Mixing
Help With Statistics and Distributions!!
up vote
1
down vote
favorite
So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?
At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!
probability statistics probability-distributions binomial-distribution
asked 2 days ago
mathsatbath
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?
At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!
probability statistics probability-distributions binomial-distribution
asked 2 days ago
mathsatbath
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?
At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!
probability statistics probability-distributions binomial-distribution
asked 2 days ago
mathsatbath
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?
At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!
probability statistics probability-distributions binomial-distribution
probability statistics probability-distributions binomial-distribution
asked 2 days ago
mathsatbath
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
mathsatbath
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
mathsatbath
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
mathsatbath
112
asked 2 days ago
mathsatbath
112
112
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago
add a comment |
What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago
What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago
What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.
Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:
If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).
If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.
Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?
answered 2 days ago
Jack M
18.3k33778
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.
Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:
If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).
If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.
Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?
answered 2 days ago
Jack M
18.3k33778
add a comment |
up vote
2
down vote
First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.
Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:
If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).
If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.
Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?
answered 2 days ago
Jack M
18.3k33778
add a comment |
up vote
2
down vote
up vote
2
down vote
First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.
Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:
If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).
If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.
Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?
answered 2 days ago
Jack M
18.3k33778
First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.
Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:
If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).
If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.
Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?
answered 2 days ago
Jack M
18.3k33778
answered 2 days ago
Jack M
18.3k33778
answered 2 days ago
Jack M
18.3k33778
answered 2 days ago
Jack M
18.3k33778
18.3k33778
add a comment |
add a comment |
mathsatbath is a new contributor. Be nice, and check out our Code of Conduct.
mathsatbath is a new contributor. Be nice, and check out our Code of Conduct.
mathsatbath is a new contributor. Be nice, and check out our Code of Conduct.
mathsatbath is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005364%2fhelp-with-statistics-and-distributions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago