Edexcel Statistics help please

I'm struggling with question one on this solomon paper

http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S1/Solomon/Solomon%20B.pdf

I've found the new mean but don't know how to find the new standard deviation and the sum of x^2.

Please give detailed answer, thanks

Reply 1

CTArsenal

8

Original post by Efemena15

I'm struggling with question one on this solomon paper

http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S1/Solomon/Solomon%20B.pdf

I've found the new mean but don't know how to find the new standard deviation and the sum of x^2.

Please give detailed answer, thanks

You know the formula for variance in S1 as the (Σx² / n) - x̅².

You know the variance is Standard Deviation squared, so

7.4² = (Σx² / n) - x̅².

It's all a matter of plugging in numbers to find the initial values. The next part of the question, n becomes 15 (as a new student joins). Their age is 42 so you just add on the 42 to Σx, and 42² to Σx² and you can find your new standard deviation and mean.

Hope this helps.

Reply 2

elldeegee

19

Original post by Efemena15

I'm struggling with question one on this solomon paper

http://papers.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S1/Solomon/Solomon%20B.pdf

I've found the new mean but don't know how to find the new standard deviation and the sum of x^2.

Please give detailed answer, thanks

Instead of giving you an answer to your question, I will instead provide a worked example using other numbers (mainly because I am unsure how to just give hints and not give a full blown explanation).

35 people shop in Sainsbury's in one hour. The amount spent in the shop at this time have a mean of £28.90 and a standard deviation of £5.60.

After this was calculated, it was discovered that one other person had visited the store and spent £39.40. Calculate the correct mean and standard deviation of the 36 shoppers.

I realise you have said you have calculated your mean but as I need it to work out my new S.D, I have provided the working out for it under the spoiler.

New mean =

50.487\dot{2}

Spoiler

Standard Deviation

\sqrt{\dfrac{\sum(x_i - \bar{x})^2}{n}}

The original S.D = 5.6 using n = 35 and

\bar{x}= 50.487\dot{2}

Say that some of the people who had shopped had spent £26.75, 32.91, £62, £14.60, £13.54 etc then you would calculate

( 26.75- 50.487\dot{2} )^2 + ( 32.91 - 50.487\dot{2} )^2 + ( 62 - 50.487\dot{2} )^2 +( 14.60- 50.487\dot{2} )^2 +( 13.54 - 50.487\dot{2} )^2 + ...

(so on and so forth)
As you now have a new value of

x_i

(39.40) , you need to get to a point where you can add on the new

(x_i - \bar{x})^2.

To the work out the new Standard Deviation, you would divide by n (REMEMBER that n is now 36), and then square root.

Unparseable latex formula:

\sqrt{\dfrac{\sum(x_i - 50.487\dot{2})^2}{35}} = 5.6 [br]\dfrac{\sum(x_i - 50.487\dot{2})^2}{35} = 5.6^2 [br]\dfrac{\sum(x_i - 50.487\dot{2})^2}{35} = 31.36 [br]\sum(x_i - 50.487\dot{2})^2} = 31.36 x 35[br]\sum(x_i - 50.487\dot{2})^2} = 1097.6[br][br]\sqrt{\dfrac{1097.6}{36}}

= 5.12 (to 2.d.p) = new S.D

Reply 3

elldeegee

19

Original post by CTArsenal

You know the formula for variance in S1 as the (Σx² / n) - x̅².

No,

\dfrac{\sum(x - \bar{x})^2}{n}

So therefore the S.D =

\sqrt{\dfrac{\sum(x - \bar{x})^2}{n}}

Reply 4

Mattp9850

11

Original post by elldeegee

No,

\dfrac{\sum(x - \bar{x})^2}{n}

So therefore the S.D =

\sqrt{\dfrac{\sum(x - \bar{x})^2}{n}}

Both are correct he just stated the computational formula -
[sum of x^2 - n(x-bar^2)]/n-1 or they way he expressed it. Both are correct but computational is faster to calculate. :smile:

sorry for no LaTeX.

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Edexcel Statistics help please

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