Question

The population proportion is 0.50. What is the probability that a sample proportion will be within ±0.04 of the population proportion for each of the following sample sizes? Round your answers to 4 decimal places. Use z-table.

A.) n=100
B.) n= 200
C.) n=500
D.) n=1,000

Answer 1

We know that $p\sim N(P,\frac{P(1-P)}{n})\sim N(0.5,\frac{0.5(1-0.5)}{n})\sim N(0.5,\frac{0.25}{n})$

A)

Required probability = $P(|p-P|<0.04)=P(|Z|<\frac{0.04}{\sqrt{\frac{0.25}{100}}})=P(|Z|<0.8)$

$=P(-0.8<Z<0.8)=P(Z<0.8)-P(Z<-0.8)$

$=2P(Z<0.8)-1=2(0.7881)-1=0.5762$

B)

Required probability = $P(|p-P|<0.04)=P(|Z|<\frac{0.04}{\sqrt{\frac{0.25}{200}}})=P(|Z|<1.13)$

$=P(-1.13<Z<1.13)=P(Z<1.13)-P(Z<-1.13)$

$=2P(Z<1.13)-1=2(0.8708)-1=0.7416$

C)

Required probability = $P(|p-P|<0.04)=P(|Z|<\frac{0.04}{\sqrt{\frac{0.25}{500}}})=P(|Z|<1.79)$

$=P(-1.79<Z<1.79)=P(Z<1.79)-P(Z<-1.79)$

$=2P(Z<1.79)-1=2(0.9633)-1=0.9266$

D)

Required probability = $P(|p-P|<0.04)=P(|Z|<\frac{0.04}{\sqrt{\frac{0.25}{1000}}})=P(|Z|<2.53)$

$=P(-2.53<Z<2.53)=P(Z<2.53)-P(Z<-2.53)$

$=2P(Z<2.53)-1=2(0.9943)-1=0.9886$