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From the given data Da 0.17 N=500 a) The sampling distribution of the sample mean Ŕ has mean p= 0.17 Standard deviatich - PC1-P) = 0.17 C1-0-17) 500 => 0.016798809 a 0.0168 (-4-decimals) b) Probability that the sample proportion will be within £0:02 of population proportion -PCL-P) 20-02) +0:02 => I 1.19 - 017(1-0.17) Vornasco V pete = P(-1-19 <2<119) = 2PCZL-119) CScanned wiñrcadošian 2. (0-1170] => 0.766
c) sample of 1000 - = P(IP-Pk 0:02) ze P- - 10.02 5 7 1-68 0-17 (10:17) > V 1000 PCI-P) vn 0-1710-17) = P(-1-68 <24168) -1-2PC2<1:68) - 1-2 (0-0465) = 0.907. CScanned with CamScanner
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