A repairable components fails at times t = 3, 4, 6, 9, 11.No further failures are observed during the observation period which ends at t = 19.Determine (to 3 decimal places) the standard deviation of the time between failures.

A repairable components fails at times t = 1, 4, 7, 10, 11.No further failures are observed during the observation period which ends at t = 18.Determine (to 3 decimal places) the mean time between failures (MTBF).

The commulative distribution of a component is given as F(t) = 0.2e^(-0vt) of the range of the random variable T= 3 ≤ t ≤ αbr Determine the mean time to failure MTTF What is the probability that the component will fail 10hrs

We observe 11 components, starting from time t = 0. The component life times are independent. Observation ends immediately following the 9th failure, which occurs at time t = 13.48. Give (correct to 2 decimal places) an appropriate estimate for F(13.48), using the Mean ranking metho

Suppose that the lifetimes of old-fashioned TV tubes are normally distributed with a standard deviation of 1.2 years. Suppose also that exactly 35% of the TV tubes die before 4.5 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Suppose that the lifetimes of tires of a certain brand are normally distributed with a mean of 74,000 miles and a standard deviation of σ miles. These tires come with a 55,000-mile warranty. The manufacturer of the tires can adjust σ during the production process, but the adjustment of σ is quite costly. The manufacturer wants to set σ once and for all so that only 5% of the tires will fail before warranty expires. Find the standard deviation to be set. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.miles