How many significant figures for an arithmetic mean?
$begingroup$
I am doing an analysis of a group of health centers scattered through a city. I'm trying to find, for a group of zip codes, how many of the health centers are within 5 miles of each zip code; then, I want to take an average. I'd like to say, for example, "On average, there are N centers within 5 miles of a given zip code in group A, and K within 5 miles of those in group B."
My question is, if (for example) in a group of 5 zip codes I come up with 3, 2, 0, 1, and 1 "nearby" centers, should I say that there are on average 1.4 centers near a given zip code, or should I say that there is 1?
arithmetic significant-figures
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
$endgroup$
add a comment |
$begingroup$
I am doing an analysis of a group of health centers scattered through a city. I'm trying to find, for a group of zip codes, how many of the health centers are within 5 miles of each zip code; then, I want to take an average. I'd like to say, for example, "On average, there are N centers within 5 miles of a given zip code in group A, and K within 5 miles of those in group B."
My question is, if (for example) in a group of 5 zip codes I come up with 3, 2, 0, 1, and 1 "nearby" centers, should I say that there are on average 1.4 centers near a given zip code, or should I say that there is 1?
arithmetic significant-figures
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
$endgroup$
add a comment |
$begingroup$
I am doing an analysis of a group of health centers scattered through a city. I'm trying to find, for a group of zip codes, how many of the health centers are within 5 miles of each zip code; then, I want to take an average. I'd like to say, for example, "On average, there are N centers within 5 miles of a given zip code in group A, and K within 5 miles of those in group B."
My question is, if (for example) in a group of 5 zip codes I come up with 3, 2, 0, 1, and 1 "nearby" centers, should I say that there are on average 1.4 centers near a given zip code, or should I say that there is 1?
arithmetic significant-figures
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
$endgroup$
I am doing an analysis of a group of health centers scattered through a city. I'm trying to find, for a group of zip codes, how many of the health centers are within 5 miles of each zip code; then, I want to take an average. I'd like to say, for example, "On average, there are N centers within 5 miles of a given zip code in group A, and K within 5 miles of those in group B."
My question is, if (for example) in a group of 5 zip codes I come up with 3, 2, 0, 1, and 1 "nearby" centers, should I say that there are on average 1.4 centers near a given zip code, or should I say that there is 1?
arithmetic significant-figures
arithmetic significant-figures
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
asked Dec 24 '18 at 3:44
Matt GuttingMatt Gutting
209112
209112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Owing to the rules of significant figures, since the fraction you'd get is $7/5$, with the $5$ as a given constant, you'd go to a precision of one place.
Thus, it'd be $1$ instead of $1.4$.
Though personally, this doesn't seem like the kind of problem that you'd use significant figures for, since significant figures are primarily meant to contend with inaccuracies and human errors in measurement. But I'll assume you know more about the context of this problem than I do.
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
$endgroup$
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
2
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
add a comment |
Your Answer
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%2f3050914%2fhow-many-significant-figures-for-an-arithmetic-mean%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
$begingroup$
Owing to the rules of significant figures, since the fraction you'd get is $7/5$, with the $5$ as a given constant, you'd go to a precision of one place.
Thus, it'd be $1$ instead of $1.4$.
Though personally, this doesn't seem like the kind of problem that you'd use significant figures for, since significant figures are primarily meant to contend with inaccuracies and human errors in measurement. But I'll assume you know more about the context of this problem than I do.
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
$endgroup$
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
2
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
add a comment |
$begingroup$
Owing to the rules of significant figures, since the fraction you'd get is $7/5$, with the $5$ as a given constant, you'd go to a precision of one place.
Thus, it'd be $1$ instead of $1.4$.
Though personally, this doesn't seem like the kind of problem that you'd use significant figures for, since significant figures are primarily meant to contend with inaccuracies and human errors in measurement. But I'll assume you know more about the context of this problem than I do.
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
$endgroup$
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
2
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
add a comment |
$begingroup$
Owing to the rules of significant figures, since the fraction you'd get is $7/5$, with the $5$ as a given constant, you'd go to a precision of one place.
Thus, it'd be $1$ instead of $1.4$.
Though personally, this doesn't seem like the kind of problem that you'd use significant figures for, since significant figures are primarily meant to contend with inaccuracies and human errors in measurement. But I'll assume you know more about the context of this problem than I do.
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
$endgroup$
Owing to the rules of significant figures, since the fraction you'd get is $7/5$, with the $5$ as a given constant, you'd go to a precision of one place.
Thus, it'd be $1$ instead of $1.4$.
Though personally, this doesn't seem like the kind of problem that you'd use significant figures for, since significant figures are primarily meant to contend with inaccuracies and human errors in measurement. But I'll assume you know more about the context of this problem than I do.
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
answered Dec 24 '18 at 3:53
Eevee TrainerEevee Trainer
10.7k31843
10.7k31843
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
2
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
add a comment |
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
2
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
$begingroup$
The reason I was asking actually has a bit less to do with significant figures than with the application - it seemed odd to me to say that "on average there are 1.4 centers near the zip code". Perhaps I should think of it as "the average number of centers near the zip code is 1.4"?
$endgroup$
– Matt Gutting
Dec 24 '18 at 4:04
2
2
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
$begingroup$
Personally both are equivalent, so I guess whichever phrasing makes sense to you. I just have kinda learned through experience that averages don't mean it's necessarily exact, i.e. you wouldn't see $1.4$ centers anywhere because what in the world is $0.4$ of a center? You'd just see mostly $1$ or $2$, maybe sometimes $0$ or $3$ rarely, depending on the distribution - that's basically what the statement "$1.4$ centers on average" means. So I think you're just kinda overthinking it and it seems perfectly natural to say there's an average of $1.4$ centers.
$endgroup$
– Eevee Trainer
Dec 24 '18 at 4:07
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.
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%2f3050914%2fhow-many-significant-figures-for-an-arithmetic-mean%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
