Question

On average, there are 12 potholes per mile on a particular stretch of the state highway. Suppose the potholes follow a Poisson distribution on the highway. a. Find the probability of finding fewer than two potholes in a quarter-mile stretch of the highway. (Do not round intermediate calculations. Round your final answer to 4 decimal places.) b. Find the probability of finding more than one pothole in a quarter-mile stretch of the highway. (Do not round intermediate calculations. Round your final answer to 4 decimal places.)

Answer 1

Answer:

a) 0.1992 = 19.92% probability of finding fewer than two potholes in a quarter-mile stretch of the highway.

b) 0.8008 = 80.08% probability of finding more than one pothole in a quarter-mile stretch of the highway.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

a. Find the probability of finding fewer than two potholes in a quarter-mile stretch of the highway.

Mean of 12 potholes per mile, which means that in a quarter-mile stretch, the mean is $\mu = \frac{12}{4} = 3$

This probability is:

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*(3)^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*(3)^{1}}{(1)!} = 0.1494$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992$

0.1992 = 19.92% probability of finding fewer than two potholes in a quarter-mile stretch of the highway.

b. Find the probability of finding more than one pothole in a quarter-mile stretch of the highway.

This is

$P(X > 1) = 1 - P(X < 2)$

We have that $P(X < 2) = 0.1992$

So

$P(X > 1) = 1 - P(X < 2) = 1 - 0.1992 = 0.8008$

0.8008 = 80.08% probability of finding more than one pothole in a quarter-mile stretch of the highway.