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\( X \) is a normally distributed random variable with mean 25 and standard deviation 17 . What is the probability that \( X \) is between 8 and 76 ? 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 Perkins Harmon. in the United States
Jan 24,2025

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The probability that \( X \) is between 8 and 76 is **0.837**.

Solución

To find the probability that \( X \) is between 8 and 76 for a normally distributed random variable \( X \) with mean \( \mu = 25 \) and standard deviation \( \sigma = 17 \), we can use the \( 68-95-99.7 \) rule (also known as the empirical rule). ### Step 1: Determine the Number of Standard Deviations First, determine how many standard deviations away from the mean the values 8 and 76 are: - **Lower Bound (8):** \[ \frac{8 - 25}{17} = \frac{-17}{17} = -1 \ \text{standard deviation} \] - **Upper Bound (76):** \[ \frac{76 - 25}{17} = \frac{51}{17} = 3 \ \text{standard deviations} \] So, the interval [8, 76] corresponds to \( \mu - 1\sigma \) to \( \mu + 3\sigma \). ### Step 2: Apply the Empirical Rule The empirical rule states: - **68%** of the data lies within \( \mu \pm 1\sigma \). - **95%** within \( \mu \pm 2\sigma \). - **99.7%** within \( \mu \pm 3\sigma \). For the interval \( \mu - 1\sigma \) to \( \mu + 3\sigma \): 1. **Probability up to \( \mu + 3\sigma \) (76):** Approximately **99.7%** or **0.997**. 2. **Probability below \( \mu - 1\sigma \) (8):** Since 68% is within \( \mu \pm 1\sigma \), the remaining 32% is split equally in the two tails. Thus, the lower tail is **16%** or **0.16**. ### Step 3: Calculate the Probability Between 8 and 76 Subtract the lower tail probability from the upper cumulative probability: \[ P(8 \leq X \leq 76) = P(X \leq 76) - P(X \leq 8) = 0.997 - 0.16 = 0.837 \] Rounded to the nearest thousandth, the probability is **0.837**. **Answer:** 0.837

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Beyond the Answer

To find the probability that \( X \) is between 8 and 76 for a normally distributed random variable with a mean of 25 and a standard deviation of 17, we first calculate the z-scores for both values. For \( X = 8 \): \[ z_1 = \frac{8 - 25}{17} = \frac{-17}{17} = -1 \] For \( X = 76 \): \[ z_2 = \frac{76 - 25}{17} = \frac{51}{17} \approx 3 \] Now, we can use the empirical rule or the \( 0.68-0.95-0.997 \) rule to find the probabilities corresponding to these z-scores. A z-score of -1 indicates that 68% of the data lies within 1 standard deviation of the mean, which covers from \( 8 \) to \( 25 \). The z-score of approximately 3 tells us that almost all (99.7%) the data lies within 3 standard deviations of the mean, covering from \( -26 \) to \( 76 \). Thus, the probability that \( X \) is between 8 and 76 is essentially \( 1 - \frac{0.5 - 0.68}{2} \), which gives us: \[ P(8 < X < 76) \approx 0.95 + 0.5 = 0.98 \] However, calculating more precisely using the standard normal distribution tables or computation gives us: \[ P(X < 76) = 0.9987 \quad \text{and} \quad P(X < 8) = 0.1587 \] So, \[ P(8 < X < 76) = P(X < 76) - P(X < 8) \approx 0.9987 - 0.1587 = 0.84 \] Therefore, the answer rounded to the nearest thousandth is: \[ \text{Probability} \approx 0.840 \]

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