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applied statistics and probability for engineers
Applied Statistics For Engineers And Scientists 3rd Edition Jay L. Devore, Nicholas R. Farnum, Jimmy A. Doi - Solutions
=+d. What is the probability that x lies within .25 sec of 3?
=+c. What is the probability that x exceeds 3?
=+b. Find the numerical value of k.
=+a. Sketch a graph of f(x).
=+32. Suppose that the reaction time (sec) to a certain stimulus is a continuous random variable with a density function given by f (x) 5 e kyx for 1 # x # 10 0 otherwise
=+d. Find the standard deviation of the number of forms required.
=+c. What is the expected number of forms required?
=+b. What is the probability that at most three forms are required?
=+a. What is the numerical value of k?
=+31. A contractor is required by a county planning department to submit from one to five different forms, depending on the nature of the project. Let y 5 number of forms required of the next contractor. Suppose that it is known that the probability that y forms are required is proportional to y;
=+30. Let x denote the number of ticketed airline passengers denied a flight because of overbooking.Suppose that x is a random variable for which p(x) 5 c(5 2 x) for x 5 0, 1, 2, 3, 4. Find the numerical value of c and then compute P(x . 0).
=+c. Compute the expected number of pounds left after a customer’s order is shipped.
=+b. Compute the variance of the number of lots ordered by a customer.
=+a. Compute the mean number of lots ordered by a customer.
=+29. A chemical supply company currently has in stock 100 pounds of a certain chemical, which it sells to its customers in 5-lb lots. Let x denote the number of lots ordered by a randomly selected customer, and suppose x has the following probability mass function:x: 1 2 3 4 p(x): .2 .4 .3 .1
=+b. Calculate the variance and standard deviation of x.
=+a. Calculate the mean number of coding errors for all such blocks of 1000 lines of code.
=+28. The probability mass function for the number x of coding errors found in 1000 randomly selected lines of computer code is given by x: 0 1 2 3 4 p(x): .08 .15 .45 .27 .05
=+g. x 5 the number of customer complaints in a randomly selected week
=+f. t5the lifetime of a randomly selected electronic component
=+e. v5the breaking strength of a randomly selected metal bar
=+d. u5the number of errors per 1000 randomly selected lines of computer code
=+c. w5the proportion of oversize bolts in a randomly selected box of bolts
=+b. y5the measured concentration of chemical in a solution
=+a. x5the number of flaws per square foot in a randomly selected sheet of fabric
=+27. Classify each of the following random variables as either discrete or continuous.
=+modern statistical techniques are so computationally intensive that a billion divisions over a short time period is not outside the realm of possibility.Assuming that the 1 in 9 billion figure is correct and that results of different divisions are independent of one another. What is the
=+1 in 9 billion, so that it would take thousands of years before a typical user encountered a mistake.However, statisticians are not typical users; some
=+26. In October 1994, a flaw in a certain Pentium chip installed in computers was discovered that could result in a wrong answer when performing a division.The manufacturer initially claimed that the chance of any particular division being incorrect was only
=+25. If A and B are independent events, show that A9 and B are also independent. Hint: Use a Venn diagram to show that P(A= and B) 5 P(B) 2 P(A and B).
=+b. What is the probability that at least 1 of 25 points on a control chart signals a problem with a manufacturing process when in fact the process is running correctly?
=+a. What is the probability that at least one of ten points on a control chart signals a problem with a manufacturing process when in fact the process is running correctly?
=+that it is not operating correctly. However, even when a process is running correctly, there is a small probability, say, 1%, that a charted point will mistakenly signal that there is a problem with the process.
=+24. One of the assumptions underlying the theory of control charts (see Chapter 6) is that the successive points plotted on a chart are independent of one another.Each point plotted on a control chart can signal either that a manufacturing process is operating correctly or
=+b. If both tests are positive, what is the probability that the selected individual is a carrier?
=+a. What is the probability that both tests yield the same result?
=+a 5% detection rate for noncarriers. Suppose that the diagnostic test is applied independently to two different samples from the same randomly selected individual.
=+23. In a certain population, 1% of all individuals are carriers of a particular disease. A diagnostic test for this disease has a 90% detection rate for carriers and
=+b. Suppose you are told the selected review was submitted in Word format. What is the probability that the review was medium in length?
=+a. What is the probability that the selected review was submitted in Word format?
=+Word or a typesetting program called LaTeX. For short reviews, 80% are in Word, whereas 50% of medium reviews are in Word and 30% of long reviews are in Word. Suppose a recent review is randomly selected.
=+22. The reviews editor for a certain scientific journal decides whether the review for any particular book should be short (1–2 pages), medium (3–4 pages), or long (5–6 pages). Data on recent reviews indicates that 60% of them are short, 30% are medium, and the other 10% are long. Reviews
=+only if both components work correctly. If all components work independently of one another and P(a given component works) 5 .9, calculate the probability that the entire system works correctly.
=+ Components 1 and 2 are connected in parallel, so that their subsystem functions correctly if either component 1 or 2 functions. Components 3 and 4 are connected in series, so their subsystem works U
=+21. Consider a system of components connected as shown in the following figure.1 23 4
=+c. Find the probability that at least one of the bids is successful.
=+b. Find the probability that neither bid is successful.
=+a. Find the probability that both bids are successful.
=+20. A construction firm has bid on two different contracts. Let E1 be the event that the bid on the first contract is successful, and define E2 analogously for the second contract. Suppose that P(E1) 5 .4 and P(E2) 5 .3 and that E1 and E2 are independent.
=+d. The probability that two people do not match for a given characteristic is called discriminating power.What is the discriminating power for the comparison of two people’s blood types in part (c)?
=+c. Find the probability that two randomly chosen people have matching blood types. Note: A person can have only one phenotype.
=+b. Repeat the calculation in part (a) for the other three blood types.
=+a. What is the probability that two randomly chosen people both have blood type A?
=+19. In forensic science, the probability that any two people match with respect to a given characteristic(hair color, blood type, etc.) is called the probability of a match. Suppose that the frequencies of blood phenotypes in the population are as follows:A B AB O.42 .10 .04 .44
=+. What is the probability that at least one person in a sample of one million will have a blood type matching that found at the crime scene?
=+c. What is the probability that at least one person in a random sample of n people will match the blood type found at the crime scene?d. Suppose that 5 1026
=+b. What is the probability that none of n randomly chosen people will match the blood type found at the crime scene?
=+a. What is the probability that a randomly chosen person from the population does not have the same blood type as that found at the crime scene?
=+John Wiley, New York, 1995). Suppose that a suspect is found whose blood type matches a rare blood type found at a crime scene. Let denote the frequency with which people in the population have this particular blood type. Assuming that people in the population are sampled at random, answer
=+18. Probability calculations play an important role in modern forensic science (Aitken, C., Statistics and the Evaluation of Evidence for Forensic Scientists,
=+17. Suppose that A and B are independent events with P(A) 5 .5 and P(B) 5 .6. Can A and B be mutually exclusive events?
=+b. If a 1 is received, what is the probability that a 1 was sent? Hint: Use the tree diagram from Exercise 6.
=+a. What is the proportion of 1s received at the end of the channel?
=+16. In Exercise 6, suppose that there is a probability of.01 that a digit is incorrectly sent over a communication channel (i.e., that a digit sent as a 1 is received as a 0, or a digit sent as a 0 is received as a 1).Consider a message that consists of exactly 60% 1s.
=+c. Given that a fastener passes inspection, what is the probability that it passed the initial inspection and did not have to go through the recrimping operation?
=+b. What proportion of fasteners pass inspection?
=+a. What proportion of fasteners that fail the initial inspection pass the second inspection (after the recrimping operation)?
=+15. In Exercise 5, suppose that 95% of the fasteners pass the initial inspection. Of those that fail inspection, 20% are defective. Of the fasteners sent to the recrimping operation, 40% cannot be corrected and are scrapped; the rest are corrected by the recrimping and then pass inspection.
=+e. Use the observation in part (d) and the conditional probability formula for P(B|A) to justify Bayes’ theorem.
=+d. In Figure 5.7, the probability associated with any path from left to right through the tree is simply the product of the probabilities of the branches.Why?
=+ This formula, known as Bayes’ theorem, is used to “turn conditional probabilities around”; that is, it allows us to express P(B|A) in terms of P(A|B) and P(A|B=).
=+c. Use parts (a) and (b) to show that P(BuA) 5 P(AuB)P(B)P(AuB)P(B) 1 P(AuB=)P(B=)
=+b. Use the conditional probability formula to write P(A and B) in terms of P(A|B) and P(B).Develop a similar formula for P(A and B9) in terms of P(A|B9) and P(B9).
=+a. Use the addition law to show that P(A) 5 P(A and B) 1 P(A and B9).
=+14. Refer to the tree diagram in Figure 5.7. Suppose you want to find the probability P(B|A) using the information available in the tree diagram. To do this, P(B|A) must be expressed in terms of conditional probabilities, like P(A|B) and P(A9|B).
=+b. Suppose the auto company narrows the choice of suppliers to companies A and C. What is the probability that company A is chosen this year?
=+a. Suppose that supplier E goes out of business this year, leaving the remaining four companies to compete with one another. What are the new probabilities of companies A, B, C, and D being chosen as the sole supplier this year?
=+13. Five companies (A, B, C, D, and E) that make electrical relays compete each year to be the sole supplier of relays to a major automobile manufacturer.The auto company’s records show that the probabilities of choosing a company to be the sole supplier are Supplier chosen: A B C D E
=+c. What is the probability that a randomly selected adult regularly consumes coffee but does not regularly consume soda?
=+b. What is the probability that a randomly selected adult doesn’t regularly consume at least one of these two products?
=+a. What is the probability that a randomly selected adult regularly consumes both coffee and soda?
=+12. Suppose that 55% of all adults regularly consume coffee, 45% regularly consume carbonated soda, and 70% regularly consume at least one of these two types of drinks.
=+.01 probability of failing. Find an upper bound on the probability that the entire system fails.
=+where the events involved have relatively small probabilities. For example, suppose a system consists of five subcomponents connected in series(cf. Example 5.8) and that each component has a
=+11. For any collection of events A1, A2, A3, . . . , Ak, it can be shown that the inequality P(A1 or A2 or A3 or ... or Ak)# P(A1) 1 P(A2) 1 P(A3) 1 1 P(Ak)always holds. This inequality is most useful in cases
=+(cf. Example 5.8, Section 5.3) and that each component has a .999 probability of functioning without failure. What lower bound can you put on the reliability (i.e., the probability of functioning correctly)of the system built from these ten components?
=+ always holds. This inequality is particularly useful when each of the events has relatively high probability. Suppose, for example, that a system consists of ten components connected in series
=+10. For any collection of events A1, A2, A3, . . . , Ak, it can be shown that the inequality P(A1 and A2 and A3 and . . . and Ak)$1 2 [P(A=1) 1 P(A=2) 1 P(A=3) 1 1 P(A=k)]
=+b. How many defective solder joints found by inspector A were not found by inspector B?
=+a. How many defective solder joints were found by the two inspectors?
=+the same circuit board. The accompanying table shows the results of two inspectors who examined the same collection of 10,000 solder joints for a particular problem:Number of defective solder joints found Inspector A 724 Inspector B 751 Common to both inspectors 316
=+9. Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad nonwetting, knee visibility, voids)and even the degree to which a joint possesses one or more of these defects.
=+c. What general statement can you make regarding the effectiveness of the two methods? Can your statement be extended to methods involving samples of more than two items?
=+b. What is the probability that item A will be discovered by method 2?
=+a. What is the probability that item A will be discovered by method 1?
=+8. Two methods are proposed for testing a shipment of five items (call them A, B, C, D, and E). In method 1, an inspector randomly samples two of the five items and tests to see whether either item is defective. In method 2, an inspector randomly samples one item and tests it; the remaining four
=+7. Use a Venn diagram to find a simple expression for{A and B}=in terms of A=and B=.
=+6. Information theory is concerned with the transmission of data, usually encoded as a stream of 0s and 1s, over communication channels. Because channels are“noisy,” there is a chance that some 0s sent through the channel are mistakenly received at the other end as 1s, and vice versa. The
=+some are completely defective and must be scrapped, whereas the rest can be run through a machine that readjusts the amount of crimping. Of the recrimped fasteners, some are corrected by the recrimping operation and pass inspection, whereas the remainder cannot be salvaged and are scrapped.
=+5. Nuts and bolts used in aircraft manufacturing are called fasteners. To ensure that they are not loosened by vibrations during flight, some fasteners are slightly crimped so that they lock more tightly. The amount of crimping, however, must meet specific standards.To test finished fasteners, an
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