At most 42 successes in 100 trials

Question

A process has 100 independent trials, each a success with probability p = 0.5. Let X be the number of successes. Using the normal approximation, estimate P(X \le 42).

Accepted Answer

Match a normal to the binomial: np = 100 	imes 0.5 = 50 and np(1-p) = 100 	imes 0.5 	imes 0.5 = 25, so X \approx N(50, 25) with standard deviation \sqrt{25} = 5. Standardize 42: z = \frac{42 - 50}{5} = \frac{-8}{5} = -1.6. The lower-tail probability below z = -1.6 is P(Z \le -1.6) \approx 0.0548, about 5.5%. Takeaway. Mean 50, sd 5, and z = -1.6 give P(X \le 42) \approx 5.5\% under the normal approximation.

At most 42 successes in 100 trials

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