Question

a review of an airlines operations revealed that historyically the airline had an average of 6.42 mishandled bags per 1000 passengers.

a. what is the chance that for the next 2000 passengers the airline will have more than 11 mishandled bags?

b. after a conuslting company evaluated the airlines operations the airline overhauled its bag-monitoring computer system. as a result the number of mishandled bags per 1000 passengers decreased by 30%. what is the probability that for the next 3000 passengers the airline will have fewer than 10 mishandled bags?

Answer 1

As it follows the poisson distribution:

P(X = k) = λke-λ/k!

Now for every 1000 customer we have 6.42 mishandled bags than for every 2000 person 12.84 (6.42*2000/1000) mishandled bags

so λ = 12.84

and if we want to find more than 11 bags got mishandled can be derived by  

P(X>11) = 1 - P(X =0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11)

= 1 - 12.840e-12.84/0! + 12.841e-12.84/1! + 12.842e-12.84/2! +12.843e-12.84/3! +12.844e-12.84/4! +12.845e-12.84/5! +12.846e-12.84/6! +12.847e-12.84/7! +12.848e-12.84/8! +12.849e-12.84/9! +12.8410e-12.84/10! +12.8411e-12.84/11!

= 1-0.3695

= 0.6304

So probability of missing more than 11 bags is 0.6304

Part B:

With new system we can expect 30% less chance to mishandled than we can expect per 1000 bags there are 4.494 (6.42*0.7) mishandled bags

So per 3000 bags we can expect mishandled bags would be 13.482

To calculate the few than 10 bags we can use the

λ = 13.482

P(X<10) = P(X =0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9)

=13.4820e-13.482/0! + 13.4821e-13.482/1! + 13.4822e-13.482/2! +13.4823e-13.482/3! +13.4824e-13.482/4!

+13.4825e-13.482/5! +13.4826e-13.482/6! +13.4827e-13.482/7! +13.4828e-13.482/8! +13.4829e-13.482/9!

= 0.1362

So probability of less than 10 mishandled bags would be 0.1362 per 3000 bags.