HW 4 Solution - STAT 350 Assignment 4 Solutions (70 points)...

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STAT 350 Assignment 4 Solutions (70 points) CHAPTER 5 1. (a) (3) Sampling without replacement means that no repeated items will occur in any sample. There are 10 possible such samples of size 3: {a,b,c}, {a,b,d}, {a,b,e}, {a,c,d}, {a,c,e}, {a,d,e}, {b,c,d}, {b,c,e}, {b,d,e}, {c,d,e}. As you learned in the discussion of the binomial distribution, the number of ways to choose a sample of 3 distinct items from a list of 5 items is given by 3 5 = 10, which shows that the list above is indeed complete. (b) (2) A contains 3 samples from the list in (a); i.e., A = { {a,b,c}, {a,c,d}, {a,c,e}} (c) (2) The complement of A is A = {{a,b,d}, {a,b,e}, { {a,d,e}, {b,c,d}, {b,c,e}, {b,d,e}, {c,d,e}} 2. A Venn diagram of these three events is (a) (2) The event ‘at least one plant is completed by the contract date’ is represented by the shaded area covered by all three circles: (b) (2) The event ‘all plants are completed by the contract date’ is the shaded area where all three circles overlap: (c) (2) The event ‘none of the plants is completed by the contract date’ is the complement of the shaded area in (a). (d) (2) The event ‘only the plant at site 1 is completed by the contract date’ is shown shaded: (e) (2) The event ‘exactly one of the three plants is completed by the contract date is: (f) (2) The event ‘either the plant at Site 1 or Site 2 or both plants are completed by the contract date’ is:
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7. (3) The event A and B is the shaded area where A and B overlap in the following Venn diagram. Its complement consists of all events that are either not in A or not in B (or not in both). That is, the complement can be expressed as A or B . 9. (a) (2) The total of 724 +751 solder joints overstates the actual number found, since 316 solder joints were found by both inspectors. To avoid such double-counting (i.e., the 316 is part of both the 724 joints found by inspector A and the 751 joints found by Inspector B), 316 should be subtracted from the raw totals, which means that 724 + 751 - 316 = 1,159 distinct joints were identified by the inspectors together. The important point to note in this problem is that the events ‘Inspector A finds a defective solder joint’ and ‘Inspector B finds a defective solder joint’ are not necessarily mutually exclusive, so we can not simply add the numbers of joints (or, equivalently, the probabilities of finding defective joints) for both inspectors. (b) (2) A and B contains 724 - 316 = 408 solder joints. 10. (2) For i = 1, 2, 3, . .. , 10, let A i denote the event ‘component i functions correctly’. The problem indicates that P(A i ) = .999 for each component which, by the law of complementary events, means that each P(A i ) = 1-.999 = .001. For a series system built from these 10 components to function correctly, all ten
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This note was uploaded on 09/21/2011 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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HW 4 Solution - STAT 350 Assignment 4 Solutions (70 points)...

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