The results of a study showed that heterosexual women, during ovulation, were significantly better at correctly identifying the sexual orientation of a man from a photograph of his face than women who were not ovulating. Near ovulation, on average women correctly identified the orientation of about \( 66 \% \) of the 100 men shown to them. Assume that the sample distribution for this study is unimodal and symmetric and that the samples are collected randomly. If this is the probability of correctly identifying the orientation of a man in any given photograph, what is the probability a woman would correctly classify 78 or more of the men? The probability is \( \square \). (Round to five decimal places as needed.)

Let \( X \) be the number of men correctly classified. The problem tells us that \[ X \sim \text{Binomial}(n=100, p=0.66) \] where \( n=100 \) and \( p=0.66 \). **Step 1. Find the mean and standard deviation.** The mean is \[ \mu = np = 100 \times 0.66 = 66. \] The variance is \[ \sigma^2 = np(1-p) = 100 \times 0.66 \times 0.34 = 22.44, \] and the standard deviation is \[ \sigma = \sqrt{22.44} \approx 4.74. \] **Step 2. Use the normal approximation to the binomial distribution.** We approximate \( X \) by a normal variable \( Y \sim N(\mu, \sigma^2) \). To find the probability of getting 78 or more correct classifications, we use the continuity correction. We want \[ P(X \ge 78) \approx P\left(Y \ge 77.5\right). \] **Step 3. Standardize the variable.** Convert \( Y \) to the standard normal variable \( Z \) via \[ Z = \frac{Y - \mu}{\sigma}. \] For \( Y = 77.5 \), \[ Z = \frac{77.5 - 66}{4.74} \approx \frac{11.5}{4.74} \approx 2.4265. \] **Step 4. Find the tail probability.** We need to find the tail probability \[ P(Z \ge 2.4265). \] Using a standard normal table or calculator, \[ P(Z \ge 2.4265) \approx 0.00759. \] Thus, the probability that a woman would correctly classify 78 or more of the men is \[ \boxed{0.00759}. \]