Answer:
The variance of the times between accidents is of 1849 days squared.
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean 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 variance of the exponential distribution is:
[tex]Var = \frac{1}{\mu^2}[/tex]
Assume that the mean time between accidents is 43 days.
This means that [tex]m = 43, \mu = \frac{1}{43}[/tex]
What is the variance of the times between accidents?
[tex]Var = \frac{1}{(\frac{1}{43})^2} = 43^2 = 1849[/tex]
The variance of the times between accidents is of 1849 days squared.
