Question

Occasionally an airline will lose a bag. Suppose a small airline has found it can reasonably model the number of bags lost each weekday using a Poisson model with a mean of 2.2 bags.

a. What is the probability that the airline will lose no bags next Monday?
b. What is the probability that the airline will lose 0, 1, or 2 bags on next Monday?
c. Suppose the airline expands over the course of the next 3 years, doubling the number of flights it makes, and the CEO asks you if it’s reasonable for them to continue using the Poisson model with a mean of 2.2. What is an appropriate recommendation? Explain.

Answer 1

Answer:

a) The probability that the airline will lose no bags next monday is 0.1108

b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227

c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2

Step-by-step explanation:

The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula

$P(X=k) = \frac{e^{-2.2} * {2.2}^k }{k!}$

a)

$ P(X=0) = \frac{e^{-2.2} * {2.2}^0 }{0!} = 0.1108$

The probability that the airline will lose no bags next monday is 0.1108.

b) Note that $P(X \in \{0,1,2\} = P(X=0) + P(X=1) + P(X=2)$ . And

$P(X=0)+P(X=1)+P(X=2) = e^{-2.2} * (1 + 2.2 + 2.2^2/2) = 0.6227$

Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.

c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.

Answer 2

Using the Poisson distribution, it is found that:

a) There is a 0.1108 = 11.08% probability that the airline will lose no bags next Monday.

b) There is a 0.6227 = 62.27% probability that the airline will lose 0, 1, or 2 bags on next Monday.

c) The mean is appropriate for a standard number of flights. If the number of flights doubles, the mean should also double, hence a mean of 4.4 would be an appropriate recommendation.

We are given only the mean, hence, the Poisson distribution is used.

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

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

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

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

In this problem, the mean is of 2.2 bags, hence $\mu = 2.2$

Item a:

The probability is P(X = 0), hence:

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

$P(X = 0) = \frac{e^{-2.2}2.2^{0}}{(0)!} = 0.1108$

There is a 0.1108 = 11.08% probability that the airline will lose no bags next Monday.

Item b:

The probability is:

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

Hence:

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

$P(X = 0) = \frac{e^{-2.2}2.2^{0}}{(0)!} = 0.1108$

$P(X = 1) = \frac{e^{-2.2}2.2^{1}}{(1)!} = 0.2438$

$P(X = 2) = \frac{e^{-2.2}2.2^{2}}{(2)!} = 0.2681$

Then:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1108 + 0.2438 + 0.2681 = 0.6227$

There is a 0.6227 = 62.27% probability that the airline will lose 0, 1, or 2 bags on next Monday.

Item c:

The mean is appropriate for a standard number of flights. If the number of flights doubles, the mean should also double, hence a mean of 4.4 would be an appropriate recommendation.

You can learn more about the Poisson distribution at https://brainly.com/question/25283929