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\( X \) is a normally distributed random variable with mean 57 and standard deviation 9. What is the probability that \( X \) is between 48 and 84 ? 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 Pierce Lindsey. in the United States
Jan 24,2025

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Answer

The probability that \( X \) is between 48 and 84 is **0.839**.

Solution

To determine the probability that \( X \) is between 48 and 84, we'll use the **68-95-99.7 (Empirical) Rule** for normally distributed data. Here's a step-by-step breakdown: 1. **Understand the Distribution:** - **Mean (\( \mu \))**: 57 - **Standard Deviation (\( \sigma \))**: 9 2. **Identify the Z-scores:** - **Lower Bound (48):** \[ Z = \frac{48 - 57}{9} = -1 \] - **Upper Bound (84):** \[ Z = \frac{84 - 57}{9} = 3 \] 3. **Apply the Empirical Rule:** - **Within 1 Standard Deviation (\( \mu \pm 1\sigma \)):** Approximately 68% of the data falls between 48 and 66. - **Within 3 Standard Deviations (\( \mu \pm 3\sigma \)):** Approximately 99.7% of the data falls between 30 and 84. 4. **Calculate the Desired Probability:** - **Probability below 84 (\( Z \leq 3 \)):** Approximately **99.7%** or **0.997**. - **Probability below 48 (\( Z \leq -1 \)):** Approximately **15.87%** or **0.1587**. To find the probability between 48 and 84: \[ P(48 \leq X \leq 84) = P(X \leq 84) - P(X \leq 48) = 0.997 - 0.1587 = 0.8383 \] 5. **Round to the Nearest Thousandth:** \[ P(48 \leq X \leq 84) \approx 0.838 \] **Final Answer:** The probability that \( X \) is between 48 and 84 is **0.839**.

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To solve this problem, we can utilize the \( 0.68-0.95-0.997 \) rule, which helps us understand the distribution of data in a normal curve based on standard deviations. 1. Calculate the z-scores for the values 48 and 84. - For \( x = 48 \): \[ z = \frac{48 - 57}{9} = \frac{-9}{9} = -1 \] - For \( x = 84 \): \[ z = \frac{84 - 57}{9} = \frac{27}{9} = 3 \] 2. According to the \( 0.68-0.95-0.997 \) rule: - Approximately 68% of the data lies within 1 standard deviation (i.e., between \( 48 \) and \( 66 \)). - Approximately 95% of the data lies within 2 standard deviations (i.e., between \( 39 \) and \( 75 \)). - Approximately 99.7% of the data lies within 3 standard deviations (i.e., between \( 30 \) and \( 84 \)). Since 48 and 84 are within the range of 1 to 3 standard deviations, we can find that \( X \) being between 48 and 84 covers almost the full extent of \( 99.7\% \) of the distribution: \[ P(48 < X < 84) \approx 0.997 \] Thus, rounding to the nearest thousandth, the probability that \( X \) is between 48 and 84 is \( \text{0.997} \).

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