This problem can be addressed by applying the normal approximation to a binomial distribution.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
