Problems

1. A system has four components, A, B, C, D. The probability that each component will work is p(A) =.90, p(B) = .70, p(C) =.95, and p(D) = .60; all components must operate in order for the system to work. Since the probabilities associated with components B and D are low, backup components, B-B and D-D, with the same probabilities are provided:

1. Draw the box and line diagram for this system.
2. What is the probability that each component will not work?
3. What is the probability that the system will work without the backup components?
4. List the ways in which the system can work with the backup components.
5. What is the probability that the system will work with the backup components?
6. How much improvement in reliability do the backup components provide?

2. (One step beyond) A simple system consists of three components, A. B, C, as shown below, with the probability that each component will work. Component A costs $20,000 each, B costs $10,000 each, and C costs $6,000 each.

1. What is the probability that the system will work?
2. Management wishes to have at least a .95 probability that the system will work, and proposes to achieve this goal by adding redundant component C's to the system, which are identical to the existing component C. Each new C would switch on if the preceding C's failed. How many C's will be required?
3. Draw the diagram of the new system.
4. How much will it cost to achieve a .95 probability by adding C's?

 

3. The mean operating life (MTBF) of TV picture tubes is 4,000 hours, and the failure rate of the tubes can be modeled by a negative exponential distribution. Use Table 4S-1 in your textbook, or use the <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image152::/sites/dl/free/0072443901/24520/Image152.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image152 (0.0K)</a>Image152 or In x key on your pocket calculator to solve these problems.

1. Determine the probability that a picture tube will fail within 3,200 hours.
2. Determine the probability that a picture tube will last at least 6,400 hours.
3. The manufacturer wishes to provide a warranty on which he will be obligated to make a replacement of only 2% of the picture tubes sold. For how many hours should the picture tubes be warranted?
4. Does the MTBF come in the middle of this distribution?

4. The service life of automobile tires is modeled by a normal curve; the Mean Time Between Failures (MTBF) is 20,000 miles and the standard deviation is 800 miles. Use Table A in the Appendix to solve these problems.

1. Determine the probability that a tire will fail before 22,000 miles.
2. Determine the probability that a tire will last at least 19,000 miles.
3. The manufacturer wishes to provide a warranty on which he will be obligated to make a replacement of only 1% of the tires sold. For how many miles should he warrant the tires?
4. Does the MTBF come in the middle of this distribution?
5. What is the difference between modeling the service life with the negative exponential distribution or with the normal curve?

5. (One step beyond) A life test has shown that coffeemakers fail in use at the rate of 0.04257 per month. (That is: of all of the millions of coffeemakers in use, 4.257% can be expected to fail in any month.) The manufacturer warrants that a coffeemaker will last for 12 months of ordinary usage. What is the reliability of a coffeemaker?

Solutions

1.

2. Let p( <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image153::/sites/dl/free/0072443901/24520/Image153.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image153 (0.0K)</a>Image153 ) be the probability that component A does not work. 
   p( <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image154::/sites/dl/free/0072443901/24520/Image154.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image154 (0.0K)</a>Image154 ) = 1 - p(A) = 1 - .90 = .10.
   p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image155::/sites/dl/free/0072443901/24520/Image155.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image155 (0.0K)</a>Image155 ) = 1 - p(B) = 1 - .70 = .30.
   p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image156::/sites/dl/free/0072443901/24520/Image156.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image156 (0.0K)</a>Image156 ) = 1 - p(C) = 1 - .95 = .05.
   p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image157::/sites/dl/free/0072443901/24520/Image157.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image157 (0.0K)</a>Image157 ) = 1 - p(D) = 1 - .60 = .40.
3. p(system works) = p(A)p(B)p(C)p(D) = (.90)(.70)(.95)(.60) = .3591.
4. 1) A works; B works; C works; D works.
   2) A works; B fails; B-B works; C works; D works.
   3) A works; B works; C works; D fails; D-D works.
   4) A works; B fails; B-B works; C works; D fails; D-D works.
5. The probability will be the sum of the probabilities for each of the four cases above.
   1) p(A)p(B)p(C)p(D) = .3591.
   2) p(A)p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image158::/sites/dl/free/0072443901/24520/Image158.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image158 (0.0K)</a>Image158 )p(B-B)p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image159::/sites/dl/free/0072443901/24520/Image159.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image159 (0.0K)</a>Image159 )p(D) = .1077.
   3) p(A)p(B)p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image160::/sites/dl/free/0072443901/24520/Image160.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image160 (0.0K)</a>Image160 )p( <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image161::/sites/dl/free/0072443901/24520/Image161.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image161 (0.0K)</a>Image161 )p(D-D) = .1436.
   4) p(A)p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image162::/sites/dl/free/0072443901/24520/Image162.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image162 (0.0K)</a>Image162 )p(B-B)p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image163::/sites/dl/free/0072443901/24520/Image163.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image163 (0.0K)</a>Image163 )p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image164::/sites/dl/free/0072443901/24520/Image164.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image164 (0.0K)</a>Image164 )p(D-D) = .0431.
   

   The probability that the system will work is .6535.

6. The improvement is .6535 - .3591 = .2944. The backups have nearly doubled the reliability of the system, but it is still very low. Even with the backups, the system can be expected to fail about 1 time in 3.

2.

1. system works) = p(A)p(B)p(C) = .7351.
2. For one additional C:
   p(works) .7351
   p(A)p(B)p( <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image165::/sites/dl/free/0072443901/24520/Image165.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image165 (0.0K)</a>Image165 )p(C) = .1838
   p(works) .9189

   For 2 additional C's:

   p(works) = .9189
   p(A)p(B)p(<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image166::/sites/dl/free/0072443901/24520/Image166.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image166 (0.0K)</a>Image166 )p( <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image167::/sites/dl/free/0072443901/24520/Image167.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image167 (0.0K)</a>Image167)p(C) = .0460
   p(works) .9649 (> .95; stop)

   Two additional C's will be required.

4. Two C's will cost $12,000.

3.

1. <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image168 ::/sites/dl/free/0072443901/24520/Image168.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image168 (1.0K)</a>Image168
2. <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image169::/sites/dl/free/0072443901/24520/Image169.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image169 (1.0K)</a>Image169
3. p(failure before time, T) = .02.
   <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image170::/sites/dl/free/0072443901/24520/Image170.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image170 (1.0K)</a>Image170
   <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image171::/sites/dl/free/0072443901/24520/Image171.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image171 (0.0K)</a>Image171
   <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image172::/sites/dl/free/0072443901/24520/Image172.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image172 (1.0K)</a>Image172; <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image173::/sites/dl/free/0072443901/24520/Image173.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image173 (1.0K)</a>Image173 .
   T/4000 = 0.02.
   T = 80 hours
4. No, over half of the area falls to the left of the MTBF.

4.

1. Convert from the T scale to the z scale.
   z = (T - MTBF)/s = (22000 - 20000)/800 = 2.50.
   p(T < 22000) = p(z < 0) + p(0 < z < 2.50) = .5000 + .4938 = .9938.
2. Convert from the T scale to the z scale.
   z = (T - MTBF)/s = (19000 - 20000)/800 = -1.25.
   p(T > 19000) = p(-1.25 < z < 0) + p(z > 0) = .3944 + .5000 = .8944.
3. Let T' = the warranted service life. Then,
   p(T < T') = .01; p(T' < T < MTBF) = .5000 - .0100 = .4900. Find the area of .4900 in the table of the standardized normal curve, read the z value, and convert to the T scale.
   p(T' < T < MTBF) = p(z' < z < 0); z' = -2.33.
   T' = MTBF - z' = 20000 - 2.33(800) = 18,100 miles.
4. Yes; half of the area is on each side of the MTBF.
5. The negative exponential distribution is skew positive, and the normal model is symmetrical. The normal model indicates that half of the items will fail before the MTBF, while the exponential model indicates that over half of the items will fail before the MTBF.

5. P(coffeemaker works for 1 month) = 1- 0.04257 = 0.95743.
P(coffeemaker works for at least12 months) = <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image174::/sites/dl/free/0072443901/24520/Image174.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image174 (1.0K)</a>Image174 = the reliability.