In an certain species of newt, offspring are born either green or black. Suppose that $45 \%$ of these newts are born green. If we sample 134 of these newts at random, the probability distribution for the proportion of green newts in the sample can be modeled by the normal distibution pictured below. Complete the boxes accurate to two decimal places.

Note: The left box is 2 standard deviations below the mean. The middle box is the mean. And, the right box is 2 standard deviations above the mean.

Step 1 ：Given that the proportion of green newts, denoted as p, is 0.45 and the sample size, denoted as n, is 134.

Step 2 ：The mean of the normal distribution is the expected proportion of green newts, which is 0.45.

Step 3 ：The standard deviation can be calculated using the formula for the standard deviation of a sample proportion, which is \(\sqrt{p(1-p)/n}\), where p is the proportion of green newts and n is the sample size. Substituting the given values, we get the standard deviation as approximately 0.043.

Step 4 ：The left box is 2 standard deviations below the mean, the middle box is the mean, and the right box is 2 standard deviations above the mean.

Step 5 ：Calculating these values, we get the left box as approximately 0.36, the middle box as 0.45, and the right box as approximately 0.54.

Step 6 ：Final Answer: The values for the boxes are approximately \(\boxed{0.36}\), \(\boxed{0.45}\), and \(\boxed{0.54}\), respectively.