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.

We are given that the random variable \( X \) is normally distributed with \( \mu = 75 \) and \( \sigma = 5 \), meaning:

\[
X \sim N(75, 5^2)
\]

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

We first standardize the value of 85 by converting it to the standard normal form \( Z \), where:
\[
Z = \frac{X - \mu}{\sigma}
\]

So, for \( X = 85 \):
\[
Z = \frac{85 - 75}{5} = \frac{10}{5} = 2
\]

Now, we need to find \( P(X > 85) = P(Z > 2) \).

Using standard normal tables or a calculator:
\[
P(Z > 2) = 1 - P(Z \leq 2) \approx 1 - 0.9772 = 0.0228
\]

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

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

We need to find the value of \( k \) such that \( P(X < k) = 0.997 \). This means finding the corresponding \( Z \)-value for which the cumulative probability is 0.997.

Using standard normal tables, we find that:
\[
P(Z < 2.747) \approx 0.997
\]

Now, convert the \( Z \)-value back to \( X \) using:
\[
k = Z \cdot \sigma + \mu
\]

So:
\[
k = 2.747 \cdot 5 + 75 = 13.735 + 75 = 88.735
\]

Thus, the value of \( k \) is approximately 88.74. This means that 99.7% of the distribution of \( X \) lies below 88.74.

### 3. **Find the two values where the middle 68.27% of the distribution lies**

The middle 68.27% of the distribution corresponds to one standard deviation above and below the mean in a normal distribution.

For the standard normal distribution, this means:
\[
\mu - \sigma \quad \text{to} \quad \mu + \sigma
\]

Thus:
\[
75 - 5 = 70 \quad \text{and} \quad 75 + 5 = 80
\]

So, the middle 68.27% of the distribution lies between **70 and 80**.

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

Would you like more details or clarification on any part of this? 

Here are some related questions to expand on this topic:

1. How do you compute probabilities for other values of \( X \)?
2. What is the 95th percentile for this distribution?
3. How do changes in \( \mu \) and \( \sigma \) affect the distribution?
4. How can you interpret the result for \( P(X > 85) \) in practical terms?
5. What is the probability that \( X \) is between 80 and 85?

**Tip:** When dealing with normal distributions, always standardize using the formula \( Z = \frac{X - \mu}{\sigma} \) to simplify the calculations!

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

Find P(X > 85).

Find k such that P(X < k) = 0.997. Interpret your result.

Find the two values where the middle 68.27% of the distribution of X lie.

Learn how to solve a normal distribution problem with µ = 75 and σ = 5. Find P(X > 85), calculate the value k such that P(X < k) = 0.997, and determine the two values where the middle 68.27% of the distribution lies. Explore detailed steps using the standard normal distribution and the empirical rule.

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