Pregunta
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Let's suppose that heights of Americans are normally distributed with a mean of 67.1 inches and a standard deviation of 5.3 inches. If a particular American is shorter than only \( 3 \% \) of all Americans, how tall is that person in inches? 76.1 inches 77.1 inches 57.1 inches 78.1 inches

Ask by Coles Rodriguez. in the United States
Mar 12,2025

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The person is approximately 57.1 inches tall.

Solución

To find the height of an American who is shorter than only \( 3\% \) of all Americans, we need to determine the height corresponding to the \( 3\% \) percentile of the normal distribution with a mean (\( \mu \)) of \( 67.1 \) inches and a standard deviation (\( \sigma \)) of \( 5.3 \) inches. ### Step 1: Find the z-score for the \( 3\% \) percentile The z-score corresponding to the \( 3\% \) percentile can be found using the standard normal distribution table or a calculator. The z-score represents how many standard deviations an element is from the mean. ### Step 2: Calculate the height using the z-score formula The formula to convert a z-score to a height in a normal distribution is: \[ X = \mu + z \cdot \sigma \] Where: - \( X \) is the height we want to find, - \( \mu = 67.1 \) inches (mean), - \( z \) is the z-score for the \( 3\% \) percentile, - \( \sigma = 5.3 \) inches (standard deviation). ### Step 3: Find the z-score for the \( 3\% \) percentile Let's find the z-score for the \( 3\% \) percentile. Using a standard normal distribution table or calculator, the z-score for \( 0.03 \) (or \( 3\% \)) is approximately \( -1.88079 \). ### Step 4: Calculate the height Now we can substitute the values into the formula: \[ X = 67.1 + (-1.88079) \cdot 5.3 \] Let's calculate this. \[ X = 67.1 - 9.973 \] \[ X \approx 57.127 \] Thus, the height of the person who is shorter than only \( 3\% \) of all Americans is approximately \( 57.1 \) inches. ### Conclusion The correct answer is \( 57.1 \) inches.

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The Deep Dive

To determine the height of a person who is shorter than only 3% of all Americans, we need to find the z-score corresponding to the 3rd percentile of the normal distribution. The z-score for the 3rd percentile is approximately -1.88. We can plug this into the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where: - \( z \) is the z-score, - \( X \) is the height, - \( \mu \) is the mean (67.1 inches), and - \( \sigma \) is the standard deviation (5.3 inches). Rearranging the formula to solve for \( X \): \[ X = \mu + z \cdot \sigma \] Substituting in the values: \[ X = 67.1 + (-1.88) \cdot 5.3 \] \[ X = 67.1 - 9.954 \] \[ X \approx 57.146 \] Therefore, the height of a person who is shorter than only 3% of Americans is approximately 57.1 inches. The correct answer is **57.1 inches.**

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