Question

e Question Help Suppose a simple random sample of size n=150 is obtained from a population whose size is N=25,000 and whose population proportion with a specified characteristic is p=0.2 Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). H = 0.2 (Round to one decimal place as needed) Determine the standard deviation of the sampling distribution of p. (Round to six decimal places as needed) (b) What is the probability of obtaining x = 39 or more individuals with the characteristic? That is what is P20.26) PC 20.26) = (Round to four decimal places as needed.) (c) What is the probability of obtaining x = 27 or fower individuals with the characteristic ? That is what is PÔSO.18)? P 0.18) - (Round to four decimal places as needed) CA Click to select your answer(s) This course stato Summer 2020 based on Sullivan Statistics informed Decisions Using Data, be Chp

Answer 1

Sol Given, Sample size, n= 150 Population size, N= 25,000 P = 0.2 Here, no.05 N { 150 < 0.05 * 25,000 > 150 $ 1250 co? np(1-P), 10 {(150) (0.2) (1-0.2) > 10] of sample proportion is So, The sampling distribution approximately normal with, mear, Me=p=0.2 P (CP) Standard deviation, o (0.2) (1-0.2) = Iso =0.0326598 6313 0.0 32 660 Ŝ wN CME = 0.2,56 = 0.03266) Ê - 0.2 follows standard Normal. Then PM Z= 0.032660

x 39 no 150 39 150 0.26 علما Here, we need to find P(Å 20.26) PCÊ,0.26)= (B-H 0.26 -0.2 ) P 7 0.03266. P(z > 1.8371) P(ZC 1.8371) - 0.966902 (from Sländard Normal lable 0.033098 0.033) x=27 n=150 27 150 = 0.18 Here, we need to find P(Ês 0.18) P(Ê $0.18) P(A-ME < 0.18 – 0:2) 0.03266 Plze-0. 6124) 0.270137 (from Standard Normal table] 0.270