Question

A population proportion is 0.70. A sample of size 200 will be taken and the sample proportion p will be used to estimate the population proportion. (Round your answers to four decimal places.)

(a)What is the probability that the sample proportion will be within ±0.03 of the population proportion?

(b)What is the probability that the sample proportion will be within ±0.05 of the population proportion?

Answer 1

a)
Here, μ = 0.7, σ = sqrt(0.7*0.3/200) = 0.0324,
x1 = 0.67 and x2 = 0.73. We need to compute P(0.67<= X <= 0.73). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.67 - 0.7)/0.0324 = -0.93
z2 = (0.73 - 0.7)/0.0324 = 0.93

Therefore, we get
P(0.67 <= X <= 0.73) = P((0.73 - 0.7)/0.0324) <= z <= (0.73 - 0.7)/0.0324)
= P(-0.93 <= z <= 0.93) = P(z <= 0.93) - P(z <= -0.93)
= 0.8238 - 0.1762
= 0.6476

b)
z = (x - μ)/σ
z1 = (0.65 - 0.7)/0.0324 = -1.54
z2 = (0.75 - 0.7)/0.0324 = 1.54

Therefore, we get
P(0.65 <= X <= 0.75) = P((0.75 - 0.7)/0.0324) <= z <= (0.75 - 0.7)/0.0324)
= P(-1.54 <= z <= 1.54) = P(z <= 1.54) - P(z <= -1.54)
= 0.9382 - 0.0618
= 0.8764