To solve this problem, we need to find the probability $P(X > 3)$ where $X \sim N(11, 4)$. The normal distribution $X$ has:

• Mean $\mu = 11$
• Standard deviation $\sigma = \sqrt{4} = 2$

We proceed with the steps:

Step 1: Standardize the variable

Convert $X$ to the standard normal distribution $Z$ using the formula: $Z = \frac{X - \mu}{\sigma}$ For $X = 3$, $Z = \frac{3 - 11}{2} = \frac{-8}{2} = -4$

Step 2: Find $P(Z > -4)$

From the standard normal table or using a calculator, the cumulative probability $P(Z \leq -4)$ is extremely small (essentially 0). Therefore: $P(Z > -4) \approx 1 - P(Z \leq -4) = 1$

Step 3: Interpret $P(X > 3)$

Since $Z > -4$ includes essentially the entire distribution, the result is close to 1.

Thus: $P(X > 3) = 1.0000$

Let me know if you have further questions or would like to expand on related topics!

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5 Related Questions:

1. How do you calculate $P(X < 3)$ for the same distribution?
2. What is the probability $P(9 \leq X \leq 15)$?
3. How does the cumulative distribution function (CDF) help solve these problems?
4. Can you explain the properties of the standard normal distribution?
5. What is the significance of $Z$-scores in normal probability calculations?

Tip:

Always draw a quick sketch of the normal distribution and shade the area of interest to visualize the problem better.