Question

A review of an airline operations revealed tat, historically, the airline had an average of 6.42 mishandled bags per 1.000 passengers. A mishandled bag is luggage that was not sent the passengers plane. It was either lost or arrived late.

a)What ise the chance that for the next 2.000 passengers,the airline will have more than 11 mishandled bags is?

b)After a consulting company evaluated the airlines operations, the airline overhauled its bag- monitoring compute system. As a result the number of mishandled bags or 1000 passengers decreased 30% . Given that new compute system, what is probability that for the next 2.000 passengers, the airline will have fever than 10 mishandled bags?

c)Given the new computer system, the probability that for the nex 3,000 passengers, the airline will have fewer 10 mishandled bags is?

Answer 1

Answer: A review of an airline's operations revealed that, historically, the airline had an average of 6.42 mishandled bags per 1,000 passengers. A "mishandled bag" is luggage that was not sent together with the passenger's plane, i.e., it was either lost or arrived late.

Solution:

p = 6.42/1000 = 0.00642

Using binomial approximation to normal distribution.

a) What is the chance that for the next 2.000 passengers, the airline will have more than 11 mishandled bags?

n = 2000

Mean, μ = np = 2000 * 0.00642 = 12.84

Standard deviation, σ = √np(1-p)

σ = √2000 * 0.00642 * 0.99358

σ = 3.5718

P(X > 11) = P(Z > 11 - 12.84 / 3.5718

P(X > 11) = P(Z > - 0.5151)

P(X > 11) = 0.6985.

Therefore, the chance that for the next 2,000 passengers, the airline will have more than 11 mishandled bags is 0.6985.

b) After a consulting company evaluated the airline's operations, the airline overhauled its bag-monitoring computer system. As a result, the number of mishandled bags per 1,000 passengers decreased by 30%. Given that new computer system, what is the probability that for the next 2,000 passengers, the airline will have fewer than 10 mishandled bags?

The Number of mishandled bags reduced by 30%

So, number of mishandled bags = 6.42*(1-0.30) = 4.494

New Probability of having a mishandled bag = 4.5/1000 = 0.0045

Mean, μ = np = 2000 * 0.0045 = 9

Standard deviation, σ = √np(1-p)

σ = √2000 * 0.0045 * 0.9955

σ = 2.9932 = 3

P(X < 10) = P(Z < 10-9/3)

P(X < 10) = P(Z < 0.3333)

P(X < 10) = 0.3707

Therefore, probability that for the next 2.000 passengers, the airline will have fever than 10 mishandled bags is 0.3707

c) Given the new computer system, the probability that for the next 3,000 passengers, the airline will have fewer than 10 mishandled bags is

Mean, μ = np = 3000 * 0.0045 = 13.5

Standard deviation, σ = √np(1-p)

σ = √3000 * 0.0045 * 0.9955

σ = 3.6659 = 3.666

P(X < 10) = P(Z < 10-13.5/3.666)

P(X < 10) = P(Z < 0.955)

P(X < 10) = 0.1685

Therefore, Probability that for the next 3000 passengers, the airline will have fewer than 10 mishandled bags is 0.1685.