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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
=+b. Next, calculate the percentage of people living less than 10 years.
=+a. Use this model to calculate the percentage of people living over 150 years.
=+48. Is the exponential distribution a reasonable one for modeling human lifetimes? To answer this question, suppose the mean lifetime is about 75 years and that lifetimes follow an exponential distribution with parameter 5 1y75 5 .0133.
=+d. Write the hazard function. Is the failure rate increasing, decreasing, or flat?
=+c. Calculate the reliability at t 5 5 600,000 cycles. Note that R() is always the same number for any Weibull distribution.
=+b. Calculate the reliability at t 5 800,000 cycles.
=+47. Reliability of mechanical springs is measured in terms of how many times (i.e., cycles) the spring can be compressed and released. Suppose that the lifetime of a certain type of spring can be modeled by a Weibull distribution with shape parameter 5 4 and scale parameter 5 600,000.a.
=+c. Suppose a manufacturer of such systems wants to increase their reliability by specifying that R(100) 5 .95. What is the mean time to failure(in cycles) for such a system?
=+b. Find the number of cycles, t, for which the reliability equals .95 (i.e., 95%).
=+a. What is the probability that a given canula connection will last for at least 100 cycles?
=+Suppose the number of times (i.e., cycles) you can connect and disconnect such a system until a membrane wears out is modeled by an exponential distribution with a mean time to failure of 5 500 cycles.
=+the surface of the rubber membrane closes up, resealing the end of the tube. However, after many such connections and disconnections, the rubber membrane may eventually wear out and fail to close properly.
=+46. Intravenous (IV) tubes that deliver liquids (drugs, saline solution, food, etc.) to medical patients are connected by inserting a plastic prong (called a canula) from one tub through a rubber membrane in the connector of another tube. Canula systems are needleless and therefore eliminate the
=+b. Examine the chart in part (a) for any out-ofcontrol points.
=+). The following table shows the number of surface flaws found on 20 successive panels:Panel Number of flaws Area of panel Panel Number of flaws Area of panel 1 3 .8 11 1 .6 2 2 .6 12 3 .8 3 3 .8 13 5 .8 4 2 .8 14 4 1.0 5 5 1.0 15 6 1.0 6 5 1.0 16 12 1.0 7 10 .8 17 3 .8 8 12 1.0 18 3 .6 9 4 .6
=+45. Painted metal panels are examined after baking at high temperatures to harden the paint. Because the manufacturer produces panels of several different sizes, inspectors simply record the number of blemishes found along with the known area of the panel(ft2
=+a. Construct a control chart for the number of pinholes per dashboard.
=+44. Forty consecutive automobile dashboards are examined for signs of pinholes in the plastic molding.The numbers of pinholes found are (read across)6 2 3 2 5 2 2 3 2 4 9 4 0 5 0 6 5 4 2 3 3 1 4 1 7 3 3 5 7 3 6 7 6 4 5 3 8 5 4 3
=+Number Number Day of flaws Day of flaws 19 63 23 42 20 42 24 39 21 45 25 38 22 43 Construct an appropriate control chart for this data, and examine it for any evidence of a lack of statistical control.
=+plant, a 250-lb sample is taken out of every batch of aspirin, and the number of off-color flaws is counted.The following table shows the numbers of flaws per 250-lb sample for a period of 25 days:Number Number Day of flaws Day of flaws 1 46 10 44 2 51 11 47 3 56 12 51 4 57 13 46 5 37 14 49 6 51
=+43. Off-color flaws in aspirin are caused by extremely small amounts of iron that change color when wet aspirin comes into contact with the sides of drying containers (“People: The Only Thing That Makes Quality Work,” Quality Progress, Sept. 1988: 63–67).Such flaws are not harmful but are
=+b. Check the chart for any out-of-control signals and, if necessary, eliminate such points from the data and reconstruct the c chart.
=+a. Construct a c chart for this data.
=+42. The following observations are the number of defects in 25 1-square-yard specimens of woven fabric(read across):3 7 5 3 4 2 8 4 3 3 6 7 2 3 2 4 7 3 2 4 4 1 4 5 6
=+c. Are there any signs of out-of-control conditions in this data?
=+b. Construct the control limits for the p chart.
=+Number tested Number failed Day Number tested Number failed 21 2011 38 24 2375 20 22 2059 13 25 2029 30 23 2045 11a. Calculate the centerline of a p chart for this data.
=+25 successive manufacturing days is given here.Day Number tested Number failed Day Number tested Number failed 1 2186 28 11 2141 31 2 2131 21 12 2019 18 3 2158 22 13 2027 27 4 2307 14 14 2376 25 5 2262 17 15 2118 27 6 2379 27 16 2251 14 7 2069 18 17 2068 31 8 2264 20 18 2242 23 9 2383 18 19 2089
=+41. After assembly and wiring of the individual keys, computer keyboards are tested by an automated test station that pushes each key several times. Daily records are kept of the number of keyboards inspected and the number that fail the inspection. Data from
=+vc. If your answer to part (b) is “yes,” then eliminate the out-of-control point(s) from the data and recompute the centerline and control limits of the p chart.
=+b. The highest number of failures on a given day was 39 and the lowest number was 13. Would either of these points indicate an out-of-control condition?
=+a. From this information, calculate the centerline and control limits for a p chart.
=+40. On each of 25 days, 100 printed circuit boards are subjected to thermal cycling; that is, they are subjected to large changes in temperature, a procedure known to cause failures in boards with weak circuit connections. Of the boards tested, a total of 578 fail to work properly after the
=+b. Interpret the chart in part (a).
=+a. Construct a control chart for the proportion of nonconforming items per lot.
=+39. The following data shows the number of nonconforming items found in 30 successive lots, each of size 50, of a finished product:4 3 0 2 2 2 0 1 1 0 3 2 1 1 0 0 2 4 2 5 0 0 1 1 0 3 2 1 2 4
=+38. Control limits for attributes charts are never negative, and it is desirable that they be positive. For a c chart, what values of the centerline c will ensure that the lower control limit is positive?
=+37. For a fixed subgroup size n, find the smallest value of p that will give a positive lower control limit on a p chart.
=+36. Explain the difference in the actions taken on a process when a point on a p chart exceeds the upper control limit versus the actions taken when a point falls below the lower control limit.
=+35. Using the data of Exercise 22, if the specification limits for the process are 60 ± .4, calculate the Cp and Cpk indexes. What conclusions can you draw about the capability of the process?
=+34. Based on your analysis in Exercise 18, if the specification limits for the process are 13 6 .2, calculate the Cp and Cpk indexes. What conclusions can you draw about the capability of the process?
=+b. Calculate the Cp, Cpu, Cpl, and Cpk indexes from the transformed data.
=+a. Describe a procedure for calculating capability indexes from the transformed data.
=+33. The data of Exercise 21 was analyzed by first transforming it into deviations from the nominal value and then running x and R charts on the transformed data. Suppose the specification limits on the process are .254 6 .01 inch.
=+32. Using the data of Exercise 16, we estimated the process standard deviation in Exercise 23. If the specification limits for the process are .40 6 .08, calculate the Cp and Cpk indexes. What conclusions can you draw about the capability of the process?
=+31. Use the data in Exercise 6 to calculate the Cp, Cpu, Cpl, and Cpk indexes. What do the indexes indicate about the capability of the process?
=+30. Use the formula given in Exercise 29 to calculate the proportion of the process that is out of specification in Exercise 28.
=+ proportion out of specification 5 P(z $ 3Cpk) 1 P(z $ 6Cp 2 3Cpk)Use this equation with the Cp and Cpk from Exercise 27 to estimate the proportion of the process that is not within the specification limits.
=+29. It can be shown that the following equation always holds for processes that can be described by a normal distribution (Farnum, N. R., Modern Statistical Quality Control and Improvement, Duxbury, Belmont, CA, 1994: 235):
=+28. A process with specification limits of 5 6 .01 has a Cp of 1.2 and a Cpk of 1.0. What is the estimated process average x from which these indexes are calculated?
=+27. A computer printout shows that a certain process has a Cp of 1.6 and a Cpk of .9. Assuming that the process is in control, what do these indexes say about the capability of this process?
=+26. A process has a Cp index of 1.2 and is centered on its nominal value. What proportion of the specification range is used by the process measurements?
=+25. Why must a process be in a state of statistical control before its capability can be measured?
=+b. If the specification limits for the process are.40 6 .08, estimate the proportions of the process measurements above the USL and below the LSL. Compare your results to those in Exercise 23(b)
=+24. Reconsider the results from Exercise 17(a).
=+b. Suppose the specification limits on the process are .40 6 .08. Assuming that a normal distribution can be used to describe the process measurements, estimate the proportion of the process measurements above the USL and below the LSL.
=+23. Reconsider the data from Exercise 16.a. Estimate the process standard deviation.
=+where the outer diameter of each casting (in mm) was recorded. The target diameter of each casting was 60 mm. The corresponding data is given here:Batch x1 x2 1 59.664 59.675 2 59.661 59.648 3 59.679 59.652 4 59.665 59.654 5 59.667 59.678 6 59.673 59.657 7 59.676 59.661 8 59.648 59.651 Batch x1
=+create a wide variety of products such as medical implants, furniture, tools, and even jewelry. The article“Improving the Process Capability of a Boring Operation by the Application of Statistical Techniques”(MIT Intl. J. Mech. Engr., 2012: 31–38) considered the production process of
=+22. Three-dimensional (3D) printing is a manufacturing technology that allows the production of three-dimensional solid objects through a meticulous layering process performed by a 3D printer. 3D printing has rapidly become a time-saving and economical way to
=+c. For comparison, create the x and R charts of the untransformed data, and evaluate these charts as in part (b).
=+b. Transform the data in this problem as described, and then create x and R charts of the transformed data. Use the extended list of out-of-control conditions (Figure 6.5) to evaluate these charts.
=+a. Use the formulas for the control limits of x and R charts to explain why the signals given by charting the deviations from nominal values will always be identical to the signals given by charting the untransformed process data.
=+data, for example, a reading of .258 would be converted to .2582.254 (the deviation from the nominal value), and then multiplied by 1000. Thus .258 transforms into 4, .254 transforms into 0, .256 becomes 2, .251 becomes 23, and so forth.
=+To simplify control chart calculations for such data, practitioners often code the data by transforming the measurements into deviations from the nominal value and then multiplying by a suitable power of 10 to eliminate decimal points. In the foregoing
=+21. Because processes are designed to produce products with fixed nominal dimensions, it is quite common to find that most of the variation in sample data occurs in the rightmost one or two decimal places. For example, the following data comes from a process making parts whose nominal length is
=+b. Do the charts in part (a) indicate that there are any other out-of-control signals present?
=+a. Recompute the control limits for both the x and s charts after removing the data from day 22.
=+20. In Exercise 19, suppose that an assignable cause was found for the unusually high average refractive index in subgroup 22.
=+19. The following table gives sample means and standard deviations, each based on subgroups of six observations of the refractive index of fiber-optic cable:Day x s Day x s 1 95.47 1.30 13 97.02 1.28 2 97.38 .88 14 95.55 1.14 3 96.85 1.43 15 96.29 1.37 4 96.64 1.59 16 96.80 1.40 5 96.87 1.52 17
=+c. If there are any out-of-control conditions found in parts (a) or (b), recalculate and interpret the revised x and R charts after eliminating these subgroups. (doing this assumes that assignable causes for out-of-control subgoups can be found prior to their elimination).
=+b. Construct an x chart for this data. Are there any out-of-control signals present?
=+Subgroup x1 x2 x3 x4 13 13.00 12.96 12.99 12.90 14 12.88 12.94 13.05 13.00 15 12.96 12.96 13.04 12.98 16 12.99 12.94 13.00 13.05 17 13.05 13.02 12.88 12.96 18 13.08 13.06 13.10 13.05 19 13.02 13.05 13.04 12.97 20 12.96 12.90 12.97 13.05 21 12.98 12.99 12.96 13.00 22 12.97 13.02 12.96 12.99 23
=+Operation by the Application of Statistical Techniques” (Intl. J. Sci. Engr. Research, Vol. 3, Issue 5, May 2012) investigated the production process of a particular bath faucet manufactured in India. The article reported the threaded stem diameter (target value being 13 mm) of each faucet in
=+18. When installing a bath faucet, it is important to properly fasten the threaded end of the faucet stem to the water-supply line. The threaded stem dimensions must meet product specifications, otherwise malfunction and leakage may occur. Authors of“Improving the Process Capability of a Boring
=+b. Construct an x chart for this data. Why are the control limits of this chart different from those in Exercise 16(b)?
=+17. Refer to the data of Exercise 16.a. Construct an s chart for this data, and check for special causes.
=+b. Construct an x chart for this data, and check for signs of special causes.
=+a. Construct an R chart for this data. Are any outof-control signals indicated by this chart?
=+16. Hourly samples of size 3 are taken from a process that produces molded plastic containers, and a critical dimension is measured. Data from the most recent 20 samples is given here:Hour x1 x2 x3 Hour x1 x2 x3 1 .36 .39 .36 11 .36 .32 .36 2 .33 .35 .30 12 .38 .47 .35 3 .51 .41 .42 13 .29 .45
=+85.2. Calculate the centerline and control limits of an R chart for this data.
=+15. Subgroups of four power units are selected once each hour from an assembly line, and the highvoltage output of each unit is measured. Suppose that the sum of the ranges of 30 such subgroups is
=+14. The control limits on x charts become closer together as the subgroup size n is increased (i.e., the A2 factor decreases as n increases). For a process that is in statistical control, does this imply that a control chart point is more likely to fall outside the control limits of an x chart
=+13. Using the extended list of “out-of-control” rules in Figure 6.5 (page 254), determine whether the processes that give rise to the adjacent control charts appear to be in statistical control. Circle any points at which an out-of-control condition is first signaled.
=+12. Suppose that the measuring instrument used to obtain data from a certain process is out of calibration, so that each of its reported measurements is off by 1 units from the true value. What effect does this have on the signals given by the x and R charts?
=+b. Using the normal distribution, what is the probability that a single control chart point falls above the UCL in a 3.09-sigma control chart?
=+a. Using the normal distribution, what is the probability that a single control chart point falls above the UCL in a 3-sigma control chart?
=+11. U.S. companies commonly use 3-sigma limits to establish control limits. Some other countries (e.g., Great Britain) use control limits that are 3.09 sigmas from the chart’s centerline.
=+10. When a process is in a state of statistical control, all of the points on a control chart should fall within the control limits. However, it is undesirable that all of the points should fall extremely near, or exactly on, the centerline of the control chart. Why?
=+Which method of choosing rational subgroups would be better able to detect when one of the machines is not in statistical control?
=+b. Method 2: Before reaching the conveyor system, a sample of five parts is taken from the output of machine 1; an hour later, five parts are taken from the output of machine 2; an hour later, five parts are sampled from machine 1; and so forth.
=+a. Method 1: Five parts per hour are sampled from the finished parts on the conveyor system each hour.
=+9. Two identical machines are used to make a particular metal part. The finished parts from both machines are mixed together on a conveyor system that moves the parts to a subsequent assembly operation. Consider the following two methods for generating rational subgroups for a control chart of
=+b. If instead the mean value is 3.00 and the standard deviation is .05, is the probability of conforming to specification smaller or larger than it was in part (a)?
=+a randomly selected cork will conform to specification?
=+used are x (subgroup mean), R (subgroup range), s (subgroup standard deviation), p(proportion nonconforming), c (number of nonconformities), and u (nonconformities per unit). A control chart derives its name from the name of the particular statistic
=+Control charts are constructed by taking successive samples from the output of a process, making measurements on the sampled items, and then plotting summary statistics of these results. Figure 6.4 shows a typical control chart. The samples, also called subgroups, of size n are taken at regular
=+cause variation is considered to be expected, but uncontrollable variation. Controllable variation, on the other hand, is variation for which we can find definite assignable causes, also called special causes. Assignable causes are frequently found when there are changes in brands of raw
=+Shewhart envisioned two types of variation that, when combined, account for all the variation in a process. The first type, common cause variation, is the result of the myriad imperceptible changes, or common causes, that occur in the everyday operation of a process.Common causes are
=+The recognition that variation is unavoidable in every repetitive process was well understood by the early pioneers of statistical quality control. To identify, and, when possible, eliminate sources of process variation, W. A. Shewhart introduced the control chart method in 1924.
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