.29 - 1.27^2 s^2 = 3.29 - 1.6129 s^2 = 1.6771 SD = sqrt(s^2) SD = sqrt(1.6771) SD = 1.3
Transcript
5.3 The manager of a large computer network has developedthe following probability distribution of the number ofinterruptions per day:
a. Compute the expected number of interruptions per day.
Mean (µ) of a probability distribution is ∑[x*p(x)] = 1.27
b. Compute the standard deviation.Variance (s^2) = ∑(x^2 * p[x]) - µ^2 s^2 = 3.29 - 1.27^2s^2 = 3.29 - 1.6129s^2 = 1.6771
SD = sqrt(s^2)SD = sqrt(1.6771)SD = 1.3
a. 0.0768b. 0.91296c. 0.33696d. 0.66304
5.13 When a customer places an order with Rudy’s On-Line Office Supplies, a computerized accounting informationsystem (AIS) automatically checks to see if the customerhas exceeded his or her credit limit. Past recordsindicate that the probability of customers exceeding theircredit limit is 0.05. Suppose that, on a given day, 20 customersplace orders. Assume that the number of customersthat the AIS detects as having exceeded their credit limit isdistributed as a binomial random variable.a. What are the mean and standard deviation of the numberof customers exceeding their credit limits?
b. What is the probability that 0 customers will exceed theirlimits?0.358486
c. What is the probability that 1 customer will exceed his orher limit?0.377354
d. What is the probability that 2 or more customers willexceed their limits?
0.26416
5.17 Assume a Poisson distribution.a. If = 2.0, find P(X > 2).0.593994
b. If = 8.0, find P(X > 3).0.986246
c. If = 0.5, find P(X 1).(missing symbol)
5.19 Assume that the number of network errors experiencedin a day on a local area network (LAN) is distributedas a Poisson random variable. The mean number of networkerrors experienced in a day is 2.4. What is the probabilitythat in any given daya. zero network errors will occur?0.090718
b. exactly one network error will occur?0.217723
c. two or more network errors will occur?0.691559
d. less than three network errors will occur?0.569709
5.35 The mean cost of a phone call handled by an automatedcustomer-service system is $0.45. The mean cost ofa phone call passed on to a “live” operator is $5.50.However, as more and more companies have implementedautomated systems, customer annoyance with such systemshas grown. Many customers are quick to leave the automatedsystem when given an option such as “Press zero totalk to a customer-service representative.” According to theCenter for Client Retention, 40% of all callers to automatedcustomer-service systems automatically opt to go to a liveoperator when given the chance (J. Spencer, “In Search ofthe Operator,” The Wall Street Journal, May 8, 2002,p. D1).If 10 independent callers contact an automated customer servicesystem, what is the probability thata. 0 will automatically opt to talk to a live operator?0.006047
b. exactly 1 will automatically opt to talk to a live operator?0.040311
c. 2 or less will automatically opt to talk to a live operator?0.16729
d. all 10 will automatically opt to talk to a live operator?0.000105
6.1 Given a standardized normal distribution (with a meanof 0 and a standard deviation of 1, as in Table E.2), what isthe probability thata. Z is less than 1.57?0.941792
b. Z is greater than 1.84?0.032884
c. Z is between 1.57 and 1.84?0.025323
d. Z is less than 1.57 or greater than 1.84?0.974677
6.5 Given a normal distribution with = 100 and= 10, what is the probability thata. X > 75?
0.99379
b. X < 70?0.00135
c. X 110?missing symbol
d. 80% of the values are between what two X values (symmetricallydistributed around the mean)?Low: 87.18448High: 112.8155
6.9 The breaking strength of plastic bags used for packagingproduce is normally distributed, with a mean of 5 pounds persquare inch and a standard deviation of 1.5 pounds per squareinch. What proportion of the bags have a breaking strength ofa. less than 3.17 pounds per square inch?0.111232
b. at least 3.6 pounds per square inch?0.824676
c. between 5 and 5.5 pounds per square inch?0.130559
d. 95% of the breaking strengths will be contained betweenwhat two values symmetrically distributed around themean?Low: 2.060054High: 7.939946
6.11 A statistical analysis of 1,000 long-distance telephonecalls made from the headquarters of the Bricks and ClicksComputer Corporation indicates that the length of thesecalls is normally distributed, with seconds andseconds.a. What is the probability that a call lasted less than 180 seconds?
b. What is the probability that a call lasted between 180 and300 seconds?
c. What is the probability that a call lasted between 110 and180 seconds?
d. What is the length of a call if only 1% of all calls areshorter?
6.13 Many manufacturing problems involve the matching of machine parts, such as shafts that fit into a valve hole. Aparticular design requires a shaft with a diameter of 22.000 mm, but shafts with diameters between 21.900 mm and22.010 mm are acceptable. Suppose that the manufacturing process yields shafts with diameters normally distributed,with a mean of 22.002 mm and a standard deviation of 0.005 mm. For this process, what is
a. the proportion of shafts with a diameter between 21.90mm and 22.00 mm?0.308538
b. the probability that a shaft is acceptable?0.99379
c. the diameter that will be exceeded by only 2% of the shafts?22.01227
d. What would be your answers in (a) through (c) if the standarddeviation of the shaft diameters was 0.004 mm?
a 0.308538b 0.97725c 22.01021
6.13 Many manufacturing problems involve the matching of machine parts, such as shafts that fit into a valve hole. Aparticular design requires a shaft with a diameter of 22.000 mm, but shafts with diameters between 21.900 mm and22.010 mm are acceptable. Suppose that the manufacturing process yields shafts with diameters normally distributed,
6.29 An industrial sewing machine uses ball bearings thatare targeted to have a diameter of 0.75 inch. The lower andupper specification limits under which the ball bearings canoperate are 0.74 inch and 0.76 inch, respectively. Past experiencehas indicated that the actual diameter of the ball bearingsis approximately normally distributed, with a mean of0.753 inch and a standard deviation of 0.004 inch. What isthe probability that a ball bearing isa. between the target and the actual mean?0.273373
b. between the lower specification limit and the target?0.22605
