A drug tester claims that a drug cures a rare skin disease 69% of the time. To verify this claim, the drug is tested on 100 patients. If at least 63 patients are cured, the claim will be accepted.

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

Given:

N= 100P= 0.69Q= 1 – P = 0.31## Calculate the mean and standard deviation:

Mean (μ)= N × P = 100 × 0.69 = 69Standard deviation (σ)= √(N × P × Q) = √(100 × 0.69 × 0.31) = √21.39 ≈ 4.62## Find the Z-score:

To determine the probability that the claim will be rejected, we need to calculate the Z-score for 63 patients:

Z= (X – μ) / σ = (63 – 69) / 4.62 = -6 / 4.62 ≈ -1.30## Calculate the probability:

We need to find

P(Z ≥ -1.30):P(Z ≥ -1.30)= 1 –P(Z < -1.30)Using the standard normal distribution table,

P(Z < -1.30)≈ 0.0968.So:

P(Z ≥ -1.30)= 1 – 0.0968 = 0.9032Therefore, the probability that the claim will be rejected, assuming the manufacturer’s claim is true, is

0.9032or90.32%.