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. The proportion of bags that contains more than 1225 chocolate chips is (Round to four decimal places as needed.)

Step 1 ：The problem is asking for the proportion of bags that contain more than 1225 chocolate chips. This is a problem of normal distribution.

Step 2 ：We need to find the z-score of 1225. The z-score is calculated as \((X - μ) / σ\), where X is the value we're interested in, μ is the mean, and σ is the standard deviation.

Step 3 ：Given that the mean (μ) is 1252, the standard deviation (σ) is 129, and X is 1225, we can substitute these values into the z-score formula to get \(z = (1225 - 1252) / 129\).

Step 4 ：Calculating the above expression, we get \(z = -0.20930232558139536\).

Step 5 ：After finding the z-score, we can use the standard normal distribution table or a function to find the proportion of bags that have a greater z-score.

Step 6 ：The proportion of bags that have a z-score greater than -0.20930232558139536 is approximately 0.582893881783751.

Step 7 ：This means that about 58.29% of the bags contain more than 1225 chocolate chips.

Step 8 ：Final Answer: The proportion of bags that contain more than 1225 chocolate chips is approximately \(\boxed{0.5829}\).