1. Suppose that the time until failure in months for a type of car battery is ex
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Question
1. Suppose that the time until failure in months for a type of car battery is exponentially distributed with 0.018 ar Compute the mean and standard deviation for the life of a batteries of this type. b. What is the probability that a battery of this type will last less than four years? c. If a wheelchair requires two of these batteries, what is the probability that neither will last four years? d. What is the probability that a battery of this type will not last 7 years given that it lasts 5 years? 2. The number of ears arriving at a toll booth follows a Poisson process with an average of 20 cars per hour. a. What is the probability that three cars arrive in a ten minute period? b. What is the probability that the first car arrives within 6 minutes? c. What is the probability that the arrival time for the first car will be between 8 and 10 minutes? d. For what time period is the probability that at least one car will arrive is 0.85? e. What is the probability that 6 or more cars will arrive in less than 15 minutes? Give the exact simplified form for the pdf and use the fnlnt key to approximate the probability.Explanation / Answer
Question
= 0.018 years-1
Here pdf of mean life of battery
f(t) = 0.018 e-0.018t
F(t) = 1 - e-0.018t
(a) Here mean life = 1/0.018 = 55.555 months
Standard deviation of life = sqrt (1/0.018) = 7.4536 months
(b) Here four years means forty eight months
Pr(x < 48 hours) = 1 - e-0.018t = 1 - e-0.018 * 48 = 1 - 0.4215 = 0.5785
(c) Here we will use two of these batteries, so we have to find that neither of the will last four years
Pr(Neither of them will not four years) = Pr(first one will not last ) * Pr(second one will not last) = 0.5785 * 0.5785 = 0.3347
(d) Pr(t < 7 years l 5 yers) = Pr(t < 2 years) as exponential distribution is memoryless distribution.
so Pr(t < 2 years) = 1 - e-2 * 12 * 0.018 = 0.3508
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