On one of its routes across Asia, Alpha Airlines flies an aircraft with checked-in luggage capacity of 8500 lbs. There are 121 seats on the flight.

The average (per passenger) weight of checked-in luggage is 68 lbs with a standard deviation of 11 lbs.

What is the probability that on a randomly selected full flight on this route the checked-in luggage capacity will be exceeded?

the probability is P=0.012 (1.2%)

Step-by-step explanation:

for the random variable X = weight of checked-in luggage, then if X is approximately normal. then the random variable X₂ = weight of N checked-in luggage = ∑ Xi, distributes normally according to the central limit theorem.

Its expected value will be:

μ₂ = ∑ E (Xi) = N*E (Xi) = 121 seats * 68 lbs/seat = 8228 lbs

for N = 121 seats and E (Xi) = 68 lbs/person * 1 person/seat = 68 lbs/seat

the variance will be

σ₂² = ∑ σ² (Xi) = N*σ² (Xi) → σ₂ = σ * √N = 11 lbs/seat * √121 seats = 121 Lbs

then the standard random variable Z

Z = (X₂ - μ₂) / σ₂ =

Zlimit = (8500 Lbs - 8228 lbs) / 121 Lbs = 2.248

P (Z > 2.248) = 1 - P (Z ≤ 2.248) = 1 - 0.988 = 0.012

P (Z > 2.248) = 0.012

then the probability that on a randomly selected full flight, the checked-in luggage capacity will be exceeded is P (Z > 2.248) = 0.012 (1.2%)