$$ X $$ is a normally distributed random variable with mean 39 and standard deviation 11 . What is the probability that $$ X $$ is greater than 17 ? Use the $$ 0.68-0.95-0.997 $$ rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

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To determine the probability that $$ X $$ is greater than 17, where $$ X $$ is normally distributed with a mean ($$ \mu $$) of 39 and a standard deviation ($$ \sigma $$) of 11, we can use the $$ 0.68-0.95-0.997 $$ empirical rule.

1. **Calculate the Z-score:**
   $$
   Z = \frac{X - \mu}{\sigma} = \frac{17 - 39}{11} = \frac{-22}{11} = -2
   $$

2. **Interpret the Z-score using the empirical rule:**
   - The empirical rule states that approximately 95% of the data lies within $$ \mu \pm 2\sigma $$.
   - Therefore, $$ P(\mu - 2\sigma \leq X \leq \mu + 2\sigma) \approx 0.95 $$.

3. **Determine the probability $$ P(X > 17) $$:**
   - $$ \mu - 2\sigma = 39 - 2 \times 11 = 17 $$.
   - The probability that $$ X $$ is less than 17 is approximately $$ 2.5\% $$ (since 5% is split equally in the two tails of the distribution).
   - Hence, the probability that $$ X $$ is greater than 17 is:
     $$
     P(X > 17) = 1 - P(X < 17) = 1 - 0.025 = 0.975
     $$

**Final Answer:**  
The probability that $$ X $$ is greater than 17 is **0.975**.
The probability that $$ X $$ is greater than 17 is **0.975**.

is a normally distributed random variable with mean 39 and standard deviation 11 .
What is the probability that is greater than 17 ?
Use the rule and write your answer as a decimal. Round to the nearest
thousandth if necessary.

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is a normally distributed random variable with mean 39 and standard deviation 11 . What is the probability that is greater than 17 ? Use the rule and write your answer as a decimal. Round to the nearest thousandth if necessary.