A 90% confidence interval for the true percentage of college students who like Brussels sprouts is (1.8%, 4.6%). What is the point estimator of the true percentage of college students who like Brussels sprouts?

And replacing the info from the problem we have:

So then the best estimator for the true proportion p is given by or equivalent to 3.2 %

Step-by-step explanation:

We want to find a confidence interval for a proportion p who represent the parameter of interest.

The confidence interval would be given by this formula:

For this case the 90% confidence interval is given by (1.8%=0.018, 4.6%=0.046) after apply the last formula

Since the confidence interval is symmetrical we can estimate the point estimator of the true percentage with this formula:

And replacing the info from the problem we have:

So then the best estimator for the true proportion p is given by or equivalent to 3.2 %

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.