The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips. A bag that contains 1425 chocolate chips is in the $\square$ st percentile. (Round to the nearest integer as needed.)

Step 1 ：The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and a standard deviation of 129 chips.

Step 2 ：We are asked to find the percentile of a bag that contains 1425 chocolate chips.

Step 3 ：The percentile of a value in a distribution is the percentage of values in the distribution that are less than or equal to that value.

Step 4 ：To find the percentile, we can use the z-score formula, which is \((X - μ) / σ\), where X is the value we're interested in, μ is the mean, and σ is the standard deviation. The z-score tells us how many standard deviations away from the mean our value is.

Step 5 ：Substituting the given values into the z-score formula, we get \((1425 - 1252) / 129 = 1.3410852713178294\).

Step 6 ：Once we have the z-score, we can use a z-table or a statistical function to find the percentile. The z-score of 1.3410852713178294 corresponds to the 91st percentile.

Step 7 ：Final Answer: A bag that contains 1425 chocolate chips is in the \(\boxed{91}\)st percentile.