Answer:
The correct option is f.
Step-by-step explanation:
The (1 - α)% confidence interval for the population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
It is provided that an experimenter flips a coin 100 times and gets 58 heads.
That is the sample proportion of heads is, [tex]\hat p=0.58[/tex].
The critical value of z for 90% confidence level is, z = 1.645.
*Use a z-table.
Compute the 90% confidence interval for the probability of flipping a head with this coin as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=0.58\pm 1.645\cdot\sqrt{\frac{0.58\times(1-0.58)}{100}}\\\\=0.58\pm 0.0812\\\\=(0.4988, 0.6612)\\\\\approx (0.499, 0.661)[/tex]
Thus, the 90% confidence interval for the probability of flipping a head with this coin is (0.499, 0.661).
The correct option is f.
