An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable x with a mean value of 48 lb and a standard deviation of 22 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With n = 100, the total weight exceeds the limit when the average weight x exceeds 6000/100.) (Round your answer to four decimal places.)

Accepted Solution

Answer:[tex]4.42*10^{-54}[/tex]Step-by-step explanation:SO for the total baggage weights onboard to exceed the 6000lb limit, with n = 100 passengers. Each of the passenger must exceed the weigh allowance of 6000/100 = 60lb limit as well.The probability of that to happen with normal distribution of 48lb and standard deviation of 22lb is:[tex]P(x > 60, \mu = 48, \sigma = 22) = 1 - 0.707 = 0.293[/tex]For all 100 passengers to exceed this limit, the probability for that to happen is[tex]0.293^{100} = 4.42*10^{-54}[/tex]which is very low