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 49 lb and a standard deviation of 18 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.)

Is a very improbable event.

Step-by-step explanation:

We want to calculate the probability that the total weight exceeds the limit when the average weight x exceeds 6000/100=60.

If we analyze the situation we this:

If represent the 100 random beggage weights for the n=100 passengers . We assume that for each for each the distribution assumed is normal with the following parameters .

Another important assumption is that the each one of the random variables are independent.

1) First way to solve the problem

The random variable S who represent the sum of the 100 weight is given by:

The mean for this random variable is given by:

And the variance is given by:

And the deviation:

So we have this distribution for S

On this case we are working with the total so we can find the probability on this way:

2) Second way to solve the problem

We know that the sample mean have the following distribution:

If we are interested on the probability that the population mean would be higher than 60 we can find this probability like this:

And with both methods we got the same probability. So it's very improbable that the limit would be exceeded for this case.

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