The standard deviation of the exponential distribution is of 0.1.
-----------------
The exponential probability distribution, with mean and standard deviation m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
-----------------
The distribution is described by:
[tex]f(x) = 10e^{-10x}[/tex]
Thus, the mean and standard deviation are given by:
[tex]m = \frac{1}{\mu} = \frac{1}{10} = 0.1[/tex]
The standard deviation is of 0.1.
A similar problem is given at https://brainly.com/question/17039711
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
Read more:
https://brainly.com/question/18562832
