Help Save & Exit Submi 2, 4, 4S, 5) mine a reasonable free-service period for a
ID: 418401 • Letter: H
Question
Help Save & Exit Submi 2, 4, 4S, 5) mine a reasonable free-service period for a model it will A manufacturer of programmable calculators is attempting to deterr introduce shortly. The manager of product testing has indicated that the calculators have an expected life of 39 months. Assume product life can be described by an exponential distribution. 2.60 2.70 2.80 2.90 3.00 3.10 3.28 3.38 3.48 3.50 3.68 3.70 3.80 3.90 4.00 4.10 4.20 0.8661 .9055 8. 0050 0.0045 0.0041 8.0037 0.0033 8.6030 .9827 9.8825 .8022 0.8020 8.8818 0.8017 0.0015 0.0014 0.0012 5.10 5.20 5.30 0.0743 0.0672 , 0608 .9558 0.0498 8.8450 8.8488 8.8369 0.0334 0.8302 0.8273 0.8247 0.0224 0.8202 0.0183 .0166 0.0150 0.9848 .8187 8.7488 8.6703 0.2 0.30 8.48 0.58 8.60 8.70 8.80 8.90 1.80 .6865 0.5488 8.4966 8.4493 8.4066 0.3679 .3329 0.3012 0.2725 8.2466 8.2231 0.2019 0.1827 5.50 5.68 5.70 5.88 5.98 6.00 6.10 6.20 6.30 6.40 6.58 6.60 6.70 1.2 1.30 1.40 1.50 1.60Explanation / Answer
1. (a) The items which are expected to fail during service period
Here T/ MTBF = 1.0 that corresponds to exp ( -T/MTBF) = 0.3679
Expected percentage = 100 ( 1-0.3679) = 0.6321 x100 = 63.21%
2. (b) In this case, exp ( -T/MTBF) = (1-0.2429) = 0.7571
which corresponds to T/ MTBF = 0.277
Service period = 39 months x0.277 = 10.8 months =11 months
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