Question

The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6 trips per year. What is the probabilityof randomly selecting an American family and finding the following?a.The family did not make a trip to an amusement park last year.b.The family took exactly one trip to an amusement park last year.c.The family took two or more trips to amusement parks last year.d.The family took three or fewer trips to amusement parks over a three-year period.e.The family took exactly four trips to amusement parks during a six-year period.

Answer 1

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

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.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$, in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$.

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

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$.

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

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$.

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$.

This is P(X = 4). So

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

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

Answer 2

Probabilities are used to determine the chances of events.

The given parameters are:

$p = 0.6$

(a) The family did not make a trip to an amusement park last year

The distribution is given as a Poisson distribution.

So, we have:

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

Where:

$\mu = np$

Last year means that:

$n = 1$ --- the number of years.

No trip, means that:

$x = 0$

So, we have:

$\mu = np$

$\mu = 1 \times 0.6$

$\mu = 0.6$

The probability becomes

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

$P(0) = \frac{e^{-0.6} \times 0.6^0}{0!}$

$P(0) = \frac{0.5488 \times 1}{1}$

$P(0) = 0.5488$

Hence, the probability that the family did not make a trip to an amusement park last year is 0.5488

(b) The family took exactly one trip to an amusement park last year

This means that:

x = 1.

So, we have:

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

$P(1) = \frac{e^{-0.6} \times 0.6^1}{1!}$

$P(1) = \frac{0.5488 \times 0.6}{1}$

$P(1) = 0.3293$

Hence, the probability that the family took exactly one trip to an amusement park last year is 0.3293

(c) The family took two or more trips to amusement parks last year

This means that:

x = 2,3...

So, we make use of the following complement rule:

$P(x\ge 2) = 1 - P(x < 2)$

This gives

$P(x\ge 2) = 1 - P(0) - P(1)$

So, we have:

$P(x\ge 2) = 1 - 0.5488 - 0.3293$

$P(x\ge 2) = 0.1219$

Hence, the probability that the family took two or more trips to an amusement park last year is 0.1219

(d) The family took three or fewer trips to amusement parks over a three-year period.

For a three-year period, we have:

$n = 3$

So, the mean of the distribution is:

$\mu = np$

$\mu = 3 \times 0.6$

$\mu = 1.8$

The probability is then represented as:

$P(x \le 3) = P(0) + P(1) + P(2) + P(3)$

Calculate P(0) to P(3) using:

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

So, we have:

$P(0) = \frac{e^{-1.8} \times 1.8^0}{0!}$

$P(0) = \frac{0.1653 \times 1}{1}$

$P(0) = 0.1653$

$P(1) = \frac{e^{-1.8} \times 1.8^1}{1!}$

$P(1) = \frac{0.1653 \times 1.8}{1}$

$P(1) = 0.2975$

$P(2) = \frac{e^{-1.8} \times 1.8^2}{2!}$

$P(2) = \frac{0.1653 \times 3.24}{2}$

$P(2) = 0.2678$

$P(3) = \frac{e^{-1.8} \times 1.8^3}{3!}$

$P(3) = \frac{0.1653 \times 5.832}{6}$

$P(3) = 0.1607$

So, we have:

$P(x \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(x \le 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607$

$P(x \le 3) = 0.8913$

Hence, the probability that the family took three or fewer trips to amusement parks over a three-year period is 0.8913

e. The family took exactly four trips to amusement parks during a six-year period.

A six-year period means that:

$n = 6$

So, the mean of the distribution is:

$\mu = np$

$\mu = 6 \times 0.6$

$\mu = 3.6$

The probability is then calculated as:

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

So, we have:

$P(4) = \frac{e^{-3.6} \times 3.6^4}{4!}$

$P(4) = \frac{0.0273 \times 167.9616}{24}$

$P(4) = 0.1911$

Hence, the probability that the family took exactly four trips to amusement parks over a six-year period is 0.1911

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