More than 30 away from the average in 900 flips

Question

A fair coin is flipped 900 times. Let X be the number of heads. Using the normal approximation, estimate P(X - 450 30), the chance the count lands more than 30 away from its average.

Accepted Answer

Match a normal to the binomial: np = 900 	imes 0.5 = 450 and np(1-p) = 900 	imes 0.5 	imes 0.5 = 225, so X \approx N(450, 225) with standard deviation \sqrt{225} = 15. A deviation of 30 is exactly z = \frac{30}{15} = 2 standard deviations. The question asks for both tails (more than 2 sd above or below), so double the one-tail probability: P(X - 450 30) \approx 2 	imes P(Z \ge 2) = 2 	imes 0.0228 = 0.0455, about 4.6%. Takeaway. Mean 450, sd 15, so a 30-unit deviation is \pm 2 sd; both tails together give 2 	imes 0.0228 \approx 4.6\%.

More than 30 away from the average in 900 flips

Your answer

More Statistics questions