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 :

Answer:

[tex]p(x\geq 520)  = 0.0293[/tex]

Step-by-step explanation:

Given data:

random sample size n = 1000

Population size is N - 2,000,000

P = 0.49

We know

[tex]\sigma_p = \sqrt{\frac{p*(1-p)}{n}}[/tex]

[tex]\sigma_ p = \sqrt{\frac{0.49*(1-0.49)}{1000}} = 0.0158[/tex]

Probability for having X =520

sample proportion [tex]\hat p = \frac{520}{1000} = 0.52[/tex]

[tex]p(x\geq 520)  = P(\hat p\geq) [/tex]

                       [tex]= P(Z\geq \frac{0.52 - 0.49}{0.0158})[/tex]

                       [tex]= P(Z\geq 1.89) = 0.0293[/tex]

[tex]p(x\geq 520)  = 0.0293[/tex]

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

[tex]\sigma_p = \sqrt{\dfrac{p(1-p)}{1000}}[/tex]

[tex]\sigma_p = \sqrt{\dfrac{0.49(1-0.49)}{1000}}[/tex]

[tex]\sigma_p =0.0158[/tex]

for x = 520

p = 520/1000 = 0.52

P(X≥520) = P(p≥0.52)

P(p≥0.52) = [tex]P(Z\geq \dfrac{0.52-0.49}{0.0158})[/tex]

P(p≥0.52) = [tex]P(Z\geq 1.898)[/tex]

using z-table

P(p≥0.52) = 0.02938

P(X≥520) =0.02938

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