To find the probability that the height of a randomly chosen ten-year-old child in Heightlandia is between 56.3 and 66.4 inches, we can use the properties of the normal distribution. We will first determine the z-scores for both height values and then use the standard normal distribution to find the probabilities.
The z-score formula is: \[ z = \frac{(X - \mu)}{\sigma} \] where:
- \(X\) is the value for which we are calculating the z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
Given:
- Mean (\(\mu\)) = 57 inches
- Standard deviation (\(\sigma\)) = 5.2 inches
Step 1: Calculate the z-scores
For 56.3 inches: \[ z_1 = \frac{(56.3 - 57)}{5.2} = \frac{-0.7}{5.2} \approx -0.1346153846 \]
For 66.4 inches: \[ z_2 = \frac{(66.4 - 57)}{5.2} = \frac{9.4}{5.2} \approx 1.8076923077 \]
Step 2: Look up the z-scores in the standard normal distribution table
Using a standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores.
For \(z_1 \approx -0.135\): The cumulative probability (area to the left) for \(z_1\) is approximately: \[ P(Z < -0.135) \approx 0.4477 \]
For \(z_2 \approx 1.808\): The cumulative probability for \(z_2\) is approximately: \[ P(Z < 1.808) \approx 0.9643 \]
Step 3: Calculate the probability between the two z-scores
The probability that a child's height is between 56.3 inches and 66.4 inches is calculated by: \[ P(56.3 < X < 66.4) = P(Z < z_2) - P(Z < z_1) \] \[ = P(Z < 1.808) - P(Z < -0.135) = 0.9643 - 0.4477 = 0.5166 \]
Final Answer
Therefore, the probability that the height of a randomly chosen child is between 56.3 and 66.4 inches is approximately 0.5166. Rounding to three decimal places, we get: \[ \text{Probability} \approx 0.517 \]
So, the final round answer is:
Answer: 0.517