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\( X \) is a normally distributed random variable with mean 55 and standard deviation 18 . What is the probability that \( X \) is between 73 and 91 ? 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 Bernard Johnston. in the United States
Jan 24,2025

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The probability that \( X \) is between 73 and 91 is approximately 0.136.

Solution

To find the probability that \( X \) is between 73 and 91 for a normally distributed random variable with mean \( \mu = 55 \) and standard deviation \( \sigma = 18 \), we can follow these steps: 1. **Standardize the values**: We will convert the values 73 and 91 into z-scores using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] 2. **Calculate the z-scores**: - For \( X = 73 \): \[ z_1 = \frac{(73 - 55)}{18} \] - For \( X = 91 \): \[ z_2 = \frac{(91 - 55)}{18} \] 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 \( z_2 \): \[ z_1 = \frac{(73 - 55)}{18} = \frac{18}{18} = 1 \] \[ z_2 = \frac{(91 - 55)}{18} = \frac{36}{18} = 2 \] ### Step 2: Find probabilities Using the empirical rule: - Approximately 68% of the data falls within 1 standard deviation of the mean. - Approximately 95% of the data falls within 2 standard deviations of the mean. From the z-scores: - The probability that \( X \) is less than 73 (which corresponds to \( z_1 = 1 \)) is approximately \( 0.8413 \). - The probability that \( X \) is less than 91 (which corresponds to \( z_2 = 2 \)) is approximately \( 0.9772 \). ### Step 3: Calculate the probability between 73 and 91 To find the probability that \( X \) is between 73 and 91, we subtract the probability of \( X < 73 \) from the probability of \( X < 91 \): \[ P(73 < X < 91) = P(X < 91) - P(X < 73) = 0.9772 - 0.8413 \] Now, let's calculate this: \[ P(73 < X < 91) = 0.9772 - 0.8413 = 0.1359 \] ### Final Answer Thus, the probability that \( X \) is between 73 and 91 is approximately \( 0.136 \) when rounded to the nearest thousandth.

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To find the probability that \( X \) is between 73 and 91 using the \( 0.68-0.95-0.997 \) rule, we first standardize the values of \( X \) using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 73 \): \[ Z(73) = \frac{73 - 55}{18} = 1.000 \] For \( X = 91 \): \[ Z(91) = \frac{91 - 55}{18} = 2.000 \] Now, we know that about 68% of the data falls within 1 standard deviation of the mean, and about 95% falls within 2 standard deviations. Therefore, we want to find the probability between these two Z-scores. The area between \( Z = 1 \) and \( Z = 2 \): - The area to the left of \( Z = 2 \) is about 0.9772 (from Z-tables). - The area to the left of \( Z = 1 \) is about 0.8413. Now, subtract the two probabilities: \[ P(73 < X < 91) = P(Z < 2) - P(Z < 1) \] \[ P(73 < X < 91) = 0.9772 - 0.8413 = 0.1359 \] Thus, the probability that \( X \) is between 73 and 91 is approximately **0.136**, rounded to the nearest thousandth.

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