Math Problem Statement

The U.S. Department of transportation maintains statistics for mishandled bags. Assume​ that, in one​ year, an airline had 0.74 mishandled bags per month. Complete parts​ (a) through​ (c) below. Question content area bottom Part 1 Bold a. nbsp What is the probability that in the next​ month, the airline will have no mishandled​ bags? While a table of Poisson​ probabilities, technology, or the equation shown below can be used to find the probability that Xequalsx events in an area of opportunity given lambda​, where lambda is the expected number of​ events, e is the mathematical​ constant, and x is the number of events ​(xequals​0, ​1, 2,...), use technology. ​P(Xequalsx ​| lambda​)equalsStartFraction e Superscript negative lambda Baseline lambda Superscript x Over x exclamation mark EndFraction Part 2 The probability that in the next​ month, there will be no mishandled bags is the same as finding Upper P left parenthesis Upper X equals 0 right parenthesis for lambda equals 0.74 . Use technology to find Upper P left parenthesis Upper X equals 0 right parenthesis​, rounding to four decimal places. ​P(Xequals​0)equals0.4771 Part 3 ​Therefore, the probability that there will be no mishandled bags in the next month is approximately 0.4771 . Part 4 Bold b. nbsp What is the probability that in the next​ month, the airline will have at least one mishandled​ bag? The probability that in the next​ month, there will be at least one mishandled bag is the same as finding Upper P left parenthesis Upper X greater than or equals 1 right parenthesis. Part 5 To find the probability that Xgreater than or equals​1, remember that the Poisson distribution is a discrete probability distribution and the sum of the probabilities for every value of X must always equal 1. First complete the equation below. Upper P left parenthesis Upper X greater than or equals 1 right parenthesis equals 1 minus Upper P left parenthesis Upper X less than 1 right parenthesis equals 1 minus Upper P left parenthesis Upper X equals 0 right parenthesis Part 6 Substitute the value found in part​ (a) for Upper P left parenthesis Upper X equals 0 right parenthesis. Upper P left parenthesis Upper X greater than or equals 1 right parenthesis equals 1minus​P(Xequals​0) equals 1minus0.4771 equals 0.5229 Part 7 ​Therefore, the probability that at least one mishandled bag in the next month is approximately 0.5229. Part 8 Bold c. nbsp What is the probability that in the next​ month, the airline will have at least two mishandled​ bags? The probability that in the next​ month, there will be at least two mishandled bags is the same as finding Upper P left parenthesis Upper X greater than or equals 2 right parenthesis. Part 9 Use the fact that the sum of the probabilities for every value of X must always equal 1 to restate the probability. Upper P left parenthesis Upper X greater than or equals 2 right parenthesis equals 1 minus Upper P left parenthesis Upper X less than 2 right parenthesis equals 1 minus left bracket Upper P left parenthesis Upper X equals 0 right parenthesis plus Upper P left parenthesis Upper X equals 1 right parenthesis right bracket Part 10 Use technology to find Upper P left parenthesis Upper X equals 1 right parenthesis​, rounding to four decimal places. Upper P left parenthesis Upper X equals 1 right parenthesisequals0.3531 Part 11 Substitute ​P(Xequals​1) and the value found in part​ (a) for Upper P left parenthesis Upper X equals 0 right parenthesis. Upper P left parenthesis Upper X greater than or equals 2 right parenthesis equals 1 minus left bracket Upper P left parenthesis Upper X equals 0 right parenthesis plus Upper P left parenthesis Upper X equals 1 right parenthesis right bracket equals 1minus​[0.4771plus0.3531​] equals 0.1698 (explain part c)

Solution

Part c Explanation:

The goal is to find the probability that the airline will have at least two mishandled bags in the next month. This is the same as finding P(X≥2)P(X \geq 2), where XX represents the number of mishandled bags.

We can approach this problem by using the complement rule, which says that the probability of an event happening is 1 minus the probability of its complement. In this case:

P(X≥2)=1−P(X<2)P(X \geq 2) = 1 - P(X < 2)

Since P(X<2)P(X < 2) means the probability of fewer than 2 mishandled bags, it includes the probabilities of X=0X = 0 and X=1X = 1. So:

P(X≥2)=1−(P(X=0)+P(X=1))P(X \geq 2) = 1 - (P(X = 0) + P(X = 1))

We already know from previous parts that:

  • P(X=0)=0.4771P(X = 0) = 0.4771
  • P(X=1)=0.3531P(X = 1) = 0.3531

Now, substitute these values into the equation:

P(X≥2)=1−(0.4771+0.3531)P(X \geq 2) = 1 - (0.4771 + 0.3531)

Calculate the sum inside the parentheses:

P(X≥2)=1−0.8302=0.1698P(X \geq 2) = 1 - 0.8302 = 0.1698

Thus, the probability that the airline will have at least 2 mishandled bags in the next month is approximately 0.1698.

Would you like more details on any of the steps? Here are some follow-up questions that you might find helpful:

  1. How is the Poisson distribution used to model real-life events?
  2. Can the Poisson distribution be applied to other airline statistics, such as flight delays?
  3. How do we calculate the probability of exactly 2 mishandled bags?
  4. What are some real-world scenarios where λ\lambda changes over time, and how does it affect probability?
  5. How does rounding affect probability outcomes in real-world applications?

Tip: The Poisson distribution is particularly useful when dealing with rare events over a fixed interval, such as accidents, customer arrivals, or machine failures.

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Math Problem Analysis

Mathematical Concepts

Probability
Poisson Distribution

Formulas

P(X = x | λ) = (e^(-λ) * λ^x) / x!
P(X ≥ 1) = 1 - P(X = 0)
P(X ≥ 2) = 1 - (P(X = 0) + P(X = 1))

Theorems

Poisson Distribution

Suitable Grade Level

Grades 11-12 / College Level

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