Respuesta :
Answer:
[tex]P(\bar x>60)=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]
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 [tex]x_1,x_2,\dots,x_100[/tex] represent the 100 random beggage weights for the n=100 passengers . We assume that for each [tex]i=1,2,3,\dots,100[/tex] for each [tex]x_i[/tex] the distribution assumed is normal with the following parameters [tex]\mu=49, \sigma=18[/tex].
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:
[tex]S=x_1 +x_2 +\dots +x_100 =\sum_{i=1}^{100} x_i[/tex]
The mean for this random variable is given by:
[tex]E(S)=\sum_{i=1}^{100} E(x_i)=100\mu = 100*49=4900[/tex]
And the variance is given by:
[tex]Var(S)=\sum_{i=1}^{100} Var(x_i)=100(\sigma)^2 = 100*(18)^2[/tex]
And the deviation:
[tex]Sd(S)=\sqrt{100(\sigma)^2} = 10*(18)=180[/tex]
So we have this distribution for S
[tex]S \sim (4900,180)[/tex]
On this case we are working with the total so we can find the probability on this way:
[tex]P(S>6000)=P(z>\frac{6000-4900}{180})=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]
2) Second way to solve the problem
We know that the sample mean have the following distribution:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}}[/tex]
If we are interested on the probability that the population mean would be higher than 60 we can find this probability like this:
[tex]P(\bar x >60)=P(\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{60-49}{\frac{18}{\sqrt{100}}})[/tex]
[tex]P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]
And with both methods we got the same probability. So it's very improbable that the limit would be exceeded for this case.
