To find the probability that a randomly selected student scores more than 90 on the exam, we can use the properties of the normal distribution.
Given:
- Mean (μ) = 82.0
- Standard deviation (σ) = 15.0
- Score (X) = 90
We can standardize the score to find the corresponding z-score using the formula:
\[ z = \frac{X - \mu}{\sigma} \]
Substituting the values:
\[ z = \frac{90 - 82}{15} = \frac{8}{15} \approx 0.5333 \]
Next, we will look up the z-score of approximately 0.53 in the standard normal distribution table, or we can use a calculator. The z-table tells us the probability that a value is less than a certain z-score.
For \( z = 0.53 \), the cumulative probability (P(Z < 0.53)) is approximately 0.7027.
To find the probability of a score greater than 90, we calculate:
\[ P(X > 90) = 1 - P(Z < 0.53) = 1 - 0.7027 \approx 0.2973 \]
Rounding this result, we find:
\[ P(X > 90) \approx 0.296 \]
Therefore, the correct answer is:
A) 0.296