given the median , find an unknown
$begingroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
$endgroup$
add a comment |
$begingroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
$endgroup$
add a comment |
$begingroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
$endgroup$
The number of you cars that students in a class have is sown in the Table above .
If median is $3$, write down an inequality satisfied by $x$ .
My workings :
Total number of students = $40+x$ students
Median position = $frac{n+1}{2} = frac{41+x}{2}$
Total students with $0$ to $3$ toys = $31+ x$ students
I do not understand how to manage this qn . Thanks for the help!
statistics
statistics
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
edited Mar 31 '16 at 19:20
V. Vancak
11k2926
11k2926
asked Mar 31 '16 at 7:52
user307640user307640
8691719
asked Mar 31 '16 at 7:52
user307640user307640
8691719
asked Mar 31 '16 at 7:52
user307640user307640
8691719
8691719
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
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',
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%2f1721456%2fgiven-the-median-find-an-unknown%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
$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
add a comment |
$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
add a comment |
$begingroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
$endgroup$
The inequality that you are looking for is $x<8$.
For your median to be equal to 3 it should be the case that at least half of the observations are $geq3$ and less than half of the observations are $geq4$ (which obviously holds here). So, you need $4+12+x<15+4+5 Leftrightarrow x<8$.
Just notice that if $x=8$ then the median would be 2.5, which you do not want.
answered Mar 31 '16 at 8:38
NikolasNikolas
628
answered Mar 31 '16 at 8:38
NikolasNikolas
628
answered Mar 31 '16 at 8:38
NikolasNikolas
628
answered Mar 31 '16 at 8:38
NikolasNikolas
628
628
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.
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%2f1721456%2fgiven-the-median-find-an-unknown%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
