To find the probability that a randomly selected test score is more than 90, we can use the properties of the normal distribution. Given that the mean score (\(\mu\)) is 79 and the standard deviation (\(\sigma\)) is 5.5, we can first calculate the z-score for a test score of 90.
The z-score is calculated using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \(X\) is the value we are interested in (90), \(\mu\) is the mean (79), and \(\sigma\) is the standard deviation (5.5).
Substituting the values:
\[ z = \frac{(90 - 79)}{5.5} = \frac{11}{5.5} \approx 2 \]
Next, we need to find the probability corresponding to this z-score. Using a standard normal distribution table or calculator, we can find the probability that \(Z\) is less than 2.
The probability that \(Z \leq 2\) is approximately 0.9772 (or 97.72%). To find the probability that a score is more than 90, we subtract this value from 1:
\[ P(Z > 2) = 1 - P(Z \leq 2) \approx 1 - 0.9772 \approx 0.0228 \]
Converting this probability into a percentage:
\[ P(Z > 2) \approx 0.0228 \times 100 \approx 2.28% \]
Therefore, the probability that a randomly selected test score is more than 90 is approximately 2.28%.
Since this does not match any of the provided response options exactly, the closest one is 5%.