Question

A population proportion is 0.2. A sample of size 150 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 ±0.04 of the population proportion?

b. What is the probability that the sample proportion will be within ±0.07 of the population proportion?

Answer 1

a)

Here, μ = 0.2, σ = 0.0327, x1 = 0.16 and x2 = 0.24. We need to compute P(0.16<= X <= 0.24). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.16 - 0.2)/0.0327 = -1.22
z2 = (0.24 - 0.2)/0.0327 = 1.22

Therefore, we get
P(0.16 <= X <= 0.24) = P((0.24 - 0.2)/0.0327) <= z <= (0.24 - 0.2)/0.0327)
= P(-1.22 <= z <= 1.22) = P(z <= 1.22) - P(z <= -1.22)
= 0.8888 - 0.1112
= 0.7776

b)

Here, μ = 0.2, σ = 0.0327, x1 = 0.13 and x2 = 0.27. We need to compute P(0.13<= X <= 0.27). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.13 - 0.2)/0.0327 = -2.14
z2 = (0.27 - 0.2)/0.0327 = 2.14

Therefore, we get
P(0.13 <= X <= 0.27) = P((0.27 - 0.2)/0.0327) <= z <= (0.27 - 0.2)/0.0327)
= P(-2.14 <= z <= 2.14) = P(z <= 2.14) - P(z <= -2.14)
= 0.9838 - 0.0162
= 0.9676