Question

Population 35,093, of those a2,429 people are registered to note In May of 2018, 20% of the registered voters in Scotland County voted in the primary election. 6. About how many people voted? (round to the nearest person) 7. Ir 100 people are randomly selected, what is the probability that at least 15% of them voted in the primary? 8. Ir 200 people are randomly selected, what is the probability that more than 25% of them voted in the primary? 9. In the state of North Carolina, there are 6,674,209 registered voters. In the May 2018 Primary election, 957,627 people voted. What is the population proportion of voters? (Round to the nearest percent) 10. If a random sample of 30 people from North Carolina is taken, what is the probability that 16% of them voted? 11. If a random sample of 100 people from North Carolina is taken, what is the probability that more than 25% of them voted? 12. If a random sample of 100 people from North Carolina is taken, what is the probability that the proportion of people in the sample who voted differs from the population proportion by more than 296? 13. If a random sample of 100 people from North Carolina is taken, what is the probability that the proportion of people in the sample who voted differs from the population proportion by less

Answer 1

$6.~No.~of~People~registered~for~vote=22429\\ No.~of~people~voted=0.20*22429=4485.8\approx 4486.\\ 7.~Let~X=no.~of~people~voted~in~a~sample~then\\\\ X\sim Binomial(100,~0.20).\\\\ Define~\hat{p}=sample~proportion~of~people~who~voted~in~the~primary=\frac{X}{100}\\\\ Using~central~limit~theorem,~\hat{p}\overset{a}{\rightarrow}N\left(0.20,~\sqrt{\frac{0.20*0.80}{100}}=0.04 \right ).\\\\ Required~probability=P(\hat{p}\geq 0.15)=1-P(\hat{p}<0.15)\\\\ =1-P(\hat{p}\leq 0.145)~(using~continuity~correction)\\\\ =1-\Phi((0.145-0.20)/0.04)=1-\Phi(-1.375)=\Phi(1.375)=0.9154\\\\ where,~\Phi(x)=cdf~of~standard~normal~distribution.\\\\ 8.~Using~same~steps~as~Question~7,\\\\ \hat{p}\overset{a}{\rightarrow}N\left(0.20,~\sqrt{\frac{0.20*0.80}{200}}=0.0283 \right ).\\ Required~probability=P(\hat{p}> 0.25)=1-P(\hat{p}\leq 0.25)\\\\=1-\Phi((0.25-0.20)/0.0283)=1-\Phi(1.7668)=0.0386.$

$9.~Population~proportion=957627/6674209=0.14\\\\ 10.~Let~Y=no.~of~people~voted~in~a~sample~then\\\\ Y\sim Binomial(30,~0.14).\\\\ P(Y=0.16*30=4.8)=P(4<Y<5)=P\left(4/30<\hat{p}<5/30 \right )\\\\ =P\left(0.1333<\hat{p}< 0.1667\right )~where,~\hat{p}=\frac{Y}{30}\\\\ =\Phi((0.1666-0.14)/0.0634)-\Phi((0.1333-0.14)/0.0634)~(using~continuity~correction)\\\\ =\Phi(0.4196)-\Phi(-0.1057)=\Phi(0.4196)+\Phi(0.1057)-1=0.2047\\\\ where~using~central~limit~theorem,~\hat{p}\overset{a}{\rightarrow}N\left(0.14,~0.0634 \right )$