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 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.)

Answer 1

`P(\bar x>60)=P(z>6.11)=1-P(z<6.11)=4.98x10^(-10)`

Is a very improbable event.

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
`x_1,x_2,\dots,x_100` represent the 100 random beggage weights for the n=100 passengers . We assume that for each
`i=1,2,3,\dots,100` for each
`x_i` the distribution assumed is normal with the following parameters
`\mu=49, \sigma=18`.

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:

`S=x_1 +x_2 +\dots +x_100 =\sum_(i=1)^(100) x_i`

The mean for this random variable is given by:

`E(S)=\sum_(i=1)^(100) E(x_i)=100\mu = 100*49=4900`

And the variance is given by:

`Var(S)=\sum_(i=1)^(100) Var(x_i)=100(\sigma)^2 = 100*(18)^2`

And the deviation:

`Sd(S)=√(100(\sigma)^2) = 10*(18)=180`

So we have this distribution for S

`S \sim (4900,180)`

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

`P(S>6000)=P(z>(6000-4900)/(180))=P(z>6.11)=1-P(z<6.11)=4.98x10^(-10)`

2) Second way to solve the problem

We know that the sample mean have the following distribution:

`\bar X \sim N(\mu,(\sigma)/(√(n))`

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

`P(\bar x >60)=P((\bar x-\mu)/((\sigma)/(√(n)))>(60-49)/((18)/(√(100))))`

`P(z>6.11)=1-P(z<6.11)=4.98x10^(-10)`

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