Math Problem Statement

Suppose the weight of pieces of passenger luggage for domestic airline flights follows a normal distribution with πœ‡ = 50 pounds and 𝜎 = 20.63 pounds. Calculate the probability that the total weight of the luggage for 80 passengers on a particular flight is between 1200 and 2400 pounds. (Assume each passenger has one piece of luggage. Round your answer to one decimal place.)

Solution

To solve this problem, we calculate the probability that the total weight of luggage for 80 passengers is between 1200 and 2400 pounds, given that the weight of luggage follows a normal distribution with a mean of 50 pounds and standard deviation of 20.63 pounds.

Step 1: Define the total weight distribution

The total weight of the luggage for 80 passengers is the sum of independent normally distributed random variables. For a normal distribution:

Mean of total weight,ΞΌtotal=nβ‹…ΞΌ=80β‹…50=4000 pounds\text{Mean of total weight}, \mu_{\text{total}} = n \cdot \mu = 80 \cdot 50 = 4000 \, \text{pounds}

Standard deviation of total weight,Οƒtotal=nβ‹…Οƒ=80β‹…20.63β‰ˆ184.23 pounds.\text{Standard deviation of total weight}, \sigma_{\text{total}} = \sqrt{n} \cdot \sigma = \sqrt{80} \cdot 20.63 \approx 184.23 \, \text{pounds}.

Step 2: Standardize the values

We standardize the weights 12001200 and 24002400 using the z-score formula:

z=xβˆ’ΞΌtotalΟƒtotalz = \frac{x - \mu_{\text{total}}}{\sigma_{\text{total}}}

For x=1200x = 1200: z1=1200βˆ’4000184.23=βˆ’2800184.23β‰ˆβˆ’15.2z_1 = \frac{1200 - 4000}{184.23} = \frac{-2800}{184.23} \approx -15.2

For x=2400x = 2400: z2=2400βˆ’4000184.23=βˆ’1600184.23β‰ˆβˆ’8.7z_2 = \frac{2400 - 4000}{184.23} = \frac{-1600}{184.23} \approx -8.7

Step 3: Calculate probabilities

From standard normal distribution tables or a calculator: P(Z<βˆ’15.2)β‰ˆ0andP(Z<βˆ’8.7)β‰ˆ0P(Z < -15.2) \approx 0 \quad \text{and} \quad P(Z < -8.7) \approx 0

Step 4: Interpret the result

The probability that the total weight of the luggage for 80 passengers is between 12001200 and 24002400 pounds is essentially 0 because the calculated z-scores are so extreme (far from the mean).

Final Answer:

The probability is approximately 0.0%.\text{The probability is approximately } 0.0 \%.

Do you want to explore similar calculations or have more details? Here are 5 related questions:

  1. How would the probability change if the standard deviation was smaller?
  2. What is the probability that the total weight exceeds 4500 pounds for 80 passengers?
  3. How is the central limit theorem applied in this problem?
  4. Can we calculate probabilities for individual luggage weights falling within a specific range?
  5. How does changing the sample size (number of passengers) affect the distribution of total weight?

Tip: Use z-score tables or statistical software for quick evaluations of probabilities for extreme z-scores.

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Central Limit Theorem

Formulas

Mean of total weight: ΞΌ_total = n Γ— ΞΌ
Standard deviation of total weight: Οƒ_total = √n Γ— Οƒ
Z-score formula: z = (x - ΞΌ) / Οƒ

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 11-12 or introductory college-level statistics