greets

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- Aug 25th 2007, 12:41 AMshirelrolling a die ...
greets

- Aug 25th 2007, 02:41 AMCaptainBlack
See the answer to your other question about the distribution of the number

of times a 5 will come up.

Now 600 times is a large number of rolls so we may as well use the normal

approximation to the binaomial distribution, in this case it will have mean:

$\displaystyle m=Np = 600 \times (1/6) = 100$

and standard deviation:

$\displaystyle

\sqrt{Np(1-p)} = \sqrt{600(1/6)(5/6)} \approx 9.13

$

So you would expect the number of 5's to be $\displaystyle 100 \pm k \times 9.13$, where

the value of $\displaystyle k$ is chossen by you depending on what confidence you want

that the actual number lies in the interval, often a value of $\displaystyle k=1.96 $ is used.

RonL