https://virtualback... ConnectED Stude... Course-Geometr... 1. Academic Calendar Avancemos Online... Mail - Michael Smi... httpsy/lifeushar Chattahoochee $干$ Tech MATH1127: Introduction to Statistics (60022) Lesson 7.6 The Central Limit Theorem for Proportions Unit 3 Chapter 7: Lesson 7.6 Assignment In an certain species of newt, offspring are born either green or black. Suppose that 35\% of these newts are born green. If we sample 213 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. Question Help: D Post to forum Submit Question IMG_2842.jpg IMG_2841.jpg IMG_2840.jpg IMG_0891.jpg EE43F43E-91AF_..jpeg

Step 1 ：Given that the proportion of green newts, p, is 0.35 and the sample size, n, is 213.

Step 2 ：We are asked to find the mean and standard deviation of the proportion of green newts in the sample.

Step 3 ：The mean of a proportion is simply the proportion itself, so the mean is \(\boxed{0.35}\).

Step 4 ：The standard deviation of a proportion can be calculated using the formula \(\sqrt{p(1-p)/n}\), where p is the proportion and n is the sample size.

Step 5 ：Substituting the given values into the formula, we get the standard deviation as \(\sqrt{0.35(1-0.35)/213} = 0.032681418533639144\).

Step 6 ：The question also asks for two standard deviations below and above the mean. This can be calculated as mean - 2*std_dev and mean + 2*std_dev respectively.

Step 7 ：Calculating these values, we get two standard deviations below the mean as \(0.35 - 2*0.032681418533639144 = \boxed{0.28}\) and two standard deviations above the mean as \(0.35 + 2*0.032681418533639144 = \boxed{0.42}\).