Note: Round your answers to 4 decimal places. The heights of adult women are normally distributed with a mean of 61 inches and a standard deviation of 3.9 inches. What is the probability that a randomly chosen adult woman will have a height of less than 56.8 inches? List the \( z \)-score needed to calculate the result. If there is more than one \( z \)-score, separate the values with a comma. \( \mathrm{z}= \) There is a than 56.8 inches. Round your answer to 4 decimal places.

To find the \( z \)-score for a height of 56.8 inches, we use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value in question (56.8 inches), \( \mu \) is the mean (61 inches), and \( \sigma \) is the standard deviation (3.9 inches). Substituting the values: \[ z = \frac{(56.8 - 61)}{3.9} = \frac{-4.2}{3.9} \approx -1.0769 \] So the \( z \)-score needed to calculate the result is: \( \mathrm{z} = -1.0769 \) Next, to find the probability that a randomly chosen adult woman will have a height of less than 56.8 inches, we can look up the \( z \)-score of -1.0769 in the standard normal distribution table, or use a calculator or software that provides the cumulative distribution function for a normal distribution. The probability corresponding to \( z = -1.0769 \) is approximately 0.1412 (rounded to four decimal places). Thus, the final answer is: There is a probability of 0.1412 that a randomly chosen adult woman will have a height of less than 56.8 inches.