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)=16 In 60 rolls of the die the expected number of times the face with 6 spots appears is given by E(X)=Mean=606=10 times.
SE(6spots)=SD=SQR(606⋅56)=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(16)10×(56)50=0.137 find six spots appears 9, 10 or 11 times
