Question

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.)

Answer 1

for normal distribution z score =(X-μ)/σ
here mean=       μ=	48
std deviation   =σ=	22.00
sample size       =n=	100
std error=σx̅=σ/√n=	2.2000

probability that the total weight of their baggage will exceed the limit =P(average weight will exceed 6000/100 =60 lb/customer):

probability =P(X>60)=P(Z>(60-48)/2.2)=P(Z>5.45)=1-P(Z<5.45)=1-(~1) = ~0.0000