Statistical use of the normal distribution problems

AI Thread Summary
The discussion revolves around a practice paper focusing on statistical problems involving the normal distribution. The user seeks help specifically with questions 11, 12, and 13, expressing uncertainty about their solutions. Another participant suggests providing a more concise summary of the doubts to facilitate assistance. The user confirms they have studied the relevant material and shares an answer of 429 grams for question 13, prompting a request for the calculations behind that answer. The conversation highlights the importance of showing work in statistical problem-solving.
Paulo2014
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Hello
This attachment is a practice paper I am doing. I know how to do everything except for questions 11, 12 and 13 so I would appreciate it if someone could please show me the process for working them out. thanks in advance.
 

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anyone?
 
It is quite a lengthy document. Perhaps you can show some of your thoughts, and post your doubts in a more concise manner?
 
Those three problems are related to the statistical use of the normal distribution. Have you had that material already?
 
Yes I have read the material. The answer I just got for question 13 was 429 grams. Is that correct?
 
Where's your work? All I see are answers.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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