The exponential distribution is defined as:

f(x|ν) = νe^(-νx) for x = 0, and 0 otherwise.

The expected value E(X) and the standard deviation σX for an exponential distribution are both equal to 1/ν.

So, we are looking for P(X ∈ [1/ν - 1/ν ; 1/ν + 1/ν]) = P(X ∈ [0 ; 2/ν]).

The cumulative distribution function (CDF) of an exponential distribution is:

F(x|ν) = 1 - e^(-νx) for x = 0, and 0 otherwise.

So, we can find the probability as follows:

P(X ∈ [0 ; 2/ν]) = F(2/ν|ν) - F(0|ν) = (1 - e^(-ν*2/ν)) - (1 - e^(-ν*0)) = 1 - e^-2 - 1 = 1 - e^-2.

So, the probability P(X ∈ [E(X) - σX ; E(X) + σX]) for a random variable X that has an exponential distribution with parameter ν is 1 - e^-2.

Question

The exponential distribution is defined as:

f(x|ν) = νe^(-νx) for x >= 0, and 0 otherwise.

The expected value E(X) and the standard deviation σX for an exponential distribution are both equal to 1/ν.

So, we are looking for P(X ∈ [1/ν - 1/ν ; 1/ν + 1/ν]) = P(X ∈ [0 ; 2/ν]).

The cumulative distribution function (CDF) of an exponential distribution is:

F(x|ν) = 1 - e^(-νx) for x >= 0, and 0 otherwise.

So, we can find the probability as follows:

P(X ∈ [0 ; 2/ν]) = F(2/ν|ν) - F(0|ν) = (1 - e^(-ν*2/ν)) - (1 - e^(-ν*0)) = 1 - e^-2 - 1 = 1 - e^-2.

So, the probability P(X ∈ [E(X) - σX ; E(X) + σX]) for a random variable X that has an exponential distribution with parameter ν is 1 - e^-2.

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