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5. For a certain machine it can be assumed that the probability that it fails in...
A machine is taken out of production either if it fails or after 5 hours, whichever comes first. By running similar machines until failure, it has been found that time to failure X has the Weibull distribution with the c.d.f. F(x) = 1-e-(F) 0 x where a 0.75 and B 8. The time until the machine is taken out of production can be represented as Y. Develop a step-by-step procedure for generating Y. A machine is taken out of production...
A machine is taken out of production either if it fails or after 5 hours, whichever comes first. By running similar machines until failure, it has been found that time to failure X has the Weibull distribution with a = 8, b = 0.75, and n = 0. Thus, the time until the machine is taken out of production can be represented as Y = min(X, 5). Develop a step-by-step procedure for generating Y.
The time between failures of a laser in a machine, X, is exponentially distributed with a mean of 25,000 hours. In other words, 1 a= (failures/hour). 25,000 Exponential Distribution (pdf): f(x) = 1.0-\x, for x > 0. (a) What is the probability that the next failure occurs in 27,000 hours? (b) What is the expected time until the third failure? (c) What is the probability that the time until the third failure exceeds 25,000 hours?
sorry it is blurry The time between failures of a laser in a machine, X, is exponentially distributed with a mean of 25,000 hours. In other words, X= (failures/hour). 25,000 Exponential Distribution (pdf): f(x) = 1.e-r, for 2 > 0. (a) What is the probability that the next failure occurs in 27,000 hours? (b) What is the expected time until the third failure? (c) What is the probability that the time until the third failure exceeds 25,000 hours?
The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standar deviation of 100 kilograms per square centimeter c) what strength is exceeded by 95% of the samples? The life of a semiconductor laser at a constant power is normally distributed with a mean of 7000 hours and a standar deviation of 600 hours a) what is the probability that a laser fails before 5000...
A manufacturing plant uses 6 specific machine tools. The life of the machine tool can be modeled by an exponential distribution with mean time between failures (MTBF) B-100 hours. Find the probability that 3 out of 6 machine tools will fail on or before 80 hours of operation. 5.
About 1% of a certain type of LED diodes fails during a 24-hour test. Failures are assumed to be independent.Consider a lightstrip consisting of 10 such LED diodes.Let X be the number of LED diodes that fail during the 24-hour test.i) What distribution does X follow?ii) What is the probability that the lightstrip will burn for the full 24-hour test with no LED diodes failure?iii) What is the probability that the lightstrip will lose at least 3 LED diodes during...
Assume the inspection time of a machine follows an exponential distribution with parameter equal to 1. (a) Calculate the probability that the inspection time lasts more than 2 hours. (b) Calculate the probability that the total inspection time lasts more than 4 hours if we have already spent 2 hours inspecting the machine. (c) Every time we need to spend more that 2 hours inspecting a machine we get paid a bonus $1000. Calculate the probability that for the next...
5. The life of a type of electric light bulb can be modeled by an exponential distribution with 0.001 (a) What is the average life of this type of light bulb? And determine the probability that a randomly picked light bulb has a life shorter than the average life. (2pt) (b) Determine the probability that a light bulb fails within 1200 hours. (2pt) (c) If 4 bulbs are installed at the same time and work independently in a house, what...
I can do the first part of the question 1a, could someone show me step by step how to do do 1b? ) Y.Ya..., Y, form a random sample from a probability distribution with cumu- lative distribution function Fy (u) and probability density function fr(u). Let Write the cumulative distribution function for Ya) in terms of Fy(y) and hence show that the probability density function for Yy is fy(1)(y) = n(1-Fr (v))"-ify(y). [8 marks] (b) An engineering system consists of...