Question

What is the standard deviation of the exponential distribution whose density function is f(x)=10*e^(-10x)?

Answer 1

We are required to find the standard deviation of the exponential distribution.

The standard deviation of the exponential distribution whose density function is f(x)=10*e^(-10x) is 0.1.

By convention, exponential probability distributions have density function in the form:

f(X) = f(x)=μ*e^(-μx)............... equation 1.

Where μ = decay parameter.

However, the decay parameter is related to the standard deviation of the distribution by the formular:

μ = 1/m

Where m = Standard deviation of the distribution.

Therefore, by comparison of equation 1 with f(x)=10*e^(-10x).

It is evident that the decay parameter, μ = 10

Therefore, the standard deviation, m = 1/μ = 1/10 = 0.1.

The standard deviation of the exponential distribution whose density function is f(x)=10*e^(-10x) is 0.1

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Answer 2

The standard deviation of the exponential distribution is of 0.1.

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The exponential probability distribution, with mean and standard deviation m, is described by the following equation:  

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

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The distribution is described by:

$f(x) = 10e^{-10x}$

• Which has $\mu = 10$

Thus, the mean and standard deviation are given by:

$m = \frac{1}{\mu} = \frac{1}{10} = 0.1$

The standard deviation is of 0.1.

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