If the coin is in fact fair and is tossed 100 times, give the smallest number N so that with a probability of 99% the number of “heads” lies within the interval [50 − N, 50 + N]. (This may involve some tedious calculations; feel free to use Mathematica.)

We can use the normal approximation to the binomial distribution to solve this problem. If we toss a fair coin 100 times, the number of heads follows a binomial distribution with parameters n = 100 and p = 0.5. The mean of this distribution is μ = np = 50, and the standard deviation is σ = sqrt(np(1-p)) = 5.

To find the smallest number N such that the probability of the number of heads lying within the interval [50 − N, 50 + N] is at least 0.99, we need to find the z-scores corresponding to the endpoints of this interval. Let z1 and z2 be the z-scores such that:

z1 = (50 - N - μ) / σ
z2 = (50 + N - μ) / σ

We want to find N such that P(z1 < Z < z2) >= 0.99, where Z is a standard normal random variable.

Using the standard normal distribution table or a calculator, we can find that the z-score corresponding to the 0.005 percentile (the lower tail probability for a 99% confidence interval) is -2.576, and the z-score corresponding to the 0.995 percentile (the upper tail probability for a 99% confidence interval) is 2.576.

Substituting these values into the equations for z1 and z2, we get:

-2.576 = (50 - N - 50) / 5
2.576 = (50 + N - 50) / 5

Simplifying these equations, we get:

N = 5 * 2.576 = 12.88

Therefore, the smallest number N such that with a probability of 99% the number of “heads” lies within the interval [50 − N, 50 + N] is 13 (rounded up from 12.88).