The number of cracks in a section of interstate highway that are significant enough to require repair is assumed to follow a Poisson distribution with a mean of two cracks per mile.

What is the probability that there are no cracks that require repair in 5 miles of highway?

Respuesta :

Answer:

4.54 x 10⁻⁵

Step-by-step explanation:

If there is a mean of two cracks per mile, there is a n average of 10 cracks in five miles.

The probability mass function of a Poisson distribution is:

[tex]P(X=x) =\frac{\lambda^x*e^{-\lambda}}{x!}[/tex]

λ = 10 cracks in five miles

For x = 0 cracks, the probability is:

[tex]P(X=0) =\frac{10^0*e^{-10}}{0!} \\P(X=0)=4.54*10^{-5}[/tex]

The probability is 4.54 x 10⁻⁵.