To solve for the probability that the random variable $X$ is greater than 3 in absolute value, where $X \sim N(1,2)$, we need to calculate:

$P(|X| > 3) = P(X > 3) + P(X < -3)$

Step 1: Standardize $X$

Since $X \sim N(1, 2)$, the mean is $\mu = 1$, and the variance is $\sigma^2 = 2$ ($\sigma = \sqrt{2}$).

For standardization: $Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$

For $X > 3$: $P(X > 3) = P\left(Z > \frac{3 - 1}{\sqrt{2}}\right) = P\left(Z > \frac{2}{\sqrt{2}}\right) = P(Z > \sqrt{2})$

For $X < -3$: $P(X < -3) = P\left(Z < \frac{-3 - 1}{\sqrt{2}}\right) = P\left(Z < \frac{-4}{\sqrt{2}}\right) = P(Z < -2\sqrt{2})$

Step 2: Use Standard Normal Distribution Table

• For $P(Z > \sqrt{2})$, where $\sqrt{2} \approx 1.414$: $P(Z > \sqrt{2}) = 1 - P(Z \leq \sqrt{2}) = 1 - 0.92135 = 0.07865$

• For $P(Z < -2\sqrt{2})$, where $-2\sqrt{2} \approx -2.828$: $P(Z < -2\sqrt{2}) = P(Z > 2\sqrt{2}) \text{ (symmetry of the normal distribution) }$ From the table: $P(Z > 2\sqrt{2}) \approx 0.00234$

Step 3: Combine Probabilities

$P(|X| > 3) = P(X > 3) + P(X < -3) = 0.07865 + 0.00234 = 0.08099$

Final Answer:

$P(|X| > 3) \approx 0.081$

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Would you like a detailed explanation of standardization or how to use normal distribution tables?

Related Questions:

1. How do you interpret the result of $P(|X| > 3)$ in practical terms?
2. How is the standardization formula derived, and why is it used in normal distributions?
3. What are the properties of a standard normal distribution $Z \sim N(0, 1)$?
4. How do you compute probabilities for a normal distribution without a table?
5. What is the impact of changing the mean or variance on $P(|X| > 3)$?

Tip:

Always sketch the normal curve and shade the relevant areas to better visualize the probabilities you are calculating.