Question

In: Math

5) The lifetime of a certain type of batteries follows an exponential distribution with the mean...


5) The lifetime of a certain type of batteries follows an exponential distribution with the mean of 12 hours.
a) What is the probability that a battery will last more than 14 hours?
b) Once a battery is depleted, it is replaced with a new battery of the same type. Assuming independence between lifetimes of batteries, what is the probability that exactly 2 batteries will be depleted within 20 hours?
c) What is the probability that it takes less than 30 hours until the fourth battery is needed?

Solutions

Expert Solution

Let X be the lifetime of the batteries      
X follows Exponential distribution with mean = 12 hours      
Hence, λ = 1/12 batteries fail per hour      
       
a) To find P(a battery will last more than 14 hours)      
that is to find P(X > 14)      
We use Excel function EXPON.DIST to find the probability      
P(X > 14) = 1 - P(X < 14)      
   = 1 - EXPON.DIST(14, 1/12, TRUE)     
   = 1 - 0.6886     
   = 0.3114     
P(a battery will last more than 14 hours) = 0.3114      
       
b) To find P(exactly 2 batteries will be depleted within 20 hours)      
Let Y be the number of batteries depleting in 20 hours      
We have 1/12 batteries depleted per hour      
Thus mean number of batteries depleting in 20 hours = 20/12      
Y ~ Poisson distribution with λ = 20/12 batteries depleting in 20 hours      
P(exactly 2 batteries will be depleted within 20 hours)      
that is to find P(Y = 2)      
We use Excel function POISSON.DIST to find the probability      
P(Y = 2) = POISSON.DIST(2, 20/12, FALSE)      
                = 0.2623      
P(exactly 2 batteries will be depleted within 20 hours) = 0.2623      
       
c) We have 1/12 batteries depleted per hour      
Thus 4 batteries will be depleted in 4*12 = 48 hours      
Let Y be the time for depletion of 4 batteries      
Y follows Exponential distribution with λ = 1/48      
To find P(Y < 30)      
We use Excel function EXPON.DIST to find the probability      
P(Y < 30) = EXPON.DIST(30, 1/48, TRUE)      
   = 0.4647     
P(it takes less than 30 hours until the fourth battery is needed) = 0.4647      
       


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