An electronic chess game has a useful life that is exponential with a mean of 30

Question

An electronic chess game has a useful life that is exponential with a mean of 30 months. Determine each of the following: T / MTBF e-T / MTBF T / MTBF e-T / MTBF T / MTBF e-T / MTBF 0.10 .9048 2.60 .0743 5.10 .0061 0.20 .8187 2.70 .0672 5.20 .0055 0.30 .7408 2.80 .0608 5.30 .0050 0.40 .6703 2.90 .0550 5.40 .0045 0.50 .6065 3.00 .0498 5.50 .0041 0.60 .5488 3.10 .0450 5.60 .0037 0.70 .4966 3.20 .0408 5.70 .0033 0.80 .4493 3.30 .0369 5.80 .0030 0.90 .4066 3.40 .0334 5.90 .0027 1.00 .3679 3.50 .0302 6.00 .0025 1.10 .3329 3.60 .0273 6.10 .0022 1.20 .3012 3.70 .0247 6.20 .0020 1.30 .2725 3.80 .0224 6.30 .0018 1.40 .2466 3.90 .0202 6.40 .0017 1.50 .2231 4.00 .0183 6.50 .0015 1.60 .2019 4.10 .0166 6.60 .0014 1.70 .1827 4.20 .0150 6.70 .0012 1.80 .1653 4.30 .0136 6.80 .0011 1.90 .1496 4.40 .0123 6.90 .0010 2.00 .1353 4.50 .0111 7.00 .0009 2.10 .1255 4.60 .0101 2.20 .1108 4.70 .0091 2.30 .1003 4.80 .0082 2.40 .0907 4.90 .0074 2.50 .0821 5.00 .0067 -------------------------------------------------------------------------------- a. The probability that any given unit will operate for at least (1) 39 months, (2) 48 months, (3) 60 months. T e-T/MTBF (1) 39 (2) 48 (3) 60 -------------------------------------------------------------------------------- b. The probability that any given unit will fail sooner than (1) 33 months, (2) 15 months, (3) 6 months. T e-T/MTBF (1) 33 (2) 15 (3) 6 -------------------------------------------------------------------------------- c. The length of service time after which the percentage of failed units will approximately equal (1) 50 percent, (2) 85 percent, (3) 95 percent, (4) 99 percent. 1?e-T/MTBF T (whole months) (1) 50% (2) 85% (3) 95% (4) 99%

Explanation / Answer

Since the mean of the exponential is 30, the pdf is f(x)=(1/30)e^(-x/30) Integrating (0 to x), we find that the cdf is F(x) = 1-e^(-x/30) Remember the cdf is the probability the unit will fail before x. I'm not going to do your homework for you, but I'll do (1) from each set. A1) The probability that the unit lasts at least 39 months is 1-(the probability the unit will fail before 39 months) 1-F(39) e^(-39/30) [I don't have a calculator, so I'll leave that to you.] B1) This is the exact definition of the cdf, so just calculate F(33): F(33) 1-e^(-33/30) C1) We've been calculating the probability/percentage this whole time. It's what the cdf equals. So, to find the length before 50 percent fail, we want to find x such that F(x)=0.5 F(x) = 0.5 1-e^(-x/30) = 0.5 e^(-x/30) = 0.5 [subtract 1 from both sides and multiply by -1] -x/30 = ln(0.5) x = -30*ln(0.5) [after you get the answer, remember it's in months]

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