not sure how to formulate this

Question

In the US the percentage of adults who have had an appendectomy is 28%. A simple random sample of 42 adults and the US is taken. Calculate the probability of the proportion of adults who have had an appendectomy in is this sample is more than 25%

Answer 1

We are asked to calculate the probability that the proportion of adults who have had an appendectomy in this simple random sample of 42 adults in the U.S. is more than 25%. In statistical terms this means that we will need to calculate

P(\hat p > 0.25).

To calculate this we will use the normal approximation to the binomial. For this approximation to be valid we will need the following two conditions to be met:

np ≥ 5

n(1 - p) ≥ 5

Since we are given that p = 0.28 and n = 42, we have that n*p = 42*.28 = 11.76 > 5 and n*(1 - p) = 42*(1 - 0.28) = 42*(0.72) = 30.24 > 5 (i.e. the normal approximation to the binomial is valid). The normal approximation to the binomial then tells us that the distribution for the sample proportion can be approximated by a normal distribution with a mean of p (i.e. 0.28) and a standard deviation of √((p*(1-p))/n) (i.e. √((0.28*(1-0.28))/42) = √((0.2016)/42) = √(0.0048)). All of this tells us that

P(\hat p > 0.25) = P(\hat p - 0.28 > 0.25 - 0.28)

= P(\hat p - 0.28 > - 0.03)

= P((\hat p - 0.28)/√(0.0048) > (- 0.03)/√(0.0048))

= P(z > (- 0.03)/√(0.0048))

≈ 0.6675,

which tells us that the probability of the proportion of adults who have had an appendectomy in this sample being more than 25% is approximately 0.6675.