Median Calculation on Grouped Interval Data
$begingroup$
I was studying for my Probability & Statistics exam. I've encountered with an example which is about finding the median of a given grouped data. Here is the table of the data;
So from the table, we are given that there are 51 children. So when we want to find median, we know that it will be in 26th position when we apply the formula as (n+1)/2 since 51 is an odd number. When we look the frequencies in the table, we can see that 26th children will be between 160 < h < 170 range. Here is the solution for finding the median which is given to me in the example;
What I didn't understand here is that how they calculated it in that way ? I understood the steps how they determined it should be between 160 < h < 170 range but I could not understand the following calculation 160 + (4/21)*10 = 162 . Can you please explain me, how this calculation works ? Thanks in advance.
probability statistics median
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
$endgroup$
add a comment |
$begingroup$
I was studying for my Probability & Statistics exam. I've encountered with an example which is about finding the median of a given grouped data. Here is the table of the data;
So from the table, we are given that there are 51 children. So when we want to find median, we know that it will be in 26th position when we apply the formula as (n+1)/2 since 51 is an odd number. When we look the frequencies in the table, we can see that 26th children will be between 160 < h < 170 range. Here is the solution for finding the median which is given to me in the example;
What I didn't understand here is that how they calculated it in that way ? I understood the steps how they determined it should be between 160 < h < 170 range but I could not understand the following calculation 160 + (4/21)*10 = 162 . Can you please explain me, how this calculation works ? Thanks in advance.
probability statistics median
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
$endgroup$
$begingroup$
The $4$ comes from the fact that after you've disposed of the first $6,$ and then $16$ values, there are still $4$ left. So you start at the left-hand end of the interval, and estimate by computing $4/21$ of the way to the next interval boundary.
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:04
$begingroup$
Actually, I would like to see a formula to use in another questions
$endgroup$
– Ozan Yurtsever
Dec 17 '18 at 21:05
$begingroup$
Maybe something like this: $$text{Est}=text{LH Endpoint}+frac{(text{Num Past LH Endpoint})cdot(text{Bin Width})}{text{Num In Bin}}.$$
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:34
add a comment |
$begingroup$
I was studying for my Probability & Statistics exam. I've encountered with an example which is about finding the median of a given grouped data. Here is the table of the data;
So from the table, we are given that there are 51 children. So when we want to find median, we know that it will be in 26th position when we apply the formula as (n+1)/2 since 51 is an odd number. When we look the frequencies in the table, we can see that 26th children will be between 160 < h < 170 range. Here is the solution for finding the median which is given to me in the example;
What I didn't understand here is that how they calculated it in that way ? I understood the steps how they determined it should be between 160 < h < 170 range but I could not understand the following calculation 160 + (4/21)*10 = 162 . Can you please explain me, how this calculation works ? Thanks in advance.
probability statistics median
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
$endgroup$
I was studying for my Probability & Statistics exam. I've encountered with an example which is about finding the median of a given grouped data. Here is the table of the data;
So from the table, we are given that there are 51 children. So when we want to find median, we know that it will be in 26th position when we apply the formula as (n+1)/2 since 51 is an odd number. When we look the frequencies in the table, we can see that 26th children will be between 160 < h < 170 range. Here is the solution for finding the median which is given to me in the example;
What I didn't understand here is that how they calculated it in that way ? I understood the steps how they determined it should be between 160 < h < 170 range but I could not understand the following calculation 160 + (4/21)*10 = 162 . Can you please explain me, how this calculation works ? Thanks in advance.
probability statistics median
probability statistics median
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
asked Dec 17 '18 at 21:01
Ozan YurtseverOzan Yurtsever
1031
1031
$begingroup$
The $4$ comes from the fact that after you've disposed of the first $6,$ and then $16$ values, there are still $4$ left. So you start at the left-hand end of the interval, and estimate by computing $4/21$ of the way to the next interval boundary.
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:04
$begingroup$
Actually, I would like to see a formula to use in another questions
$endgroup$
– Ozan Yurtsever
Dec 17 '18 at 21:05
$begingroup$
Maybe something like this: $$text{Est}=text{LH Endpoint}+frac{(text{Num Past LH Endpoint})cdot(text{Bin Width})}{text{Num In Bin}}.$$
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:34
add a comment |
$begingroup$
The $4$ comes from the fact that after you've disposed of the first $6,$ and then $16$ values, there are still $4$ left. So you start at the left-hand end of the interval, and estimate by computing $4/21$ of the way to the next interval boundary.
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:04
$begingroup$
Actually, I would like to see a formula to use in another questions
$endgroup$
– Ozan Yurtsever
Dec 17 '18 at 21:05
$begingroup$
Maybe something like this: $$text{Est}=text{LH Endpoint}+frac{(text{Num Past LH Endpoint})cdot(text{Bin Width})}{text{Num In Bin}}.$$
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:34
$begingroup$
The $4$ comes from the fact that after you've disposed of the first $6,$ and then $16$ values, there are still $4$ left. So you start at the left-hand end of the interval, and estimate by computing $4/21$ of the way to the next interval boundary.
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:04
$begingroup$
The $4$ comes from the fact that after you've disposed of the first $6,$ and then $16$ values, there are still $4$ left. So you start at the left-hand end of the interval, and estimate by computing $4/21$ of the way to the next interval boundary.
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:04
$begingroup$
Actually, I would like to see a formula to use in another questions
$endgroup$
– Ozan Yurtsever
Dec 17 '18 at 21:05
$begingroup$
Actually, I would like to see a formula to use in another questions
$endgroup$
– Ozan Yurtsever
Dec 17 '18 at 21:05
$begingroup$
Maybe something like this: $$text{Est}=text{LH Endpoint}+frac{(text{Num Past LH Endpoint})cdot(text{Bin Width})}{text{Num In Bin}}.$$
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:34
$begingroup$
Maybe something like this: $$text{Est}=text{LH Endpoint}+frac{(text{Num Past LH Endpoint})cdot(text{Bin Width})}{text{Num In Bin}}.$$
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:34
add a 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%2f3044438%2fmedian-calculation-on-grouped-interval-data%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%2f3044438%2fmedian-calculation-on-grouped-interval-data%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
$begingroup$
The $4$ comes from the fact that after you've disposed of the first $6,$ and then $16$ values, there are still $4$ left. So you start at the left-hand end of the interval, and estimate by computing $4/21$ of the way to the next interval boundary.
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:04
$begingroup$
Actually, I would like to see a formula to use in another questions
$endgroup$
– Ozan Yurtsever
Dec 17 '18 at 21:05
$begingroup$
Maybe something like this: $$text{Est}=text{LH Endpoint}+frac{(text{Num Past LH Endpoint})cdot(text{Bin Width})}{text{Num In Bin}}.$$
$endgroup$
– Adrian Keister
Dec 17 '18 at 21:34