Step-by-step Solution:

Correct Answer: B (0.191)

This problem requires us to find a probability for a normally distributed variable. The standard method is to convert the given scores into Z-scores and then use the properties of the standard normal distribution.

Step 1: Standardize the Scores (Convert to Z-scores)

First, we convert the score range [450, 500] into a Z-score range. The formula to convert a score X from a normal distribution with mean $\mu$ and standard deviation $\sigma$ to a standard normal score Z is:

$$ Z = \frac{X - \mu}{\sigma} $$

Given values are $\mu = 500$ and $\sigma = 100$.

• For a score of 450:
  $Z_1 = \frac{450 - 500}{100} = \frac{-50}{100} = -0.5$
• For a score of 500:
  $Z_2 = \frac{500 - 500}{100} = \frac{0}{100} = 0$

So, finding the probability of a student scoring between 450 and 500, $P(450 \le X \le 500)$, is equivalent to finding the probability $P(-0.5 \le Z \le 0)$.

Step 2: Use Properties of the Standard Normal Distribution

The probability $P(-0.5 \le Z \le 0)$ represents the area under the standard normal curve between $Z=-0.5$ and $Z=0$.

Due to the symmetry of the normal distribution curve about the mean (Z=0), this area is equal to the area between $Z=0$ and $Z=0.5$.

$$ P(-0.5 \le Z \le 0) = P(0 \le Z \le 0.5) $$

We can find this area using the given cumulative distribution information. The cumulative distribution gives the area to the left of a Z-score.

$$ P(0 \le Z \le 0.5) = P(Z \le 0.5) - P(Z < 0) $$

Step 3: Calculate the Final Probability

We are given the following values:

• $P(Z \le 0.5) = 0.691$ (from the problem statement).
• $P(Z < 0) = 0.5$ (since the mean of the standard normal distribution is 0, exactly half the area is to its left).

Now, we can calculate the probability:

$$ P(-0.5 \le Z \le 0) = P(Z \le 0.5) - P(Z < 0) = 0.691 - 0.5 = 0.191 $$

The probability that a randomly selected student scored between 450 and 500 is 0.191.