We are given a normally distributed random variable X with µ = 75 and σ = 5.

Find P (X > 85).
Find k such as P(X<k) = 0.997. Interpret your result.
Find the two values where the middle 68.27% of the distribution of X lie.

Let's break down and solve each part of the problem step by step.

### Given Information:
- \( \mu = 75 \) (mean)
- \( \sigma = 5 \) (standard deviation)
- The random variable \( X \) follows a normal distribution, i.e., \( X \sim N(\mu = 75, \sigma = 5) \).

### 1. **Find \( P(X > 85) \)**:

We are looking for the probability that \( X \) is greater than 85.

First, we convert this to a standard normal variable \( Z \) using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
For \( X = 85 \):
\[
Z = \frac{85 - 75}{5} = 2
\]

Now, \( P(X > 85) = P(Z > 2) \).

Using standard normal distribution tables (or a calculator), we know that:
\[
P(Z > 2) \approx 0.0228
\]

So:
\[
P(X > 85) \approx 0.0228
\]

### 2. **Find \( k \) such that \( P(X < k) = 0.997 \)**:

This means we need to find the value of \( k \) where 99.7% of the distribution lies to the left of \( k \). This corresponds to finding the 99.7th percentile of the normal distribution.

From the standard normal distribution tables or a calculator, the Z-value corresponding to a cumulative probability of 0.997 is approximately 2.747.

Now, convert this Z-value back to the original \( X \) distribution:
\[
k = \mu + Z \cdot \sigma = 75 + 2.747 \cdot 5 \approx 75 + 13.735 = 88.735
\]

Thus, \( k \approx 88.74 \).

**Interpretation**: There is a 99.7% probability that the value of \( X \) is less than or equal to approximately 88.74.

### 3. **Find the two values where the middle 68.27% of the distribution of \( X \) lie**:

The middle 68.27% of the distribution corresponds to 1 standard deviation on either side of the mean in a normal distribution (since 68.27% is the empirical rule for \( 1 \sigma \)).

So, we calculate the range of values that are within 1 standard deviation from the mean:
\[
\mu - \sigma = 75 - 5 = 70
\]
\[
\mu + \sigma = 75 + 5 = 80
\]

Therefore, the two values are 70 and 80.

**Interpretation**: 68.27% of the values of \( X \) will lie between 70 and 80.

### Summary of Results:
1. \( P(X > 85) \approx 0.0228 \)
2. \( P(X < k) = 0.997 \) gives \( k \approx 88.74 \)
3. The middle 68.27% of the distribution lies between 70 and 80.

Would you like any further details or have questions on these calculations?

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Here are 5 relative questions to expand the understanding:

1. How do we calculate probabilities for values other than 85 in a normal distribution?
2. How does changing \( \mu \) or \( \sigma \) affect the distribution and probabilities?
3. What is the interpretation of the 95% confidence interval in a normal distribution?
4. How would the probability change if we wanted \( P(X < 65) \)?
5. What is the probability that \( X \) lies between 70 and 80?

**Tip**: In normal distributions, about 68%, 95%, and 99.7% of values lie within 1, 2, and 3 standard deviations from the mean, respectively. This is known as the empirical rule.

We are given a normally distributed random variable X with µ = 75 and σ = 5. Find P (X > 85). Find k such as P(X < k) = 0.997. Interpret your result. Find the two values where the middle 68.27% of the distribution of X lie.

Solve normal distribution problems with mean µ = 75 and standard deviation σ = 5. Learn how to find P(X > 85), determine the 99.7th percentile, and locate the two values where the middle 68.27% of the distribution lies. Detailed explanations provided for high school level (Grades 10-12).

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