Question

13.Suppose that 60% of people in the population support raising taxes to pay for wetlands preservation. Suppose you randomly sample 100 people and ask them whether they support raising taxes for this purpose. If you divide the number of people in your sample who support raising taxes by the total number of people asked (100), you obtain a random variable, let’s call it p̂. Use the central limit theorem to come up with the approximate probability that p̂ is within 0.1 of the true proportion 0.6.

15. Set up, but do not evaluate, an integral that will yield the density of Z = X + Y, where X and Y are both standard normal random variable with parameters µ=0 and σ=1. You may assume that X and Y are independent.

Answer 1

We would be looking at the first question here that is Q13.

Q13) For the sample proportion, the mean is given as the true proportion which is 0.6, and Using the Cental limit thereom, the distribution of sample proportion here is given as:

Now the probability required here is:
P( 0.6 - 0.1 < p < 0.6 + 0.1)

Converting it to a standard normal variable, we get here:

Getting it from the standard normal tables, we get here:

Therefore 0.9586 is the required probability here.