The length of time until the breakdown of an essential piece of equipment is imp

Question

The length of time until the breakdown of an essential piece of equipment is important in the decision of the use of auxiliary equipment. Assume time to breakdown of a randomly chosen generator, Y, follows an exponential distribution with a mean of 15 days. What is the probability a generator will break down in the next 21 days? A company owns 7 such generators. Let X denote the random variable describing how many generators break down in the next 21 days. Assuming the breakdown of any one generator is independent of breakdowns of the other generators, what is the probability that at least 6 of the 7 generators will operate for the next 21 days without a breakdown?

Explanation / Answer

a)

The left tailed area in an exponential distribution is          
          
Area = 1 - e^(-lambda*x)          
          
As          
          
x = critical value =    21      
          
          
Then          
          
Area =    0.753403036   [ANSWER]

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b)

Here, the probability of no breakdown in the next 21 days is 1 - 0.753403036 = 0.246596964.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    7      
p = the probability of a success =    0.246596964      
x = our critical value of successes =    6      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   5   ) =    0.998758638
          
Thus, the probability of at least   6   successes is  
          
P(at least   6   ) =    0.001241362 [ANSWER]

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