Dgdrxrt

What steps are necessary to read a Modern SSD in Medieval Europe?

Is it possible to create a QR code using text?

That's an odd coin - I wonder why

Find a path from s to t using as few red nodes as possible

What did the word "leisure" mean in late 18th Century usage?

Incomplete cube

What difference does it make matching a word with/without a trailing whitespace?

How did scripture get the name bible?

Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact

Shortening a title without changing its meaning

Arrows in tikz Markov chain diagram overlap

Ising model simulation

Is it correct to say moon starry nights?

How badly should I try to prevent a user from XSSing themselves?

A hang glider, sudden unexpected lift to 25,000 feet altitude, what could do this?

Is it reasonable to ask other researchers to send me their previous grant applications?

logical reads on global temp table, but not on session-level temp table

How to pronounce fünf in 45

How should I connect my cat5 cable to connectors having an orange-green line?

Is it a bad idea to plug the other end of ESD strap to wall ground?

Is there a rule of thumb for determining the amount one should accept for a settlement offer?

Early programmable calculators with RS-232

Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico

How to show a landlord what we have in savings?

Determining an unknown value of the variable of a statistical distribution

1

$begingroup$

This question was asked in a statistics session in my university:

The respective means of the first 8 and last 8 values of a statistical series of 15 values are 8 and 14. Given that the mean of the series is 12, then the 8th value is:

A. 2
B. 3
C. 4
D. 5
E. None of the above

The answer key says that (C) is the correct answer.

I tried in vain to solve this problem, but I couldn't figure out how.

I thought that since 8 is the mean of the first 8 values, therefore the 8th value is for sure greater than 8.

Can someone please explain? Any help or comment is appreciated. Thanks.

statistics

share|cite|improve this question

asked Mar 28 at 10:50

user208973user208973

82

New contributor

user208973 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

This question was asked in a statistics session in my university:

The respective means of the first 8 and last 8 values of a statistical series of 15 values are 8 and 14. Given that the mean of the series is 12, then the 8th value is:

A. 2
B. 3
C. 4
D. 5
E. None of the above

The answer key says that (C) is the correct answer.

I tried in vain to solve this problem, but I couldn't figure out how.

I thought that since 8 is the mean of the first 8 values, therefore the 8th value is for sure greater than 8.

Can someone please explain? Any help or comment is appreciated. Thanks.

statistics

share|cite|improve this question

asked Mar 28 at 10:50

user208973user208973

82

New contributor

user208973 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

This question was asked in a statistics session in my university:

The respective means of the first 8 and last 8 values of a statistical series of 15 values are 8 and 14. Given that the mean of the series is 12, then the 8th value is:

A. 2
B. 3
C. 4
D. 5
E. None of the above

The answer key says that (C) is the correct answer.

I tried in vain to solve this problem, but I couldn't figure out how.

I thought that since 8 is the mean of the first 8 values, therefore the 8th value is for sure greater than 8.

Can someone please explain? Any help or comment is appreciated. Thanks.

statistics

share|cite|improve this question

asked Mar 28 at 10:50

user208973user208973

82

New contributor

user208973 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 28 at 10:50

user208973user208973

82

New contributor

user208973 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 28 at 10:50

user208973user208973

82

New contributor

user208973 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 10:50

user208973user208973

82

New contributor

user208973 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 10:50

user208973user208973

82

1 Answer
1

active

oldest

votes

0

$begingroup$

Let's define your 15 values as $(x_1,...,x_15)$. We know that:

$$12 = frac115sum_i = 1^15 x_i$$

Now, we have that:

$$frac115sum_i = 1^15 x_i = frac115 left ( sum_i = 1^8 x_i + sum_i = 8^15 x_i - x_8 right ) $$

where $x_8$ is subtracted in the end since otherwise you would be summing it two times, as last term in the first summation and first term in the second summation.

But since we are also told that:

$$frac18sum_i = 1^8 x_i = 8 Rightarrow sum_i = 1^8 x_i = 64$$

and that:

$$frac18sum_i = 8^15 x_i = 14 Rightarrow sum_i = 8^15 x_i = 112$$

we can conclude that:

$$12 = frac115 left ( 64 + 112 - x_8 right )$$

leading to $$x_8 = -4$$

Possibly a typo in your question?

share|cite|improve this answer

edited Mar 28 at 13:12

answered Mar 28 at 12:59

NicgNicg

514

New contributor

Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer seems logical; it is possibly a typo in the sessions book. Thank you.
  $endgroup$
  – user208973
  Mar 28 at 13:12
  

  
  
  
  

add a comment | 

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
);



);

user208973 is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165730%2fdetermining-an-unknown-value-of-the-variable-of-a-statistical-distribution%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Let's define your 15 values as $(x_1,...,x_15)$. We know that:

$$12 = frac115sum_i = 1^15 x_i$$

Now, we have that:

$$frac115sum_i = 1^15 x_i = frac115 left ( sum_i = 1^8 x_i + sum_i = 8^15 x_i - x_8 right ) $$

where $x_8$ is subtracted in the end since otherwise you would be summing it two times, as last term in the first summation and first term in the second summation.

But since we are also told that:

$$frac18sum_i = 1^8 x_i = 8 Rightarrow sum_i = 1^8 x_i = 64$$

and that:

$$frac18sum_i = 8^15 x_i = 14 Rightarrow sum_i = 8^15 x_i = 112$$

we can conclude that:

$$12 = frac115 left ( 64 + 112 - x_8 right )$$

leading to $$x_8 = -4$$

Possibly a typo in your question?

share|cite|improve this answer

edited Mar 28 at 13:12

answered Mar 28 at 12:59

NicgNicg

514

New contributor

Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer seems logical; it is possibly a typo in the sessions book. Thank you.
  $endgroup$
  – user208973
  Mar 28 at 13:12
  

  
  
  
  

add a comment | 

0

$begingroup$

Let's define your 15 values as $(x_1,...,x_15)$. We know that:

$$12 = frac115sum_i = 1^15 x_i$$

Now, we have that:

$$frac115sum_i = 1^15 x_i = frac115 left ( sum_i = 1^8 x_i + sum_i = 8^15 x_i - x_8 right ) $$

where $x_8$ is subtracted in the end since otherwise you would be summing it two times, as last term in the first summation and first term in the second summation.

But since we are also told that:

$$frac18sum_i = 1^8 x_i = 8 Rightarrow sum_i = 1^8 x_i = 64$$

and that:

$$frac18sum_i = 8^15 x_i = 14 Rightarrow sum_i = 8^15 x_i = 112$$

we can conclude that:

$$12 = frac115 left ( 64 + 112 - x_8 right )$$

leading to $$x_8 = -4$$

Possibly a typo in your question?

share|cite|improve this answer

edited Mar 28 at 13:12

answered Mar 28 at 12:59

NicgNicg

514

New contributor

Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your answer seems logical; it is possibly a typo in the sessions book. Thank you.
  $endgroup$
  – user208973
  Mar 28 at 13:12
  

  
  
  
  

add a comment | 

$begingroup$

Let's define your 15 values as $(x_1,...,x_15)$. We know that:

$$12 = frac115sum_i = 1^15 x_i$$

Now, we have that:

$$frac115sum_i = 1^15 x_i = frac115 left ( sum_i = 1^8 x_i + sum_i = 8^15 x_i - x_8 right ) $$

where $x_8$ is subtracted in the end since otherwise you would be summing it two times, as last term in the first summation and first term in the second summation.

But since we are also told that:

$$frac18sum_i = 1^8 x_i = 8 Rightarrow sum_i = 1^8 x_i = 64$$

and that:

$$frac18sum_i = 8^15 x_i = 14 Rightarrow sum_i = 8^15 x_i = 112$$

we can conclude that:

$$12 = frac115 left ( 64 + 112 - x_8 right )$$

leading to $$x_8 = -4$$

Possibly a typo in your question?

share|cite|improve this answer

edited Mar 28 at 13:12

answered Mar 28 at 12:59

NicgNicg

514

New contributor

Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited Mar 28 at 13:12

answered Mar 28 at 12:59

NicgNicg

514

New contributor

Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 28 at 12:59

NicgNicg

514

New contributor

Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 28 at 12:59

NicgNicg

514