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Probability of making it on the train if $100$ people stand in line in front of you and each person takes...
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The train is leaving in $10$ minutes and there are $100$ people standing in line before you to buy tickets. Each person buys $1.85$ tickets on average with a standard deviation of $1.5$ and $9$ people per minute get through on average. The time for each client to get through follows an exponential law (of parameter $lambda=9$ minutes). There are $200$ tickets left when you arrive. What is the probability of making it on the train (on time and with a ticket)?
The expected value for the number of people that have gone through after $10$ minutes is $10cdotlambda=90$ so I would be tempted to say that $P(text{making it})=P({S_{90}-90cdot1.85over 1.5cdotsqrt{90}}<frac{200-90cdot1.85}{1.4cdotsqrt{90}})=phi(2.3541)simeq0.9906$ assuming that each person's time and expected number of tickets bought are independent, using the central limit theorem.
But this seems false because I didn't use the (exponential) nature of the law followed by the people. But on the other hand with the central limit theorem all that matters is the expected value and standard deviation...
probability central-limit-theorem
asked Jan 28 at 19:33
H. WalterH. Walter
1047
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add a comment |
$begingroup$
The train is leaving in $10$ minutes and there are $100$ people standing in line before you to buy tickets. Each person buys $1.85$ tickets on average with a standard deviation of $1.5$ and $9$ people per minute get through on average. The time for each client to get through follows an exponential law (of parameter $lambda=9$ minutes). There are $200$ tickets left when you arrive. What is the probability of making it on the train (on time and with a ticket)?
The expected value for the number of people that have gone through after $10$ minutes is $10cdotlambda=90$ so I would be tempted to say that $P(text{making it})=P({S_{90}-90cdot1.85over 1.5cdotsqrt{90}}<frac{200-90cdot1.85}{1.4cdotsqrt{90}})=phi(2.3541)simeq0.9906$ assuming that each person's time and expected number of tickets bought are independent, using the central limit theorem.
But this seems false because I didn't use the (exponential) nature of the law followed by the people. But on the other hand with the central limit theorem all that matters is the expected value and standard deviation...
probability central-limit-theorem
asked Jan 28 at 19:33
H. WalterH. Walter
1047
$endgroup$
add a comment |
$begingroup$
The train is leaving in $10$ minutes and there are $100$ people standing in line before you to buy tickets. Each person buys $1.85$ tickets on average with a standard deviation of $1.5$ and $9$ people per minute get through on average. The time for each client to get through follows an exponential law (of parameter $lambda=9$ minutes). There are $200$ tickets left when you arrive. What is the probability of making it on the train (on time and with a ticket)?
The expected value for the number of people that have gone through after $10$ minutes is $10cdotlambda=90$ so I would be tempted to say that $P(text{making it})=P({S_{90}-90cdot1.85over 1.5cdotsqrt{90}}<frac{200-90cdot1.85}{1.4cdotsqrt{90}})=phi(2.3541)simeq0.9906$ assuming that each person's time and expected number of tickets bought are independent, using the central limit theorem.
But this seems false because I didn't use the (exponential) nature of the law followed by the people. But on the other hand with the central limit theorem all that matters is the expected value and standard deviation...
probability central-limit-theorem
asked Jan 28 at 19:33
H. WalterH. Walter
1047
$endgroup$
The train is leaving in $10$ minutes and there are $100$ people standing in line before you to buy tickets. Each person buys $1.85$ tickets on average with a standard deviation of $1.5$ and $9$ people per minute get through on average. The time for each client to get through follows an exponential law (of parameter $lambda=9$ minutes). There are $200$ tickets left when you arrive. What is the probability of making it on the train (on time and with a ticket)?
The expected value for the number of people that have gone through after $10$ minutes is $10cdotlambda=90$ so I would be tempted to say that $P(text{making it})=P({S_{90}-90cdot1.85over 1.5cdotsqrt{90}}<frac{200-90cdot1.85}{1.4cdotsqrt{90}})=phi(2.3541)simeq0.9906$ assuming that each person's time and expected number of tickets bought are independent, using the central limit theorem.
But this seems false because I didn't use the (exponential) nature of the law followed by the people. But on the other hand with the central limit theorem all that matters is the expected value and standard deviation...
probability central-limit-theorem
probability central-limit-theorem
asked Jan 28 at 19:33
H. WalterH. Walter
1047
asked Jan 28 at 19:33
H. WalterH. Walter
1047
asked Jan 28 at 19:33
H. WalterH. Walter
1047
asked Jan 28 at 19:33
H. WalterH. Walter
1047
asked Jan 28 at 19:33
H. WalterH. Walter
1047
1047
add a comment |
add a comment |
1 Answer
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You have to have all $100$ people buy their tickets and you need there to be at least one ticket left. You have computed the chance that the first $90$ people buy all the tickets, but that is not what you care about.
First, what is the chance that all $101$ (including you) have finished within $10$ minutes? The mean time is $frac 19$ minute per person, so the mean time for all $101$ is $frac {101}9approx 11.22$ minutes. The variance for one person is $frac 1{81}$ minute$^2$, so the variance for $101$ is $frac {101}{81}$ minute$^2$. You need to be a little more than one standard deviation better than average to get to try and buy a ticket.
Assuming all $100$ ahead of you buy their tickets, the mean number of tickets bought is $185$ with a standard deviation of $15$. You need the number purchased to be $199$ or less, so you can afford to be $frac {14}{15}$ standard deviation high. What is the chance of that?
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
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A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
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@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
add a comment |
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1 Answer
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$begingroup$
You have to have all $100$ people buy their tickets and you need there to be at least one ticket left. You have computed the chance that the first $90$ people buy all the tickets, but that is not what you care about.
First, what is the chance that all $101$ (including you) have finished within $10$ minutes? The mean time is $frac 19$ minute per person, so the mean time for all $101$ is $frac {101}9approx 11.22$ minutes. The variance for one person is $frac 1{81}$ minute$^2$, so the variance for $101$ is $frac {101}{81}$ minute$^2$. You need to be a little more than one standard deviation better than average to get to try and buy a ticket.
Assuming all $100$ ahead of you buy their tickets, the mean number of tickets bought is $185$ with a standard deviation of $15$. You need the number purchased to be $199$ or less, so you can afford to be $frac {14}{15}$ standard deviation high. What is the chance of that?
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
$begingroup$
@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
add a comment |
$begingroup$
You have to have all $100$ people buy their tickets and you need there to be at least one ticket left. You have computed the chance that the first $90$ people buy all the tickets, but that is not what you care about.
First, what is the chance that all $101$ (including you) have finished within $10$ minutes? The mean time is $frac 19$ minute per person, so the mean time for all $101$ is $frac {101}9approx 11.22$ minutes. The variance for one person is $frac 1{81}$ minute$^2$, so the variance for $101$ is $frac {101}{81}$ minute$^2$. You need to be a little more than one standard deviation better than average to get to try and buy a ticket.
Assuming all $100$ ahead of you buy their tickets, the mean number of tickets bought is $185$ with a standard deviation of $15$. You need the number purchased to be $199$ or less, so you can afford to be $frac {14}{15}$ standard deviation high. What is the chance of that?
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
$begingroup$
@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
add a comment |
$begingroup$
You have to have all $100$ people buy their tickets and you need there to be at least one ticket left. You have computed the chance that the first $90$ people buy all the tickets, but that is not what you care about.
First, what is the chance that all $101$ (including you) have finished within $10$ minutes? The mean time is $frac 19$ minute per person, so the mean time for all $101$ is $frac {101}9approx 11.22$ minutes. The variance for one person is $frac 1{81}$ minute$^2$, so the variance for $101$ is $frac {101}{81}$ minute$^2$. You need to be a little more than one standard deviation better than average to get to try and buy a ticket.
Assuming all $100$ ahead of you buy their tickets, the mean number of tickets bought is $185$ with a standard deviation of $15$. You need the number purchased to be $199$ or less, so you can afford to be $frac {14}{15}$ standard deviation high. What is the chance of that?
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
$endgroup$
You have to have all $100$ people buy their tickets and you need there to be at least one ticket left. You have computed the chance that the first $90$ people buy all the tickets, but that is not what you care about.
First, what is the chance that all $101$ (including you) have finished within $10$ minutes? The mean time is $frac 19$ minute per person, so the mean time for all $101$ is $frac {101}9approx 11.22$ minutes. The variance for one person is $frac 1{81}$ minute$^2$, so the variance for $101$ is $frac {101}{81}$ minute$^2$. You need to be a little more than one standard deviation better than average to get to try and buy a ticket.
Assuming all $100$ ahead of you buy their tickets, the mean number of tickets bought is $185$ with a standard deviation of $15$. You need the number purchased to be $199$ or less, so you can afford to be $frac {14}{15}$ standard deviation high. What is the chance of that?
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
edited Jan 28 at 20:58
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
answered Jan 28 at 20:24
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
$begingroup$
@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
add a comment |
$begingroup$
A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
$begingroup$
@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
$begingroup$
A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
$begingroup$
A very nice answer, +1. Allow me to remark that you should also take into the account the time it takes to buy a ticket yourself (i.e., consider 101 people buying a ticket).
$endgroup$
– jvdhooft
Jan 28 at 20:31
$begingroup$
@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
$begingroup$
@jvdhooft: Good point. I thought about it but wasn't sure, as with the fact that they need to buy $199$ tickets or less. I will update.
$endgroup$
– Ross Millikan
Jan 28 at 20:32
add a comment |
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