This problem involves a normal distribution, where scores on a standardized exam are normally distributed with a mean $\mu = 521$ and a standard deviation $\sigma = 83$. The goal is to calculate the probability (shaded area) that a score lies between 355 and 687.

Step-by-step solution:

1. Convert the raw scores to z-scores using the formula: $z = \frac{x - \mu}{\sigma}$ where:

   • $x$ is the value from the distribution,
   • $\mu = 521$ is the mean,
   • $\sigma = 83$ is the standard deviation.

2. Calculate the z-score for 355: $z_1 = \frac{355 - 521}{83} = \frac{-166}{83} = -2.00$

3. Calculate the z-score for 687: $z_2 = \frac{687 - 521}{83} = \frac{166}{83} = 2.00$

4. Find the cumulative probability associated with these z-scores:

   • From the standard normal distribution table, the cumulative probability for $z = -2.00$ is approximately $0.0228$.
   • The cumulative probability for $z = 2.00$ is approximately $0.9772$.

5. Calculate the area between the two z-scores: $\text{Area} = P(z_2) - P(z_1) = 0.9772 - 0.0228 = 0.9544$

Thus, the shaded area under the curve, which represents the probability of a score falling between 355 and 687, is 0.95 (rounded to two decimal places).

Final answer:

$0.95$

Do you need any additional details or further explanation?

Here are some relative questions for further understanding:

1. How does changing the standard deviation affect the width of the normal curve?
2. What is the significance of a z-score in a normal distribution?
3. How do you interpret probabilities in the context of a normal distribution?
4. Can you explain why the area under the normal curve totals to 1?
5. How do different values of the mean affect the center of the normal distribution?

Tip: When using a standard normal distribution table, make sure to interpret the values correctly depending on whether the z-score is positive or negative.