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A die is rolled 60 times.

Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times.
The exact chance that the face with six spots appears 9, 10, or 11 times?

my solution; in one roll of the die P(6 spots) = 1/6 In 60 rolls of the die the expected number of times the face with 6 spots appears is given by E(X)=Mean=60 * 1/6 = 10 times.

SE(6 spots)=SD=SQR(60×1/6×5/6)=2.8868

Applying the normal approximation to the chance that the face with six spots appears 10 times, I obtain the following: mean = 10

Standard deviation = 2.8868

Using the continuity correction of 0.5 the following z-scores are found: z1=(9.5−10)/2.8868=−0.1732 and z2=(10.5−10)/2.8868=0.1732 Subtracting the cumulative probabilities for the two z-scores gives 0.5688 - 0.4312 =

0.1376 which is the approximate chance that the face with six spots appears 10 times. find the six spots appears between 9,10, or 11?

The exact chance that the face with six spots appears 10 times is found from

60C10×(1/6)^10×(5/6)^50=0.137 find six spots appears 9, 10 or 11 times?

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What do the multiple question marks contribute to the exposition of the question? –  joriki May 5 '13 at 3:09

3 Answers 3

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For the "Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times". and Find the exact chance that the face with six spots appears 9, 10, or 11 times. can you please check that the answers 0.3967 and .39589 are correct.

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Both Answers of 0.3967 for "the chance that the face with six spots appears between 9, 10, or 11 times" and .39589 for "the exact chance that the face with six spots appears 9, 10, or 11 times are Correct! User73408. Thanks –  statistics-student13 May 9 '13 at 4:44

find six spots appears 9, 10 or 11 times? ANS 60C9×(1/6)^9×(5/6)^51+60C10×(1/6)^10×(5/6)^50+60C11×(1/6)^11×(5/6)^49

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Welcome to Math.SE. Thank you for your answer. It would help if you used MathJax to format your answer better, and also if you gave more explanation for why you believe this to be the right answer. –  vadim123 May 5 '13 at 16:56

Your calculations (as far as I'm able to follow them in your somewhat idiosyncratic exposition) all appear correct. However, they calculate (approximately and exactly, respectively) the probability that a $6$ is rolled $10$ times. You can apply the same approach to the probabilities that a $6$ is rolled $9$ times or $11$ times, and then add the three results. (In the case of the Gaussian approximation, you can also calculate the sum in one go by subtracting the values of the cumulative distribution function at $11.5$ and $8.5$.)

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Thanks Clearly Understood the remaining 2 steps how to get the answers. –  statistics-student13 May 5 '13 at 23:59

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