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

Answer:

$P(\bar x>60)=P(z>6.11)=1-P(z$

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 $x_1,x_2,\dots,x_100$ 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:

$Sd(S)=\sqrt{100(\sigma)^2} = 10*(18)=180$

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:

$P(S>6000)=P(z>\frac{6000-4900}{180})=P(z>6.11)=1-P(z$

2) Second way to solve the problem

We know that the sample mean have the following distribution:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}}$

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

$P(\bar x >60)=P(\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{60-49}{\frac{18}{\sqrt{100}}})$

$P(z>6.11)=1-P(z$

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