Question

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

Answer 1

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

$\bar{x}$ = 544 × 0.47 = 255.68

Standard deviation = $\sqrt{np(1-p)}$

or

Standard deviation = $\sqrt{544\times0.51(1-0.51)}$

or

Standard deviation = 11.6595

Now,

$P(\bar{x}\geq 47\%)$

=  $P(z\geq \frac{\bar{x}-\mu}{\sigma})$

=  $1- P(z\leq \frac{\bar{x}-\mu}{\sigma})$

= $1- P(z\leq \frac{255.68-277.44}{11.6595})$

=   $1- P(z\leq -1.8662)$

Now, from standard z value table, we get

= 1 -  0.031021

= 0.9689