Suppose that a new treatment is successful in curing a common ailment $66 \%$ of the time. If the treatment is tried on a random sample of 130 patients, approximate the probability that fewer than 87 will be cured. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps. (If necessary, consult a list of formulas.)

Step 1 ：Calculate the mean (μ) of the binomial distribution using the formula μ = n*p, where n is the number of trials and p is the probability of success. In this case, n = 130 (the number of patients) and p = 0.66 (the probability that a patient is cured). So, μ = 130*0.66 = 85.8.

Step 2 ：Calculate the standard deviation (σ) of the binomial distribution using the formula σ = \(\sqrt{n*p*(1-p)}\). In this case, n = 130, p = 0.66. So, σ = \(\sqrt{130*0.66*(1-0.66)}\) = \(\sqrt{29.172}\) = 5.4.

Step 3 ：Standardize the value we're interested in (87) using the z-score formula: z = (X - μ) / σ. However, since we're looking for the probability that fewer than 87 patients are cured, we need to use 86.5 (87 - 0.5) to apply the correction for continuity. So, z = (86.5 - 85.8) / 5.4 = 0.13.

Step 4 ：Find the probability that a standard normal random variable is less than 0.13. This is given by the cumulative distribution function (CDF) for the standard normal distribution. Using a standard normal distribution table or a calculator, we find that P(Z < 0.13) = 0.552.

Step 5 ：So, the probability that fewer than 87 out of 130 patients will be cured is approximately 0.552, or 55.2%. \(\boxed{0.552}\)