Question

A recent study reported that 25% of the residents of a particular community lived in poverty. Suppose a random sample of 300 residents of this community is taken. We wish to determine the probability that 29% or mon in poverty. Complete parts (a) and (b) below. a. Before doing any calculations, determine whether this probability is greater than 50% or less than 50%. Why? O A. The answer should be less than 50%, because the resulting Z-score will be negative and the sampling distribution is approximately Normal, OB. The answer should be less than 50%, because 0.29 is greater than the population proportion of 0.25 and because the sampling distribution is approximately Normal. OC. The answer should be greater than 50%, because the resulting Z-score will be positive and the sampling distribution is approximately Normal. OD. The answer should be greater than 50%, because 0.29 is greater than the population proportion of 0.25 and because the sampling distribution is approximately Normal. b. Calculate the probability that 29% or more of the sample will be living in poverty. Assume the sample is collected in such a way that the conditions for using the CLT are met. [email protected]) - (Round to three decimal places as needed.)

Answer 1

Given, population proportion p = 0.25 CLT proof m.p=10 → (300):(0.25) 210 = 75310 n11-P) 210 → (300) (1-0.29) >10. → 225 210. SO, CLT conditions are met Given probability 29% ie 0.29 is greater than the population proportion of 0.25. so, The answer should be less than 50%, because 0- 29 is greater than 0.25 and because sample dia tribution in approximately normal 0. 29 -0.25 * PLP20,29). pl 20:29) pose poate = P(Z > A) » P(2004) .25) 0.04 0.1875 300 0.025 3) Plz>1.6) → 1-P(Z 51.6) PL €20.29) = [0.055 9 1 – 0.9452 = 0.0548 70.0SS. 0.055.