Question

A manufacturer determines that, on the average, a television set is used 1.6 hours per day. A one-year warranty is offered on the picture tube having a mean time to failure of 2000 hours. If the distribution is exponential,

a) What percentage of the tube will fail during the warranty period?

b) What is the design life if a reliability of 0.95 is desired?

c) What is the probability that this tube will not fail during the first 50 days, but it will not complete 150 days without failures.

Answer 1

in exponential distribution :

λ = 1/mean = 1/2000

probability of failure within t hours

P(x<=t) = 1 - e^(-t/2000) {time of failure}

a.

fail during 1 year = fail within 365*1.6 hours = fail within 584 hours

P(fail during 1 year) = P(x<=584) = 1 - e^(-584/2000) = 0.25

P(fail during 1 year) = 0.25

b.

reliability = 0.95

P(failue) = 1-0.95 = 0.05

t for P(x<=t) = 0.05

0.05 = 1 - e^(-t/2000)

t = -2000(ln(1-0.05)) = 102.58 hours

design life for reliability 0.95 = 102.58 hours

= (102.58/1.6) days = 64.1125 days {on average 1.6 hours use per day}

c.

P(50<=x<=150) = P(x<=150) - P(x<=50)

= [ 1 - e^(-150/2000) ] - [ 1 - e^(-50/2000) ]

P(50<=x<=150) = 0.0476 = probability of failure after 50 days but within 150 days

P.S. (please upvote if you find the answer satisfactory)