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

We are given that the heights \( X \) follow a normal distribution with mean \(\mu = 55.6\) inches and standard deviation \(\sigma = 5.6\) inches. We want to calculate \[ P(62.9 < X < 64.4). \] **Step 1. Standardize the values** For any value \( x \), the z-score is calculated as \[ z = \frac{x - \mu}{\sigma}. \] Calculate the z-scores for \( x = 62.9 \) and \( x = 64.4 \). For \( x = 62.9 \): \[ z_1 = \frac{62.9 - 55.6}{5.6} = \frac{7.3}{5.6} \approx 1.30357. \] For \( x = 64.4 \): \[ z_2 = \frac{64.4 - 55.6}{5.6} = \frac{8.8}{5.6} \approx 1.57143. \] **Step 2. Find the corresponding probabilities** We need the probabilities \( \Phi(z_1) \) and \( \Phi(z_2) \) from the standard normal distribution. Assume: \[ \Phi(1.30357) \approx 0.9032 \quad \text{and} \quad \Phi(1.57143) \approx 0.9418. \] **Step 3. Calculate the probability between the two values** The probability that \( X \) is between 62.9 and 64.4 inches is: \[ P(62.9 < X < 64.4) = \Phi(1.57143) - \Phi(1.30357) \approx 0.9418 - 0.9032 \approx 0.0386. \] **Step 4. Round the final answer** Rounded to 3 decimal places, the probability is: \[ 0.039. \] Thus, the final answer is \( 0.039 \).