Question

A drug tester claims that a drug cures a rare skin disease 72% of the time. The claim is checked by testing the drug on 100 patients. If at least 66 patients are cured, the claim will be accepted.

Find the probability that the claim will be rejected assuming that the manufacturer’s claim is true. Use the normal distribution to approximate the binomial distribution if possible. Round to four decimal places.

Answer 1

The probability that the drug claim is rejected given the drug actually cures the disease 72% of the time is approximately 0.0735 or 7.35%, calculated by approximating the binomial distribution with a normal distribution and finding the Z-score.

Step-by-step explanation:

Here, we assume that the claim is true. We have 100 patients, and the claim is rejected if fewer than 66 are cured. Using the normal distribution to approximate the binomial distribution, we calculate the probability of rejection.

First, we determine the mean (μ) and the standard deviation (σ) for our binomial distribution. The mean μ = np and the standard deviation σ = √npq, where 'n' is the number of trials (100), 'p' is the probability of success (0.72), and 'q' is the probability of failure (1-p).

μ = 100 × 0.72 = 72

σ = √(100 × 0.72 × 0.28) = √(20.16) = 4.49 (approximately)

The Z-score for 65.5 (since we want fewer than 66, we add 0.5 for continuity correction) is given by:

Z = (X - μ) / σ = (65.5 - 72) / 4.49 = -1.45 (approximately)

Using a Z-table, or a calculator, we find the probability of a Z-score being less than -1.45.

The probability is approximately 0.0735. Therefore, the probability that the claim is rejected given the drug actually works is 0.0735, or 7.35%.