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Expert: - 6/20/2009


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QUESTION: Two hundred fair dice are thrown 1000 times. Use the normal approximation to the binomial distribution to find the number of times you would expect to have the following number of sixes a)30, b)53, c)more than 38, d)less than 28, e)between 28 & 38 inclusive. Thanks a lot

ANSWER: There are 200 dice, thrown 1000 times.
The expected number of 6's in each toss is 200/6 = 100/3.
The variance is 200(1/6)(5/6) = 250/9,
so the standard deviation is 5√10/3.

a) 30 in the binomial is 29.5 to 30.5 in the normal.
For each value, subtract of the average and divide by the number of standard deviations.  This will give a value on the normal distribution to look up.  This must be done for each of the following problems.

b) To find a 53, use 52.5 to 53.5 and do the same thing.

c) For more than 38, go from 37.5 and up.

d) For less than 28, go from 28.5 and down.

e) For between 28 and 38 inclusive, go from 27.5 to 38.5.

---------- FOLLOW-UP ----------

QUESTION: Thanks you very much for spending your precious time in helping me. Why the expected value not 200*6 instead of 200/6 ? Can you explain, i dont get it? Part c & d, didnt state it is inclusive so i think part c) should be p(x>38.5) & part d) p(x<27.5). Much appreciated

Answer
Since there are only 200 choices, 200*6 is 1200, which is far more than 200.  You use 200/6 because this is the number of choices for each roll of the die.  Note that if you add up this number of each possibility, you get 200/6 + 200/6 + 200/6 + 200/6 + 200/6 + 200/6 =
1200/6 = 200, which is the total number of possibilities.

Yes, the corrections on (c) and (d) are right.
Thanks for the correction and I will remember to look at statistics questions closer in the future.

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