An electronic chess game has a useful life that is exponential with a mean of 30 month. Determine each of the following:
a. The probability that any given unit will operate for at least (1) 39 months. (2) 48 men the (3) 60 months.
b. The probability that any given unit will fail sooner than (1) 33 months. (2) 15 months. (3) 6 months.
c. The length of service tune after which the percentage of tilled units will approximately equal (1) 50 percent, (2) 85 percent, (3) 95 percent, (4) 99 percent.
a)
To determine: The probability the unit will operate at least for the following times.
Introduction:
Mean time between failures (MTBF):
The mean time between failures is a term which denotes the time that is elapsed between the first failure of a product and the second failure of a product. It is calculated during the normal system operation.
Answer to Problem 12P
1) 39 months = 0.2725
2) 48 months = 0.2019
3) 60 months = 0.1353
Explanation of Solution
Given information:
MTBF = 30 months
Formula to calculate the probability of no failure before a time period:
Calculation of probability:
1) 39 months:
Hence, the probability the unit will operate at least for the 39 months is 0.2725.
2) 48 months:
Hence, the probability the unit will operate at least for the 48 months is 0.2019.
3) 60 months:
Hence, the probability the unit will operate at least for the 60 months is 0.1353.
b)
To determine: The probability the unit will fail before the following times.
Introduction:
Mean time between failures (MTBF):
The mean time between failures is a term which denotes the time that is elapsed between the first failure of a product and the second failure of a product. It is calculated during the normal system operation.
Answer to Problem 12P
1) 33 months = 0.6671
2) 15 months = 0.3935
3) 6 months = 0.1813
Explanation of Solution
Given information:
MTBF = 30 months
Formula to calculate the probability of failure before a time period:
Calculation of probability:
1) 33 months:
Hence, the probability the unit will fail before 33 months is 0.6671.
2) 15 months:
Hence, the probability the unit will fail before 15 months is 0.3935.
3) 6 months:
Hence, the probability the unit will fail before 6 months is 0.1813.
c)
To determine: The length of service time for the percentage of failed units.
Introduction:
Mean time between failures (MTBF):
The mean time between failures is a term which denotes the time that is elapsed between the first failure of a product and the second failure of a product. It is calculated during the normal system operation.
Answer to Problem 12P
1) 50 percent = 21 months.
2) 85 percent = 57 months.
3) 95 percent = 90 months.
4) 99 percent = 138 months.
Explanation of Solution
Given information:
MTBF = 30 months
Formula to calculate the probability of no failure before a time period:
Calculation of probability:
The different probabilities are obtained from the above table.
1) 50 percent:
From the table for the value of 0.50 is equivalent to:
Therefore,
Hence, the length of service time for the percentage of failed units is 21 months.
2) 85 percent:
From the table for the value of 0.15 is equivalent to:
Therefore,
Hence, the length of service time for the percentage of failed units is 57 months.
3) 95 percent:
From the table for the value of 0.05 is equivalent to:
Therefore,
Hence, the length of service time for the percentage of failed units is 90 months.
4) 99 percent:
From the table for the value of 0.01 is equivalent to:
Therefore,
Hence, the length of service time for the percentage of failed units is 138 months.
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