For the normal approximation to the binomial distribution with n = 100 and p = 1/2,a score of X = 60 corresponds to a z-score of z = 2.00.

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This statement is true. For a normal approximation to the binomial distribution with n = 100 and p = 1/2, the mean is µ = np = 100(1/2) = 50 and the standard deviation is σ = sqrt(np(1-p)) = sqrt(100(1/2)(1-1/2)) = 5. Therefore, the z-score corresponding to X = 60 is (60-50)/5 = 2.00.

For the Normal Approximation to the Binomial Distribution with N