Homework SourseAn electronic chess game has a useful life that is exponenti
An electronic chess game has a useful life that is exponential with a mean of 32 months. Determine each of the following:
The probability that any given unit will fail sooner than (1) 30 months, (2) 16 months, (3) 8 months. (Use the \"Probability\" values to 1 decimal place and other values to 4 decimal places for intermediate calculations.)
The length of service time after which the percentage of failed units will approximately equal (1) 45 percent, (2) 85 percent, (3) 90 percent, (4) 98 percent. (Use the \"Probability\" values to 1 decimal place and other values to 4 decimal places for intermediate calculations. Round your final answers to the nearest whole number)
| 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 | |
Solution
Since the mean of the exponential is 32, the pdf is f(x)=(1/32)e^(-x/32)
Integrating (0 to x), we find that the cdf is F(x) = 1-e^(-x/32)
Remember the cdf is the probability the unit will fail before x.
The probability that any given unit will fail sooner than (1) 30 months,
We have to calculate calculate F(30):
F(30)= 1-e^(-30/32) = 0.608 i.e 60.8%
The probability that any given unit will fail sooner than (2) 16 months,
We have to calculate calculate F(16):
F(16)= 1-e^(-16/32) = 0.393 i.e 39.3%
The probability that any given unit will fail sooner than (3) 8 months.
We have to calculate calculate F(08):
F(08)= 1-e^(-08/32) = 0.221 i.e 22.1%
We\'ve been calculating the probability/percentage this whole time. It\'s what the cdf equals. So, to find the length before 45 percent fail, we want to find x such that F(x)=0.45
1-e^(-x/32) = 0.45
e^(-x/32) = 0.45 [subtract 1 from both sides and multiply by -1]
-x/32 = ln(0.45)
x = -32*ln(0.45)
= 25.55
For 85%
x = -32*ln(0.85) = 5.20
For 90%
x = -32*ln(0.90) = 3.37
For 98%
x = -32*ln(0.98) = 0.646
