Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e., do not use percentages. Give your final answer with at least  3
  dp.

Compute the probability that an observation  𝑦
  drawn from a normal distribution with mean  𝜇=44.44
  and standard deviation  𝜎=7.65
  is in the interval  (35.82,49.94)
 , i.e.,  Prob(35.82<𝑦<49.94)
 .

To compute the probability that an observation \( y \) drawn from a normal distribution with mean \( \mu = 44.44 \) and standard deviation \( \sigma = 7.65 \) lies within the interval \( (35.82, 49.94) \), we follow these steps:

### Step 1: Standardize the values to the Z-distribution
We convert the boundaries of the interval to their corresponding Z-scores using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where \( X \) is the boundary value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

#### For the lower bound \( X = 35.82 \):
\[
Z_1 = \frac{35.82 - 44.44}{7.65} = \frac{-8.62}{7.65} \approx -1.1275
\]

#### For the upper bound \( X = 49.94 \):
\[
Z_2 = \frac{49.94 - 44.44}{7.65} = \frac{5.50}{7.65} \approx 0.7183
\]

### Step 2: Find the cumulative probabilities
We now find the cumulative probability corresponding to these Z-scores using the standard normal distribution table or a calculator.

- \( P(Z \leq -1.1275) \approx 0.1296 \)
- \( P(Z \leq 0.7183) \approx 0.7634 \)

### Step 3: Calculate the probability that \( y \) lies within the interval
The probability that \( y \) is between 35.82 and 49.94 is given by:
\[
P(35.82 < y < 49.94) = P(Z \leq 0.7183) - P(Z \leq -1.1275)
\]
\[
P(35.82 < y < 49.94) = 0.7634 - 0.1296 = 0.6338
\]

Thus, the probability that an observation \( y \) lies within the interval \( (35.82, 49.94) \) is **0.634** (to three decimal places).

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you interpret the Z-score in a normal distribution?
2. What is the significance of the cumulative distribution function in probability?
3. Can you find the probability for a different interval, such as \( (40, 50) \), using the same normal distribution?
4. How would the probability change if the standard deviation were smaller?
5. What is the probability that \( y \) is greater than 50 in this distribution?

**Tip:** When solving probability problems involving the normal distribution, always start by standardizing the variable to the Z-distribution to simplify your calculations.

Learn how to compute the probability that an observation drawn from a normal distribution with mean \( \mu = 44.44 \) and standard deviation \( \sigma = 7.65 \) lies within the interval \( (35.82, 49.94) \). Detailed step-by-step solution with calculations and interpretation.

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