Between 16 and 24 buyers out of 100

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Each of 100 independent callers buys with probability p = 0.2. Let X be the number who buy. Using the normal approximation, estimate P(16 \le X \le 24).

Accepted Answer

Approximate the binomial by a normal with the same mean and variance: np = 100 	imes 0.2 = 20 and np(1-p) = 100 	imes 0.2 	imes 0.8 = 16, so X \approx N(20, 16) with standard deviation \sqrt{16} = 4. Now notice the range 16 to 24 is exactly one standard deviation on each side of the mean: 16 = 20 - 4, \qquad 24 = 20 + 4. By the empirical (68-95-99.7) rule, a normal puts about 68% of its probability within one standard deviation of the mean. So P(16 \le X \le 24) \approx 0.68. Takeaway. With mean 20 and sd 4, the interval 16 to 24 is \pm 1 sd, so the normal approximation gives about 68%.

Between 16 and 24 buyers out of 100

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