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:

The probability that any given unit will operate for at least (1) 39 months, (2) 48 months, (3) 60 months. (Enter your answer to 4 decimal places using probabilities shown in the above table.)

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. (Enter your answer to the nearest whole month using the closest appropriate table value.)

T / MTBF e-T / MTBF T / MTBF e-T / MTBF T / MTBF e-T / MTBF 0.10 0.9048 2.60 0.0743 5.10 0.0061 0.20 0.8187 2.70 0.0672 5.20 0.0055 0.30 0.7408 2.80 0.0608 5.30 0.0050 0.40 0.6703 2.90 0.0550 5.40 0.0045 0.50 0.6065 3.00 0.0498 5.50 0.0041 0.60 0.5488 3.10 0.0450 5.60 0.0037 0.70 0.4966 3.20 0.0408 5.70 0.0033 0.80 0.4493 3.30 0.0369 5.80 0.0030 0.90 0.4066 3.40 0.0334 5.90 0.0027 1.00 0.3679 3.50 0.0302 6.00 0.0025 1.10 0.3329 3.60 0.0273 6.10 0.0022 1.20 0.3012 3.70 0.0247 6.20 0.0020 1.30 0.2725 3.80 0.0224 6.30 0.0018 1.40 0.2466 3.90 0.0202 6.40 0.0017 1.50 0.2231 4.00 0.0183 6.50 0.0015 1.60 0.2019 4.10 0.0166 6.60 0.0014 1.70 0.1827 4.20 0.0150 6.70 0.0012 1.80 0.1653 4.30 0.0136 6.80 0.0011 1.90 0.1496 4.40 0.0123 6.90 0.0010 2.00 0.1353 4.50 0.0111 7.00 0.0009 2.10 0.1255 4.60 0.0101 2.20 0.1108 4.70 0.0091 2.30 0.1003 4.80 0.0082 2.40 0.0907 4.90 0.0074 2.50 0.0821 5.00 0.0067

Explanation / Answer

Since the mean of the exponential is 30, the probability distribution ==> f(x)=(1/30)e^(-x/30)
Integrating (0 to x), we find that the cumulative distribution ==> F(x) = 1-e^(-x/30)
Note: cumulative distribution is the probability the unit will fail before x.

a) (i) 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)

= 0.27253 = 27.253%

(ii) The probability that the unit lasts at least 48 months is
= 1-(the probability the unit will fail before 48 months)
= 1-F(48)
= e^(-48/30)

= 0.2019 = 20.19%

(iii) The probability that the unit lasts at least 60 months is
= 1-(the probability the unit will fail before 60 months)
= 1-F(60)
= e^(-60/30)

= 0.13534 = 13.534%

b) (i) The probability that the unit lasts less than 33 months is actually the CDF itself
= F(33)
= 1 - e^(-33/30)

= 1 - 0.33287

= 0.66713 = 66.713%

(ii) The probability that the unit lasts less than 15 months is actually the CDF itself
= F(15)
= 1 - e^(-15/30)

= 1 - 0.60653

= 0.39347 = 39.347%

(iii) The probability that the unit lasts less than 6 months is actually the CDF itself
= F(6)
= 1 - e^(-6/30)

= 1 - 0.81873

= 0.18127 = 18.127%

c) (i) To find the length of time after which 50 percent fail, we will find 'x' such that F(x)=0.5

==> F(x) = 1 - e^(-x/30) = 0.5

==> e^(-x/30)=0.5

==> -x/30 = ln(0.5)

==> x = -30 ln(0.5)

= -30 * -0.6931 = 20.7944 ~ 21 months

(ii) To find the length of time after which 85 percent fail, we will find 'x' such that F(x)=0.85

==> F(x) = 1 - e^(-x/30) = 0.85

==> e^(-x/30)=0.15

==> -x/30 = ln(0.15)

==> x = -30 ln(0.15)

= -30 * -1.8971 = 56.9136 ~ 57 months

(iii) To find the length of time after which 95 percent fail, we will find 'x' such that F(x)=0.95

1==> F(x) = 1 - e^(-x/30) = 0.95

==> e^(-x/30)=0.05

==> -x/30 = ln(0.05)

==> x = -30 ln(0.05)

= -30 * -2.9957

= 89.872 ~ 90 months

(iv) To find the length of time after which 99 percent fail, we will find 'x' such that F(x)=0.99

==> F(x) = 1 - e^(-x/30) = 0.99

==> e^(-x/30)=0.01

==> -x/30 = ln(0.01)

==> x = -30 ln(0.01)

= - 30 * -4.6052

= 138.155 ~ 138 months

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