Question

1. The 2010 census found that 13.9% of adults in the us identified as Hispanic. A poll plans to interview 1500 adults at random, Cateutate mean and standard deviation of sampling distributien of Hispanics for samples of this sizes Select best answer below: x = 61.00293 u x = .139 0 6 x = .00893 5.139 & 300893 MA=.865 & 5o =, 00893 M 2. A charity receives 20a5 contributions Assume them to be independent and identically distributed with mean2125 Standard deviation 250. Calculate 90th percentile for si distribution of total contributions received

Answer 1

Question 1: According to the provided data, p=0.139 n = 1500 The mean and standard deviation of this sampling distribution is, My = p=0.139 p(1-P) 10.139(1-0.139 = 0.00893 1500 12

Question 2: Consider X be a random variable which follows normal distribution indicates the number of contributions. The mean is 3125 with a standard deviation of 250. The sample size, n is 2025. The mean and standard deviation for the sum of X is, Mgr = nji = 2025(3125) = 6328125 Ir = Vo = 2025 (250) = 11250 The required 90th percentile is. PEX sa)= 0.90 a-6328125 = 0.90 11250 a-6328125 0(0.90) (Using 2-table) 11250 a-6328125 =1.28155 11250 7 = 6328125+(1.28155x11250) 6342542