Help With Statistics and Distributions!!
up vote
2
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
add a comment |
up vote
2
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
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
Nov 19 at 19:09
add a comment |
up vote
2
down vote
favorite
up vote
2
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
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 Nov 19 at 19:03
mathsatbath
162
162
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
Nov 19 at 19:09
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
Nov 19 at 19:09
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
Nov 19 at 19:09
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
Nov 19 at 19:09
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
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?
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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?
add a comment |
up vote
3
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?
add a comment |
up vote
3
down vote
up vote
3
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?
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 Nov 19 at 19:35
Jack M
18.4k33778
18.4k33778
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Nov 19 at 19:09