A drug tester claims that a drug cures a rare skin disease $84 \%$ of the time. The claim is checked by testing the drug on 100 patients. If at least 76 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. The probability is (Round to four decimal places as needed.)

Step 1 ：Given that the drug cures the disease 84% of the time, we are asked to find the probability that less than 76 patients are cured. This is a binomial problem, but since the sample size is large, we can use the normal approximation to the binomial distribution.

Step 2 ：The mean of the binomial distribution is \(np\), and the standard deviation is \(\sqrt{np(1-p)}\), where \(n\) is the number of trials (in this case, the number of patients), and \(p\) is the probability of success (in this case, the probability that a patient is cured).

Step 3 ：Let's calculate the mean and standard deviation. Given \(n = 100\) and \(p = 0.84\), we find that the mean is \(np = 84.0\) and the standard deviation is \(\sqrt{np(1-p)} = 3.6660605559646724\).

Step 4 ：We can now use the normal distribution to find the probability that less than 76 patients are cured. We calculate the z-score using the formula \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the number of successes, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Substituting the given values, we get \(z = -2.3185650837574188\).

Step 5 ：Finally, we find the probability corresponding to this z-score from the standard normal distribution table, which is 0.0102093150867656.

Step 6 ：Thus, the probability that the claim will be rejected, assuming that the manufacturer's claim is true, is approximately \(\boxed{0.0102}\).