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 probability that a randomly selected bag contains fewer than 1000 chocolate chips is
(Round to four decinal places as needed.)

Step 1 ：We are given a problem of normal distribution with a mean (\(\mu\)) of 1252 and a standard deviation (\(\sigma\)) of 129. We are asked to find the probability that a randomly selected bag contains fewer than 1000 chocolate chips. This is equivalent to finding the cumulative probability P(X < 1000).

Step 2 ：To solve this, we need to standardize the value 1000 using the Z-score formula: \(Z = \frac{X - \mu}{\sigma}\), where X is the value we are interested in (1000 in this case), \(\mu\) is the mean, and \(\sigma\) is the standard deviation. The Z-score gives us the number of standard deviations the value X is away from the mean.

Step 3 ：Substituting the given values into the Z-score formula, we get a Z-score of approximately -1.95. This indicates that the value of 1000 is approximately 1.95 standard deviations below the mean.

Step 4 ：Once we have the Z-score, we can use a Z-table or a statistical function to find the cumulative probability associated with that Z-score. The cumulative probability is approximately 0.0254. This indicates that there is a 2.54% chance that a randomly selected bag contains fewer than 1000 chocolate chips.

Step 5 ：Final Answer: The probability that a randomly selected bag contains fewer than 1000 chocolate chips is approximately \(\boxed{0.0254}\) or \(\boxed{2.54\%}\).