Question

i do not understand why b and c are wrong can someone help me please...

Recycling trash, reducing waste, and reusing materials are eco-actions that will help the environment. According to a USA Today snapshot, 78% of respondents list recycling as the leading way to help our environment. Suppose that a random sample of n = 100 adults is selected and that the 78% figure is correct. Top Eco-acti. Talente Help the Environment Survey respondents say recycling trash, reducing waste, and reusing materials make the biggest difference in protecting the environment. Recycle Rene Materials Race Waste Purchase Organic Product Denate to avog 62 27% (a) Does the distribution of p, the sample proportion of adults who list recycling as the leading way to help the environment have an approximate normal distribution? If so, what is its mean and standard deviation? O Yes, the distribution is approximately normal. The mean is 78 and the standard deviation is 22. o Yes, the distribution is approximately normal. The mean is 0.78 and the standard deviation is 0.0414. Yes, the distribution is approximately normal. The mean is 0.78 and the standard deviation is 0.22. No, the distribution is not normal because np is less than 5. No, the distribution is not normal because ng is less than 5. (b) What is the probability that the sample proportion p is less than 72%? (Round your answer to four decimal places.) 0.0735x (c) What is the probability that p lies in the interval 0.66 to 0.72 not including the endpoints? (Round your answer to four decimal places.) 0.0716x

Answer 1

Date: 11/11/2019 Answer b) The probability is that the sample proportion less than 72% is, First, compute the mean and standard deviation of sampling distribution of sample proportion. The mean of the sampling distribution of sample proportion is, 0.78 The standard deviation of the sampling distribution of sample proportion is, |P(1- p) (0.78)(1-0.78) 100 -0.0414250.0414

The probability is that the sample proportion less than 72% is, First, compute the z score then find probability based on standard normal table. For p 0.72 converts to р-П, бр 0.72 0.78 0.041425 =-1.448 From the standard normal distribution table, the associated probability for the area to the left is shown below P(z<-1.448) 0.0738 The probability is that the sample proportion less than 72% is 0.0738|

c) The probability is that the sample proportion between 0.66 and 0.72 is, First, compute the z score then find probability based on standard normal table For p 0.66 converts to P-PP 0.66-0.78 0.041425 =-2.897 For p 0.72 converts to 0.72 0.78 0.041425 =-1.448

From the standard normal distribution table, the associated probability for z values and subtracts the probability is, p(0.66 ps 0.72) p-2.897 zs-1.448) = p(z -1.448)- p(zs-2.897) =0.0738- 0.0019 0.0719 The probability that the sample proportion is between 0.66 and 0.72 is 0.0719