Question

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter?=20. What is the probability that the number of drivers willA) Be at most 10?B) Exceed 20?C) Be Between 10 and 20, inclusive? Be strictly between 10 and 20?D) Be within 2 standard deviations of the mean value?

Answer 1

Answer:

(A) The probability that the number of drivers will  be at most 10 is 0.0108.

(B) The probability that the number of drivers will  exceed 20 is 0.4409.

(C) The probability that the number of drivers will  be between 10 and 20 (inclusive) is 0.5483 and excluding 10 and 20 is 0.4595.

(D) The probability that the number of drivers will within 2 standard deviations of the mean is 0.9549.

Step-by-step explanation:

Let X = number of drivers who travel a distance during a designated time period.

The random variable X follows a Poisson distribution with parameter λ = 20.

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

(A)

Compute the probability that the number of drivers will  be at most 10 as follows:

P (X ≤ 10) = P (X = 0) + P (X = 1) + P (X = 2) ... + P (X = 10)

               $=\frac{e^{-20}20^{0}}{0!}+\frac{e^{-20}20^{1}}{1!}+...+\frac{e^{-20}20^{9}}{9!}+\frac{e^{-20}20^{10}}{10!}\\=0.00+0.00+...+0.00291+0.00582\\=0.0108$

Thus, the probability that the number of drivers will  be at most 10 is 0.0108.

(B)

Compute the probability that the number of drivers will  exceed 20 as follows:

P (X > 20) = 1 - P (X ≤ 20)

                = 1 - [P (X = 0) + P (X = 1) + P (X = 2) ... + P (X = 19) + P (X = 20)]

                $=1-[\frac{e^{-20}20^{0}}{0!}+\frac{e^{-20}20^{1}}{1!}+...+\frac{e^{-20}20^{19}}{19!}+\frac{e^{-20}20^{20}}{20!}]\\=1-[0.00+0.00+...+0.0884+0.08884]\\=0.4409$

Thus, the probability that the number of drivers will  exceed 20 is 0.4409.

(C)

Compute the probability that the number of drivers will  be between 10 and 20 (inclusive) as follows:

P (10 ≤ X ≤ 20) = P (X ≤ 20) - P (X ≤ 10)

                        $=0.5591-0.0108\\=0.5483$

Compute the probability that the number of drivers will  be exactly between 10 and 20 as follows:

P (10 < X < 20) = P (X = 11) + P (X = 12) + ... + P (X = 19)

                        $=\frac{e^{-20}20^{11}}{11!}+\frac{e^{-20}20^{12}}{12!}+\frac{e^{-20}20^{13}}{13!}+...+\frac{e^{-20}20^{19}}{19!}\\=0.01058+0.01763+0.02712+...+0.08884\\=0.4595$

Thus, the probability that the number of drivers will  be between 10 and 20 (inclusive) is 0.5483 and excluding 10 and 20 is 0.4595.

(D)

The mean of a Poisson distribution is,

$\mu=\lambda=20$

The standard deviation of a Poisson distribution is:

$\sigma=\sqrt{\lambda}=\sqrt{20}=4.4721$

Compute the probability that the number of drivers will within 2 standard deviations of the mean as follows:

$P(\mu-2\sigma$

Thus, the probability that the number of drivers will within 2 standard deviations of the mean is 0.9549.