A random sample of the birth weights of 186 babies has a mean of 3103 grams and a standard deviation of 686 grams. Assume the distribution of birth weights is normally distributed. Construct a 95\% confidence interval of the mean birth weight for all such babies.

Step 1 ：We are given a random sample of the birth weights of 186 babies with a mean of 3103 grams and a standard deviation of 686 grams. We are asked to construct a 95% confidence interval of the mean birth weight for all such babies.

Step 2 ：To construct a 95% confidence interval for the mean birth weight, we need to use the formula for a confidence interval, which is: \[\bar{x} \pm z \frac{s}{\sqrt{n}}\] where: \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：For a 95% confidence interval, the z-score is approximately 1.96.

Step 4 ：We have all the values we need to plug into this formula: \(\bar{x} = 3103\) grams, \(s = 686\) grams, and \(n = 186\).

Step 5 ：Substituting these values into the formula, we get the margin of error as approximately 98.59 grams.

Step 6 ：We then subtract this margin of error from the mean to get the lower bound of the confidence interval, and add it to the mean to get the upper bound. This gives us a lower bound of approximately 3004.41 grams and an upper bound of approximately 3201.59 grams.

Step 7 ：Final Answer: The 95% confidence interval for the mean birth weight of all such babies is approximately \(\boxed{(3004.41, 3201.59)}\) grams.