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 $P(6 \text{ spots}) = \frac{1}{6}$ In $60$ rolls of the die the expected number of times the face with $6$ spots appears is given by $E(X)=\text{Mean}=\frac{60}{6} = 10$ times.
$SE(6 spots)=SD=SQR(\frac{60}{6}\cdot \frac{5}{6})=2.8868$
Applying the normal approximation to the chance that the face with six spots appears 10 times, I obtain the following: mean = 10
$\text{Standard deviation} = 2.8868$
Using the continuity correction of 0.5 the following z-scores are found:
$z_1=(9.5−10)/2.8868=−0.1732$
and
$z_2=(10.5−10)/2.8868=0.1732$
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
$60C10\left(\frac{1}{6}\right)^{10×}\left(\frac{5}{6}\right)^{50}=0.137$ find six spots appears 9, 10 or 11 times?
Best Answer
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 $8.5$.)