Question

13) the true proportion of people in Texas that use a courier service is .04. if 205 people are sampled, what is the probability that the sample portion will differ from the population portion by less than .03?

Answer 1

Given: 0.04,q 1-p 0.96, n 205 Taking normal approximation to binomial, the sampling distribution of sample pro- portion follows normal distribution Shape: Normal p ~ N (μ, σ2), where mean, 14p-p-0.04 Mp = 0.04 variance, σ2- standard deviation, ơn-v02- px (1-p 0.04 x096.0001873 205 0.04 x 0.96 0.013686 p0.013686 pN (0.04,0.0136862) P(l 0.040.03) Normal Distribution, μ = 0.04, σ = 0.013686, P( = P (-(0.03)-(0.04) <p < (0.03) + (0.04)) we convert this to standard normal using z- 0.04| < 0.03) = P (-(0.03) < þ 0.01 (0.04) 0.013686 ะ-2.192021 ~-2.19 007- (0 04

22 ·013686 ะ 2.192021 ะ 2.19 P-2.19< Z <2.19) Area in between -2.19 and 2.19 0.9857 -0.0143 0.9714 0.0143 .9857 2.19 -2.19 Plp_ 0.04| < 0.03)-P(-2.19 < Z < 2.19) PZ2.19)- P(Z<-2.19) 0.9857 -0.014:3 0.9714 P(Z<2.19): in a z-table having area to the left of z, locate 2.1 in the left most column. Move across the row to the right under column 0.09 and get value 0.9857 P(Z <-2.19): in a z-table having area to the left of z, locate-21 in the left most column. Move across the row to the right under column 0.09 and get value 0.0143 Plpー0.04! < 0.03) 0.9714 97.14% Note that exact answer using technology is: 0.971617586170498