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