The space shuttle can travel about 8*10 to the power of 5 centimeters per second. Is it more appropriate to report this rate as
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Answer:
a) P(t>3)=0.30
b) P(t>10|t>9)=0.67
Step-by-step explanation:
We have a repair time modeled as an exponentially random variable, with mean 1/0.4=2.5 hours.
The parameter λ of the exponential distribution is the inverse of the mean, so its λ=0.4 h^-1.
The probabity that a repair time exceeds k hours can be written as:
(a) the probability that a repair time exceeds 3 hours?
(b) the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours?
The exponential distribution has a memoryless property, in which the probabilities of future events are not dependant of past events.
In this case, the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours is equal to the probability that a repair takes at least (10-9)=1 hour.