To compute the probability that an observation $y$ drawn from a normal distribution with mean $\mu = 49.87$ and standard deviation $\sigma = 7.16$ falls within the interval $(41.99, 57.75)$, we follow these steps:

Step 1: Compute the z-scores

We need to calculate the z-scores for both bounds of the interval, $y = 41.99$ and $y = 57.75$.

The z-score formula is: $z = \frac{y - \mu}{\sigma}$

For $y = 41.99$:

$z_1 = \frac{41.99 - 49.87}{7.16} = \frac{-7.88}{7.16} \approx -1.1006$

For $y = 57.75$:

$z_2 = \frac{57.75 - 49.87}{7.16} = \frac{7.88}{7.16} \approx 1.1006$

Step 2: Find the probabilities

Now we find the probabilities corresponding to these z-scores using the standard normal distribution table or a calculator.

• The probability corresponding to $z_1 = -1.1006$ is approximately $0.1358$.
• The probability corresponding to $z_2 = 1.1006$ is approximately $0.8642$.

Step 3: Calculate the probability of the interval

The probability that $y$ lies within the interval $(41.99, 57.75)$ is the difference between the two probabilities: $P(41.99 < y < 57.75) = P(z_2) - P(z_1) = 0.8642 - 0.1358 = 0.7284$

Final Answer

The probability that an observation $y$ falls within the interval $(41.99, 57.75)$ is 0.728 (to three decimal places).

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Would you like more details or have any questions? Here are some related questions:

1. What is the probability that $y$ is less than 41.99?
2. How would the probability change if the interval were wider?
3. What happens to the probability if the mean $\mu$ is increased?
4. How would a smaller standard deviation $\sigma$ affect this probability?
5. How can you visualize this probability on a normal distribution curve?

Tip: Understanding the z-score and how it relates to areas under the standard normal curve is key to solving problems involving normal distributions.