Question

A die is rolled 60 times.Find the expected number of times the face with 6 spots appears.
Find the SE of the number of times the face with 6 spots appears.
Find the normal approximation to the chance that the face with six spots appears 10 times
Find the exact chance that the face with six spots appears 10 times.
Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times.
Find the exact chance that the face with six spots appears 9, 10, or 11 times.

Answer 1

@pandu@benjamin2123@sara98

Answer 2

@sara98

Answer 3

@benjamin2123

Answer 4

???

Answer 5

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
60 * 1/6 = 10 times.
\[SE(6\ spots)=\sqrt{60\times \frac{1}{6}\times \frac{5}{6}}=2.8868\]
Applying the normal approximation to the chance that the face with six spots appears 10 times, we obtain the following:
mean = 10
Standard deviation = 2.8868
Using the continuity correction of 0.5 the following z-scores are found:
\[z _{1}=\frac{9.5-10}{2.8868}=-0.1732\]
\[z _{2}=\frac{10.5-10}{2.8868}=0.1732\]
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.

The exact chance that the face with six spots appears 10 times is found from
\[60C10\times (\frac{1}{6})^{10}\times (\frac{5}{6})^{50}=0.137\]

Answer 6

9 10 11:
normal approximation 0.3966
exact chance 0.3959