Answered

Suppose 51 % of the banks in Switzerland are private organizations.

If a sample of 544 banks is selected, what is the probability that the sample proportion of private banks will be greater than 47 % ? Round your

answer to four decimal places.

Answer :

Answer:

0.9689

Step-by-step explanation:

Given:

banks in Switzerland are private organizations, p = 51% = 0.51

Sample size, n = 544 banks

To find:

Probability (the sample proportion of private banks will be greater than 47%)

Now,

Mean of the sample, μ = np = 544 × 0.51 = 277.44

[tex]\bar{x}[/tex] = 544 × 0.47 = 255.68

Standard deviation = [tex]\sqrt{np(1-p)}[/tex]

or

Standard deviation = [tex]\sqrt{544\times0.51(1-0.51)}[/tex]

or

Standard deviation = 11.6595

Now,

[tex]P(\bar{x}\geq 47\%)[/tex]

=  [tex]P(z\geq \frac{\bar{x}-\mu}{\sigma})[/tex]

=  [tex]1- P(z\leq \frac{\bar{x}-\mu}{\sigma})[/tex]

= [tex]1- P(z\leq \frac{255.68-277.44}{11.6595})[/tex]

=   [tex]1- P(z\leq -1.8662)[/tex]

Now, from standard z value table, we get

= 1 -  0.031021

= 0.9689

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