Compute the probability that an observation 𝑦 drawn from a normal distribution with mean 𝜇=51.41and standard deviation 𝜎=8.11 is in the interval (38.12,62.15), i.e., Prob(38.12<𝑦<62.15).

To compute the probability that an observation \( y \) drawn from a normal distribution with mean \( \mu = 51.41 \) and standard deviation \( \sigma = 8.11 \) lies in the interval \( (38.12, 62.15) \), we'll follow the same steps as before.

### Steps:

1. **Standardize the bounds**: Convert the values 38.12 and 62.15 to their corresponding z-scores.
   \[
   z_1 = \frac{38.12 - \mu}{\sigma} = \frac{38.12 - 51.41}{8.11}
   \]
   \[
   z_2 = \frac{62.15 - \mu}{\sigma} = \frac{62.15 - 51.41}{8.11}
   \]

2. **Find the CDF for the z-scores**: Use the standard normal distribution table (or a calculator) to find the probabilities corresponding to these z-scores.

3. **Compute the desired probability**:
   \[
   P(38.12 < y < 62.15) = P(z_1 < Z < z_2) = \Phi(z_2) - \Phi(z_1)
   \]
   where \( \Phi(z) \) is the CDF of the standard normal distribution.

Let's calculate this step by step.

### Step 1: Standardize the bounds
\[
z_1 = \frac{38.12 - 51.41}{8.11} \approx \frac{-13.29}{8.11} \approx -1.639
\]
\[
z_2 = \frac{62.15 - 51.41}{8.11} \approx \frac{10.74}{8.11} \approx 1.324
\]

### Step 2: Find the CDF values
Using a standard normal distribution table or calculator:

\[
\Phi(-1.639) \approx 0.0505
\]
\[
\Phi(1.324) \approx 0.9076
\]

### Step 3: Compute the probability
\[
P(38.12 < y < 62.15) = \Phi(1.324) - \Phi(-1.639) \approx 0.9076 - 0.0505 = 0.8571
\]

### Final Answer
The probability that an observation \( y \) lies in the interval \( (38.12, 62.15) \) is approximately **0.8571**.

Would you like more details on any part of this process or have any questions?

---

Here are five related questions you might explore:
1. How would you calculate the probability for a one-sided interval, such as \( P(y > 38.12) \)?
2. What is the impact of increasing or decreasing the standard deviation on the probability?
3. How would the probability change if the mean were different?
4. How does the Central Limit Theorem relate to normal distributions?
5. What are some examples of phenomena that follow a normal distribution in nature?

**Tip:** Always remember that the area under the entire standard normal curve is 1, representing the total probability.

Learn how to compute the probability that an observation drawn from a normal distribution with mean 51.41 and standard deviation 8.11 lies in the interval (38.12, 62.15). Detailed step-by-step solution using standardization, CDF values, and the normal distribution properties.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation