What are the Calculations for Exponential Distribution in Bank Arrival Times?

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SUMMARY

The discussion focuses on the calculations for exponential distribution related to bank arrival times, specifically with a rate parameter of $\lambda = 2$. The expected time between arrivals is correctly calculated as 30 minutes. The standard deviation was initially miscalculated; the correct standard deviation is $\sigma = 0.5$. The probabilities $P(X \le 4)$ and $P(2 \le X < 5)$ were computed accurately, yielding approximately 0.9996 and 0.018270, respectively.

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shamieh
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Let $X$ = the time between two successive arrivals at the drive-up window of a local bank. $X$ has an exponential distribution with $\lambda = 2$. That is the probability density of $X$ is $f(X | \lambda) = \lambda e^{-\lambda x}, X > 0 $ with $\lambda = 2$. Compute the following:

a) The expected time between two successive arrivals.

b) The standard deviation of the time between successive arrivals.

c) $P(X\le4)$

d) $(P(2\le X<5)$

I just need someone to check my work to make sure I'm doing these right.

I think I've got the first part.. would it be

a) $\mu = 1/2 => 30$ minutes or half an hour?

And for b) I got:

b) $\sigma^2 = 1/\lambda^2 = (1/2)^2 = (1/4)^2 = 1/16$
so $\sigma^2 = \sqrt{1/16} => \sigma = .25$ ?

c) $P(X \le 4) = 1 - e^{-2*4} \approx 0.9996$

d) $\int^5_2 2e^{-2x} dx \approx 0.018270$
 
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Hi shamieh,

your answers are correct except for b). For an exponential distributed variable $X$ holds indeed $\mbox{Var}(X) = \frac{1}{\lambda^2} = \frac{1}{4}$ and hence for the standard deviation $\sigma^2 = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
 

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