Question

Suppose a simple random sample of size n = 1000 is obtained from a population whose size is N = 2,000,000 and whose population proportion with a specified characteristic is p = 0.49 . What is the probability of obtaining x = 520 or more individuals with the​ characteristic?

Answer 1

Answer:

$p(x\geq 520) = 0.0293$

Step-by-step explanation:

Given data:

random sample size n = 1000

Population size is N - 2,000,000

P = 0.49

We know

$\sigma_p = \sqrt{\frac{p*(1-p)}{n}}$

$\sigma_ p = \sqrt{\frac{0.49*(1-0.49)}{1000}} = 0.0158$

Probability for having X =520

sample proportion $\hat p = \frac{520}{1000} = 0.52$

$p(x\geq 520) = P(\hat p\geq)$

                       $= P(Z\geq \frac{0.52 - 0.49}{0.0158})$

                       $= P(Z\geq 1.89) = 0.0293$

$p(x\geq 520) = 0.0293$

Answer 2

Answer:

P(X≥520) =0.02938

Step-by-step explanation:

given,

n = 1000

Population size = N = 2,000,000

Specified characteristic = P = 0.49

Probability of obtaining x = 520

$\sigma_p = \sqrt{\dfrac{p(1-p)}{1000}}$

$\sigma_p = \sqrt{\dfrac{0.49(1-0.49)}{1000}}$

$\sigma_p =0.0158$

for x = 520

p = 520/1000 = 0.52

P(X≥520) = P(p≥0.52)

P(p≥0.52) = $P(Z\geq \dfrac{0.52-0.49}{0.0158})$

P(p≥0.52) = $P(Z\geq 1.898)$

using z-table

P(p≥0.52) = 0.02938

P(X≥520) =0.02938