Question

The random sample of 400 people is selected and the population proportion is .29.

a) Describe the sampling distribution of the sample proportion (name of of distribution such as mu or beta)

b) Find the probability of obtaining the sample proportion is greater than .33

c) Find the sample proportion that has the top 7%

Answer 1

a)

mean = pcap = 0.29

std.dev = sqrt(pcap *(1-pcap)/n)
= sqrt(0.29 *(1-0.29)/400)
= 0.0227

b)

Here, μ = 0.29, σ = 0.0227 and x = 0.33. We need to compute P(X >= 0.33). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.33 - 0.29)/0.0227 = 1.76

Therefore,
P(X >= 0.33) = P(z <= (0.33 - 0.29)/0.0227)
= P(z >= 1.76)
= 1 - 0.9608 = 0.0392

c)

z value at top7% = 1.48

z = (x - mean)/s

1.48 = (x - 0.29)/0.0227

x = 0.0227 * 1.48+ 0.29

x = 0.32