Question

The proportion of individuals insured by the All-Driver Automobile Insurance Company who received at least one traffic ticket during a five-year period is .15.

a) Show the sampling distribution of p if a random sample of 150 insured individuals is used to estimate the proportion having received at least one ticket.

b) What is the probability that the sample proportion will be within +-.03 of the population proportion?

Answer 1

a)

sampling distribution of p = 0.15

std.deviation = sqrt( p *(1-p)/n)
= sqrt(0. 15 *(1-0.15)/150)
= 0.0292

b)

Here, μ = 0.15, σ = 0.0292, x1 = 0.12 and x2 = 0.18. We need to compute P(0.12<= X <= 0.18). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.12 - 0.15)/0.0292 = -1.03
z2 = (0.18 - 0.15)/0.0292 = 1.03

Therefore, we get
P(0.12 <= X <= 0.18) = P((0.18 - 0.15)/0.0292) <= z <= (0.18 - 0.15)/0.0292)
= P(-1.03 <= z <= 1.03) = P(z <= 1.03) - P(z <= -1.03)
= 0.8485 - 0.1515
= 0.6970