To find the probability \( P(119.9 < X < 122.9) \), where \( X \) is normally distributed with a mean \( \mu = 124.7 \text{ cm} \) and standard deviation \( \sigma = 2.2 \text{ cm} \), we will first convert the lengths 119.9 cm and 122.9 cm into z-scores.
The formula for calculating the z-score is: \[ z = \frac{(X - \mu)}{\sigma} \]
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Find the z-score for 119.9 cm: \[ z_1 = \frac{(119.9 - 124.7)}{2.2} = \frac{-4.8}{2.2} \approx -2.1818 \]
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Find the z-score for 122.9 cm: \[ z_2 = \frac{(122.9 - 124.7)}{2.2} = \frac{-1.8}{2.2} \approx -0.8182 \]
Next, we will find the probabilities corresponding to these z-scores using the standard normal distribution table or a calculator.
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Find the probability for \( z_1 \): Using a standard normal distribution table or calculator: \[ P(Z < -2.1818) \approx 0.0146 \]
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Find the probability for \( z_2 \): Using a standard normal distribution table or calculator: \[ P(Z < -0.8182) \approx 0.2061 \]
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Now, we can find the probability that \( X \) is between 119.9 cm and 122.9 cm: \[ P(119.9 < X < 122.9) = P(Z < -0.8182) - P(Z < -2.1818) \] \[ P(119.9 < X < 122.9) \approx 0.2061 - 0.0146 = 0.1915 \]
Thus, the final answer, represented as a number accurate to four decimal places, is: \[ \boxed{0.1915} \]
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