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The question states: Let X = time between calls to a service center, It is known that X follows an exponential distribution with a mean of 15 minutes,

part 1. What is the probability that x is greater than 20 minutes.

So far this is the only one I think I've been able to work out. Since we know that the average is 15 i.e. $ \mu$ = 15, then $\lambda$ = $\frac{1}{15}$, which means that the probability of X > 20 is:

P(X>20) = $e^{-\lambda x}$ = $e^{-1}$ or .3678 (36.78%)

the other questions are as follows:

part 2: if there are no service calls during the last 10 minutes what is the probability that there will be no service call during the next 30 minutes?

part 3: find the 90th percentile of X

part 4: what is the median time between calls to the service center.

can anyone help me solve these problems?

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1 Answer 1

up vote 2 down vote accepted

$1.$ Minor slip, you want $e^{-20/15}$. In general, $P(X>x)=e^{-\lambda x}$. Here $\lambda=1/15$ and $x=20$.

$2.$ The exponential distribution is memoryless. The fact that there have been no calls for $10$ minutes makes no difference. The answer is just $P(X>30)$, which is $e^{-30/15}$.

$3.$ You want to find the $x$ such that $P(X>x)=0.1$. So you want to solve the equation $e^{-x/15}=0.1$. Take the natural logarithm of both sides. I would prefer first rewriting the equation as $e^{x/15}=10$, in order to avoid negative numbers in the calculation.

$4.$ This is basically the same problem as ($3$), you want the $m$ such that $e^{-m/15}=1/2$. Note that the mean is a fair bit larger than the median. That's because of the long tail on the right.

Remark: In case the memorylessness of the exponential has not been proved, here is a proof. By the usual formula for conditional probabilities, we have $$P((X\gt a+x)| (X>a))=\frac{P((X\gt a+x) \cap (X \gt a))}{P(X>a)}.$$ The numerator on the right is just $P(X>a+x)$, which is $e^{-\lambda(x+a)}$. The denominator is $e^{-\lambda a}$. Divide. We get $e^{-\lambda x}$.

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haha @Andre that's not a typo that's a conceptual error. i didn't know what the answer was. so thanks! –  franklin May 8 '12 at 0:32

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