Can someone please explain this! Problem 4S-12 An electronic chess game has a us
ID: 3022212 • Letter: C
Question
Can someone please explain this!
Problem 4S-12
An electronic chess game has a useful life that is exponential with a mean of 31 months. Determine each of the following:
The probability that any given unit will operate for at least (1) 42 months, (2) 50 months, (3) 60 months. (Use the "Probability" values to 1 decimal place for intermediate calculations.)
The probability that any given unit will fail sooner than (1) 29 months, (2) 12 months, (3) 5 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) 60 percent, (2) 80 percent, (3) 90 percent, (4) 99 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)
An electronic chess game has a useful life that is exponential with a mean of 31 months. Determine each of the following:
Explanation / Answer
just substitute the value of MTBF =31 in each and every question for given respective values of t = 42,56,90......etc we can calculate all th vlues for the given problem above.
a. 1 when t =42. then e-42/31 = e-1.35 = .2725
similarly for all other values you can calculate just by keeping the values in T and MTBF =31.
for c part of the question they hve given 60% mens value of 0.6 which is to be subtacted from 1 = .4
0.4 = e-T/ 31.
applying log on both sides we get the value of T for every percentages they have given.
ln 0.4 = - T / 31.
T = - 31 X ln 0.4 = 28.4 = 28 months.