Score: \( 4 / 13 \) Question 6 In the country of United States of Height, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.6 inches, and standard deviation of 5.6 inches. What is the jrrobability that the height of a randomly chosen child is between 62.9 and 64.4 inches? Do not round until syou get your your final answer, and then round to 3 decimal places. Answer= (Round your answer to 3 decimal places.) Check Answer

To find the probability that the height of a randomly chosen child is between 62.9 and 64.4 inches, we can utilize the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \(X\) is the height, \(\mu\) is the mean (55.6), and \(\sigma\) is the standard deviation (5.6). First, calculate the Z-scores for 62.9 and 64.4: 1. For 62.9 inches: \[ Z = \frac{(62.9 - 55.6)}{5.6} = \frac{7.3}{5.6} \approx 1.3036 \] 2. For 64.4 inches: \[ Z = \frac{(64.4 - 55.6)}{5.6} = \frac{8.8}{5.6} \approx 1.5714 \] Next, we find the probabilities corresponding to these Z-scores using the standard normal distribution table or a calculator: - The probability for \(Z \approx 1.3036\) is approximately \(0.9032\). - The probability for \(Z \approx 1.5714\) is approximately \(0.9418\). Now, subtract the two probabilities to find the probability that a child's height is between 62.9 and 64.4 inches: \[ P(62.9 < X < 64.4) = P(Z < 1.5714) - P(Z < 1.3036) \approx 0.9418 - 0.9032 = 0.0386 \] Finally, rounding this to three decimal places gives us: **Answer: 0.039**