Scores on a certain standardized exam follow a $N(\mu=125, \sigma=15)$ distribution. Brittany scored a 149 on the exam.

What percentage of test takers did Brittany score better than? Round to the nearest percent, but do not include a percent sign in your answer.

Step 1 ：First, we need to standardize Brittany's score. This is done by subtracting the mean from her score and then dividing by the standard deviation. This will give us a z-score, which tells us how many standard deviations Brittany's score is from the mean. The formula for calculating the z-score is \(Z = \frac{X - \mu}{\sigma}\) where \(X\) is Brittany's score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 2 ：Substitute the given values into the formula: \(X = 149\), \(\mu = 125\), and \(\sigma = 15\). So, \(Z = \frac{149 - 125}{15} = \frac{24}{15} = 1.6\).

Step 3 ：Now, we need to find the percentage of test takers that Brittany scored better than. This is done by looking up the z-score in a standard normal distribution table, which gives us the cumulative probability up to that z-score.

Step 4 ：Looking up a z-score of 1.6 in the standard normal distribution table gives us a cumulative probability of approximately 0.9452. This means that Brittany scored better than approximately 94.52% of test takers.

Step 5 ：Rounding to the nearest percent, we get \(\boxed{95}\). So, Brittany scored better than approximately 95% of test takers.