Step 1 :The problem is asking for the proportion of bags that contains more than 1200 chocolate chips. This is a problem of finding the probability of a value in a normal distribution.
Step 2 :The formula for finding the z-score of a value in a normal distribution is \( (X - \mu) / \sigma \), where X is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 :Once we have the z-score, we can find the probability associated with it using a z-table or a function that gives the cumulative distribution function of the normal distribution.
Step 4 :However, since we want the proportion of bags that contains more than 1200 chocolate chips, we need to subtract the cumulative probability from 1.
Step 5 :Given that the mean is 1252, the standard deviation is 18, and the value is 1200, we can calculate the z-score as \( -2.888888888888889 \).
Step 6 :Using this z-score, we can find the cumulative probability as \( 0.9980669717300181 \).
Step 7 :Finally, subtracting this cumulative probability from 1 gives us the proportion of bags that contains more than 1200 chocolate chips.
Step 8 :The final answer is \(\boxed{0.9981}\).