Question

A study shows that 5% of airplane flights arrived before the scheduled arrival time and 2.3% arrived more than 30 minutes after the schedule time.

a) Calculate P (|T| <= 5)

b) On a given day, 500 flights are due to arrive at these airports. What is the probability that at least 50 planes arrive within 5 minutes of the schedule time?

Answer 1

what is the random variable T representing here? Is there a distribution you're supposed to assume for T ?

Answer 2

@kirbykirby I left the first question out because I thought the info was irrelevant, and since I solved it, I didn't need help for that, but I'll post it here for clarity!

a) Let T be the random variable 'difference between scheduled time and actual arrival time' of the flights at these airports. Assuming that T follows a normal distribution, determine the value of its parameters.

Answer 3

It's always safer to post all the necessary info, because it may be relevant :)

Regardless, you say you have solved it anyway?

Answer 4

or is it you need help with b) only?

Answer 5

I need help with a and b from the first post please :)

Answer 6

I don't understand how to calculate P(|T| <=5)

Answer 7

|T| <= 5 is the same as -5 <= T <= 5
So find P(-5 <= T <= 5)

Answer 8

And the problem gives you the information that P(T > 30) = 0.023

Answer 9

ok, I got all the way to (-5 <= T <= 5) before getting stuck. I understand that I'm supposed to look for T in the above equation right?

Answer 10

actually I have to go sorry :( .. if no one helps in the meantime I will get back with you later.

Answer 11

ok sure! thanks though!

Answer 12

@tkhunny @ganeshie8 are you guys able to assist me? :x

Answer 13

If a plane lands as scheduled, then \(T=0\), so before the scheduled time you'd have \(T<0\) and after you'd have \(T>0\).

You're given that \(P(T<0)=0.05\) and \(P(T>30)=0.023\). You can use this info to find the mean time difference and standard deviation using the standard normal distribution.
\[\frac{T-\mu}{\sigma}=Z\]
\[\begin{align*}
P(T<0)&=P(\mu+Z\sigma<0)\\
0.05&=P\left(Z<-\frac{\mu}{\sigma}\right)
\end{align*}~~~~~~~~~~\Rightarrow~~(1)~~~~-\frac{\mu}{\sigma}=1.65\]
\[\begin{align*}
P(T>30)&=P(\mu+Z\sigma>30)\\
0.023&=P\left(Z<\frac{30-\mu}{\sigma}\right)
\end{align*}~~~~\Rightarrow~~(2)~~~~\frac{30-\mu}{\sigma}=2.83\]
Solve for the mean \(\mu\) and standard deviation \(\sigma\). Or do you already have this part done?

Answer 14

Actually, (2) should be about
\[\frac{30-\mu}{\sigma}=1.99\]

Answer 15

And (1) should be
\[-\frac{\mu}{\sigma}=-1.65\]
Sorry about that!

Answer 16

Once you have the mean/std dev, you can easily compute the probability in part (b). Like @kirbykirby said,
\[\begin{align*}P(|T|\le5)&=P(-5\le T\le5)\\
&=P(T\le5)-P(T\le-5)\\
&=P\left(\frac{T-\mu}{\sigma}\le\frac{5-\mu}{\sigma}\right)-P\left(\frac{T-\mu}{\sigma}\le-\frac{5-\mu}{\sigma}\right)\\
&=P\left(Z\le\frac{5-\mu}{\sigma}\right)-P\left(Z\le-\frac{5-\mu}{\sigma}\right)
\end{align*}\]
You can simplify this a bit using symmetry, but I think the computation is pretty straightforward from here.

Answer 17

@SithsAndGiggles thanks! I already have the first part down, and I now understand the second part. I have trouble with the last question though D: (b) On a given day, 500 flights are due to arrive at these airports. What is the probability that at least 50 planes arrive within 5 minutes of the schedule time?)

Answer 18

The answer you get for \(P(|T|<5)\) is the proportion of planes that land within 5 minutes. Treat this as a binomial success probability. If \(X\) is the random variable for number of planes landing within 5 minutes, then \(p=P(|T|<5)\).

The probability that at least 50 of 500 planes will land within 5 minutes is
\[\large P(X\ge50)=\sum_{k=50}^{500}\binom{500}{k}p^{k}(1-p)^{500-k}\]

Answer 19

how did you know/ decide to use the binomial success probability? am I correct to assume then, that binomcdf is used for this question? and since we are looking for P(X=>50), it is also written as 1-P(X<=49)?

Answer 20

Yes, \(P(X\ge50)=1-P(X\le49)\). If that makes computation easier for you, then by all means.

The reason for using a binomial variable is that we're concerned with a particular set of planes that land. We're also not worried about the difference in scheduled and actual arrival time. We have a probability associated with the planes of interest (those that land within 5 minutes of schedule), but we don't have anything that suggests how many planes meet this condition, so we have to introduce a discrete random variable.

The main clue for using a binomial distribution was the fact that we're concerned with planes satisfying \(|T|<5\) as opposed to \(|T|>5\). Right away we can establish a success/fail relationship between two types of planes. Does this make sense? The reasoning is kind of second nature to me, but actually explaining it is hard to do...

Answer 21

oh wow! thank you so much! I have been trying to review my probabilities unit and binomial, poisson, normal etc and squeezing them all in 1.5 weeks has been confusing me as they all have the cdf/pdf components and I get confused which ones I should use when.

so planes who are late 'fail' and planes who arrive on time/early = 'success' in the success/fail relationship?

Answer 22

Almost! Planes that "fail" land either more than 5 minutes before or after they're expected, meaning \(T<-5\) or \(T>5\). Planes that succeed land within the interval, \(-5<T<5\).

Answer 23

ok. so if I were to plug everything into my GDC in binoncdf, the will the trial value be 50 or 500? I know the probability is 0.137 (found from before, and the X value will be 49.

Answer 24

oh nevermind; the value would be 500 because that is the sample size?

Answer 25

The parameters of a binomial distribution are the size of the population (in this case, \(n=500\)) and the success probability (\(p=P(|T|<5)\))

Merely arriving on time translates to \(T=0\), which is a point. For continuous distributions, the probability of obtaining a point is zero, \(P(T=k)=0\).

The takeaway is that you have to be careful about what you mean by "on time."

I've never actually used binomcdf and all those functions on a calculator (rather, I use the pdf and cdf formulas themselves). The total number of trials 500 though.

Answer 26

ok thank you so much! I hope you'll still be online to answer more (of my) questions :P

Answer 27

You're welcome! I'll be around...