Mixing
Stuck at a problem with an exponential distribution
$begingroup$
So I'm given an assignment in probability which states the following:
One bus in the city's public transport system in one route spends an amount of fuel which has an exponential distribution where the expected fuel consumption is 5 gallons.
a) In one day the city bus makes 50 routes. Find the least amount of fuel needed for the entire day so that the amount of fuel is enough with a probability not less than 0.985.
b) The bus has 50 gallons of fuel. How many routes can the bus make so that we will be certain with a probability of 0.985 that it will have enough fuel.
So I think I got a) solved. First of all the random variable X where the expected value is 1/λ from here we can find λ to be equaled to 1/8 but since it spends fuel for 50 routes then λ=50/8. So since X is the amount of fuel then we find X using the equation:
p(x) = $λe^{-λx}$
or:
0.985 = $frac{50}{8}e^{-frac{50}{8} x}$
From here a) is solved. However b) I don't know how to solve. I assume here x = 50 so do we need to find λ by solving the equation:
0.985 = $λe^{-50λ}$
probability
asked Jan 25 at 17:31
David MasonDavid Mason
587
$endgroup$
|
show 1 more comment
$begingroup$
So I'm given an assignment in probability which states the following:
One bus in the city's public transport system in one route spends an amount of fuel which has an exponential distribution where the expected fuel consumption is 5 gallons.
a) In one day the city bus makes 50 routes. Find the least amount of fuel needed for the entire day so that the amount of fuel is enough with a probability not less than 0.985.
b) The bus has 50 gallons of fuel. How many routes can the bus make so that we will be certain with a probability of 0.985 that it will have enough fuel.
So I think I got a) solved. First of all the random variable X where the expected value is 1/λ from here we can find λ to be equaled to 1/8 but since it spends fuel for 50 routes then λ=50/8. So since X is the amount of fuel then we find X using the equation:
p(x) = $λe^{-λx}$
or:
0.985 = $frac{50}{8}e^{-frac{50}{8} x}$
From here a) is solved. However b) I don't know how to solve. I assume here x = 50 so do we need to find λ by solving the equation:
0.985 = $λe^{-50λ}$
probability
asked Jan 25 at 17:31
David MasonDavid Mason
587
$endgroup$
$begingroup$
$p(x)=lambda e^{-lambda x}$ is the probability density function which (perhaps a bit confusing at first) is not a probability per se but a probability per unit, so solving $p(x)=p$ for some chance $0<p<1$ is misguided at best. You must instead use the cumulative distribution function $P(Xleq x)$.
$endgroup$
– LoveTooNap29
Jan 25 at 21:41
$begingroup$
Yes I am aware of that but it says the minimum amount of fuel needed for the probability of 0.985. Do you think the solution is wrong?
$endgroup$
– David Mason
Jan 25 at 22:44
$begingroup$
Of course because whatever $x$ you find that satisfies $p(x)=p$, yields a value of $p(x)$ which, I repeat, is not a probability. You need to find the smallest amount of fuel $x$ such that the amount spent, $X$, does not exceed it with probability $p=0.985$, i.e. you need to solve $min {x: P(Xleq x)=0.985}.$ Does that make sense?
$endgroup$
– LoveTooNap29
Jan 25 at 22:47
$begingroup$
Yes it does. So how do I solve that? Using the CDF?
$endgroup$
– David Mason
Jan 25 at 22:54
$begingroup$
Do you know what the CDF of an exponentially random variable is given by? Once you recall that, it will be straightforward to solve. Hint: for any continuous random variable $X$ with PDF $f_X(x)$, the CDF is defined as $F_X(x):=P(Xleq x)=int_{-infty}^x f_X(x) mathrm{d}x$.
$endgroup$
– LoveTooNap29
Jan 25 at 23:04
|
show 1 more comment
$begingroup$
So I'm given an assignment in probability which states the following:
One bus in the city's public transport system in one route spends an amount of fuel which has an exponential distribution where the expected fuel consumption is 5 gallons.
a) In one day the city bus makes 50 routes. Find the least amount of fuel needed for the entire day so that the amount of fuel is enough with a probability not less than 0.985.
b) The bus has 50 gallons of fuel. How many routes can the bus make so that we will be certain with a probability of 0.985 that it will have enough fuel.
So I think I got a) solved. First of all the random variable X where the expected value is 1/λ from here we can find λ to be equaled to 1/8 but since it spends fuel for 50 routes then λ=50/8. So since X is the amount of fuel then we find X using the equation:
p(x) = $λe^{-λx}$
or:
0.985 = $frac{50}{8}e^{-frac{50}{8} x}$
From here a) is solved. However b) I don't know how to solve. I assume here x = 50 so do we need to find λ by solving the equation:
0.985 = $λe^{-50λ}$
probability
asked Jan 25 at 17:31
David MasonDavid Mason
587
$endgroup$
So I'm given an assignment in probability which states the following:
One bus in the city's public transport system in one route spends an amount of fuel which has an exponential distribution where the expected fuel consumption is 5 gallons.
a) In one day the city bus makes 50 routes. Find the least amount of fuel needed for the entire day so that the amount of fuel is enough with a probability not less than 0.985.
b) The bus has 50 gallons of fuel. How many routes can the bus make so that we will be certain with a probability of 0.985 that it will have enough fuel.
So I think I got a) solved. First of all the random variable X where the expected value is 1/λ from here we can find λ to be equaled to 1/8 but since it spends fuel for 50 routes then λ=50/8. So since X is the amount of fuel then we find X using the equation:
p(x) = $λe^{-λx}$
or:
0.985 = $frac{50}{8}e^{-frac{50}{8} x}$
From here a) is solved. However b) I don't know how to solve. I assume here x = 50 so do we need to find λ by solving the equation:
0.985 = $λe^{-50λ}$
probability
probability
asked Jan 25 at 17:31
David MasonDavid Mason
587
asked Jan 25 at 17:31
David MasonDavid Mason
587
asked Jan 25 at 17:31
David MasonDavid Mason
587
asked Jan 25 at 17:31
David MasonDavid Mason
587
asked Jan 25 at 17:31
David MasonDavid Mason
587
587
$begingroup$
$p(x)=lambda e^{-lambda x}$ is the probability density function which (perhaps a bit confusing at first) is not a probability per se but a probability per unit, so solving $p(x)=p$ for some chance $0<p<1$ is misguided at best. You must instead use the cumulative distribution function $P(Xleq x)$.
$endgroup$
– LoveTooNap29
Jan 25 at 21:41
$begingroup$
Yes I am aware of that but it says the minimum amount of fuel needed for the probability of 0.985. Do you think the solution is wrong?
$endgroup$
– David Mason
Jan 25 at 22:44
$begingroup$
Of course because whatever $x$ you find that satisfies $p(x)=p$, yields a value of $p(x)$ which, I repeat, is not a probability. You need to find the smallest amount of fuel $x$ such that the amount spent, $X$, does not exceed it with probability $p=0.985$, i.e. you need to solve $min {x: P(Xleq x)=0.985}.$ Does that make sense?
$endgroup$
– LoveTooNap29
Jan 25 at 22:47
$begingroup$
Yes it does. So how do I solve that? Using the CDF?
$endgroup$
– David Mason
Jan 25 at 22:54
$begingroup$
Do you know what the CDF of an exponentially random variable is given by? Once you recall that, it will be straightforward to solve. Hint: for any continuous random variable $X$ with PDF $f_X(x)$, the CDF is defined as $F_X(x):=P(Xleq x)=int_{-infty}^x f_X(x) mathrm{d}x$.
$endgroup$
– LoveTooNap29
Jan 25 at 23:04
|
show 1 more comment
$begingroup$
$p(x)=lambda e^{-lambda x}$ is the probability density function which (perhaps a bit confusing at first) is not a probability per se but a probability per unit, so solving $p(x)=p$ for some chance $0<p<1$ is misguided at best. You must instead use the cumulative distribution function $P(Xleq x)$.
$endgroup$
– LoveTooNap29
Jan 25 at 21:41
$begingroup$
Yes I am aware of that but it says the minimum amount of fuel needed for the probability of 0.985. Do you think the solution is wrong?
$endgroup$
– David Mason
Jan 25 at 22:44
$begingroup$
Of course because whatever $x$ you find that satisfies $p(x)=p$, yields a value of $p(x)$ which, I repeat, is not a probability. You need to find the smallest amount of fuel $x$ such that the amount spent, $X$, does not exceed it with probability $p=0.985$, i.e. you need to solve $min {x: P(Xleq x)=0.985}.$ Does that make sense?
$endgroup$
– LoveTooNap29
Jan 25 at 22:47
$begingroup$
Yes it does. So how do I solve that? Using the CDF?
$endgroup$
– David Mason
Jan 25 at 22:54
$begingroup$
Do you know what the CDF of an exponentially random variable is given by? Once you recall that, it will be straightforward to solve. Hint: for any continuous random variable $X$ with PDF $f_X(x)$, the CDF is defined as $F_X(x):=P(Xleq x)=int_{-infty}^x f_X(x) mathrm{d}x$.
$endgroup$
– LoveTooNap29
Jan 25 at 23:04
$begingroup$
$p(x)=lambda e^{-lambda x}$ is the probability density function which (perhaps a bit confusing at first) is not a probability per se but a probability per unit, so solving $p(x)=p$ for some chance $0<p<1$ is misguided at best. You must instead use the cumulative distribution function $P(Xleq x)$.
$endgroup$
– LoveTooNap29
Jan 25 at 21:41
$begingroup$
$p(x)=lambda e^{-lambda x}$ is the probability density function which (perhaps a bit confusing at first) is not a probability per se but a probability per unit, so solving $p(x)=p$ for some chance $0<p<1$ is misguided at best. You must instead use the cumulative distribution function $P(Xleq x)$.
$endgroup$
– LoveTooNap29
Jan 25 at 21:41
$begingroup$
Yes I am aware of that but it says the minimum amount of fuel needed for the probability of 0.985. Do you think the solution is wrong?
$endgroup$
– David Mason
Jan 25 at 22:44
$begingroup$
Yes I am aware of that but it says the minimum amount of fuel needed for the probability of 0.985. Do you think the solution is wrong?
$endgroup$
– David Mason
Jan 25 at 22:44
$begingroup$
Of course because whatever $x$ you find that satisfies $p(x)=p$, yields a value of $p(x)$ which, I repeat, is not a probability. You need to find the smallest amount of fuel $x$ such that the amount spent, $X$, does not exceed it with probability $p=0.985$, i.e. you need to solve $min {x: P(Xleq x)=0.985}.$ Does that make sense?
$endgroup$
– LoveTooNap29
Jan 25 at 22:47
$begingroup$
Of course because whatever $x$ you find that satisfies $p(x)=p$, yields a value of $p(x)$ which, I repeat, is not a probability. You need to find the smallest amount of fuel $x$ such that the amount spent, $X$, does not exceed it with probability $p=0.985$, i.e. you need to solve $min {x: P(Xleq x)=0.985}.$ Does that make sense?
$endgroup$
– LoveTooNap29
Jan 25 at 22:47
$begingroup$
Yes it does. So how do I solve that? Using the CDF?
$endgroup$
– David Mason
Jan 25 at 22:54
$begingroup$
Yes it does. So how do I solve that? Using the CDF?
$endgroup$
– David Mason
Jan 25 at 22:54
$begingroup$
Do you know what the CDF of an exponentially random variable is given by? Once you recall that, it will be straightforward to solve. Hint: for any continuous random variable $X$ with PDF $f_X(x)$, the CDF is defined as $F_X(x):=P(Xleq x)=int_{-infty}^x f_X(x) mathrm{d}x$.
$endgroup$
– LoveTooNap29
Jan 25 at 23:04
$begingroup$
Do you know what the CDF of an exponentially random variable is given by? Once you recall that, it will be straightforward to solve. Hint: for any continuous random variable $X$ with PDF $f_X(x)$, the CDF is defined as $F_X(x):=P(Xleq x)=int_{-infty}^x f_X(x) mathrm{d}x$.
$endgroup$
– LoveTooNap29
Jan 25 at 23:04
|
show 1 more comment
0
active
oldest
votes
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',
autoActivateHeartbeat: false,
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%2f3087378%2fstuck-at-a-problem-with-an-exponential-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3087378%2fstuck-at-a-problem-with-an-exponential-distribution%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
$begingroup$
$p(x)=lambda e^{-lambda x}$ is the probability density function which (perhaps a bit confusing at first) is not a probability per se but a probability per unit, so solving $p(x)=p$ for some chance $0<p<1$ is misguided at best. You must instead use the cumulative distribution function $P(Xleq x)$.
$endgroup$
– LoveTooNap29
Jan 25 at 21:41
$begingroup$
Yes I am aware of that but it says the minimum amount of fuel needed for the probability of 0.985. Do you think the solution is wrong?
$endgroup$
– David Mason
Jan 25 at 22:44
$begingroup$
Of course because whatever $x$ you find that satisfies $p(x)=p$, yields a value of $p(x)$ which, I repeat, is not a probability. You need to find the smallest amount of fuel $x$ such that the amount spent, $X$, does not exceed it with probability $p=0.985$, i.e. you need to solve $min {x: P(Xleq x)=0.985}.$ Does that make sense?
$endgroup$
– LoveTooNap29
Jan 25 at 22:47
$begingroup$
Yes it does. So how do I solve that? Using the CDF?
$endgroup$
– David Mason
Jan 25 at 22:54
$begingroup$
Do you know what the CDF of an exponentially random variable is given by? Once you recall that, it will be straightforward to solve. Hint: for any continuous random variable $X$ with PDF $f_X(x)$, the CDF is defined as $F_X(x):=P(Xleq x)=int_{-infty}^x f_X(x) mathrm{d}x$.
$endgroup$
– LoveTooNap29
Jan 25 at 23:04