Question

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 1

0.9689

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 =
`√(np(1-p))`

or

Standard deviation =
`√(544*0.51(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 (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