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charikaar
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#1
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#1
How many times must a fair dice be rolled in order for there to be a less than 1% chance that the mean of all scores differs from 3.5 by more than 0.1.

could anyone please help me with this problem.

Thank you.
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jpepsred
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ghostwalker
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(Original post by jpepsred)
Bump
9 year old thread - must be a record bump.

What do you know about the distribution of the sample mean if the die is rolled n times say?
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jpepsred
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Mean(X) is given by the distribution (3.5, 2.92/n), where 2.92 is the variance of X. I get that much.

I end up with p(z < 0.0585sqrt(n)) > 0.995 which, by referring to the tables, implies that 0.0585sqrt(n) >2.6.

The solutionbank, however, gives 0.0585sqrt(n) > 2.5758

2.5758 isn't a value of z which is in the tables, and I can't understand where it comes from.

It makes a huge difference: book gives n => 1936, I get n=> 1976.

I wonder if op appreciates finally getting an answer a decade after their a levels.
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undercxver
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(Original post by jpepsred)
Bump
(Original post by ghostwalker)
9 year old thread - must be a record bump.
Omg :toofunny: :toofunny: :toofunny:
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Zacken
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(Original post by jpepsred)
...
You need to use the percentage points table which is given a page below the table in the edexcel booklet or right under the table in the OCR booklet.

Image
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jpepsred
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THANK YOU. I haven't used that since s1. Completely forgot it existed!
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Zacken
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(Original post by jpepsred)
THANK YOU. I haven't used that since s1. Completely forgot it existed!
Sure, try and avoid bumping old threads in the future, create a new thread instead. Closing this now.
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