Pregunta
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\( X \) is a normally distributed random variable with mean 38 and standard deviation 13 . What is the probability that \( X \) is between 51 and 77 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

Ask by Chavez Huff. in the United States
Jan 24,2025

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The probability that \( X \) is between 51 and 77 is approximately 0.157.

Solución

To find the probability that \( X \) is between 51 and 77 for a normally distributed random variable with mean \( \mu = 38 \) and standard deviation \( \sigma = 13 \), we can follow these steps: 1. **Standardize the values**: We will convert the values 51 and 77 into z-scores using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] 2. **Calculate the z-scores**: - For \( X = 51 \): \[ z_1 = \frac{(51 - 38)}{13} \] - For \( X = 77 \): \[ z_2 = \frac{(77 - 38)}{13} \] 3. **Use the z-scores to find probabilities**: We will then look up the probabilities corresponding to these z-scores in the standard normal distribution table or use the empirical rule. Let's calculate the z-scores first. ### Step 1: Calculate z-scores Calculating \( z_1 \) and \( z_2 \): \[ z_1 = \frac{(51 - 38)}{13} = \frac{13}{13} = 1 \] \[ z_2 = \frac{(77 - 38)}{13} = \frac{39}{13} = 3 \] ### Step 2: Find probabilities Using the empirical rule: - Approximately 68% of the data falls within 1 standard deviation of the mean (between \( \mu - \sigma \) and \( \mu + \sigma \)). - Approximately 95% of the data falls within 2 standard deviations of the mean (between \( \mu - 2\sigma \) and \( \mu + 2\sigma \)). - Approximately 99.7% of the data falls within 3 standard deviations of the mean (between \( \mu - 3\sigma \) and \( \mu + 3\sigma \)). Now, we need to find the probability that \( X \) is between 51 and 77, which corresponds to \( z_1 = 1 \) and \( z_2 = 3 \). - The probability that \( Z < 1 \) is approximately 0.8413 (from z-tables). - The probability that \( Z < 3 \) is approximately 0.9987 (from z-tables). ### Step 3: Calculate the probability between z-scores Now, we can find the probability that \( X \) is between 51 and 77: \[ P(51 < X < 77) = P(Z < 3) - P(Z < 1) = 0.9987 - 0.8413 \] Calculating this gives: \[ P(51 < X < 77) = 0.9987 - 0.8413 = 0.1574 \] ### Final Answer Thus, the probability that \( X \) is between 51 and 77 is approximately \( 0.157 \) when rounded to the nearest thousandth. \[ \boxed{0.157} \]

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The Deep Dive

To find the probability that \( X \) is between 51 and 77, first, we need to find how many standard deviations away these values are from the mean. The mean \( \mu \) is 38 and the standard deviation \( \sigma \) is 13. For \( X = 51 \): \[ Z_1 = \frac{51 - 38}{13} = \frac{13}{13} = 1 \] For \( X = 77 \): \[ Z_2 = \frac{77 - 38}{13} = \frac{39}{13} = 3 \] Now, according to the \( 0.68-0.95-0.997 \) rule: - About 68% of the data lies within 1 standard deviation (from \( 38 - 13 \) to \( 38 + 13 \)). - About 95% lies within 2 standard deviations (from \( 38 - 2 \times 13 \) to \( 38 + 2 \times 13 \)), which translates to the interval from \( 12 \) to \( 64 \). - About 99.7% lies within 3 standard deviations (from \( 38 - 3 \times 13 \) to \( 38 + 3 \times 13 \)), which translates to the interval from \( -1 \) to \( 77 \). Now, we are interested in the probability between \( 51 \) (which is 1 standard deviation away) and \( 77 \) (which is 3 standard deviations away). Since we know that: - From the mean (38) to 1 standard deviation above (51) is 34% (half of 68%). - From the mean to 3 standard deviations above (77) is 49.85% (half of 99.7%). Therefore, the probability that \( X \) is between 51 and 77 is: \[ 0.34 + 0.497 = 0.84 \] Thus, the final probability that \( X \) is between 51 and 77 is approximately \( 0.840 \).

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