Final answer:
The probability distribution of X for the number of men developing prostate cancer is binomial, which for a large sample size can be approximated by a normal distribution. To find the probability of at most 20 men developing cancer, we would use the normal approximation. The sampling distribution of ˉPˆ with three men can assume four values, each with its own probability.
Step-by-step explanation:
Regarding the given probability that 1 in 9 Canadian men will develop prostate cancer during their lifetime, we can use this information to address the student's questions.
Part (a)
The probability distribution of X, which represents the number of Canadian men in a sample of 180 who will eventually develop prostate cancer, would be a binomial distribution. This is because each man either will or will not develop prostate cancer, which constitutes a binary outcome, and each trial (or man's health outcome) can be assumed to be independent of the others.
Part (b)
For a large number of trials like 180, the binomial distribution can be approximated by the normal distribution, provided certain conditions are met, such as np and n(1-p) are both greater than 5, where 'n' is the number of trials and 'p' the probability of success.
Part (c)
To approximate the probability that at most 20 of the 180 Canadian men will eventually develop prostate cancer using the normal approximation to the binomial, we would find the z-score and use the standard normal distribution to find this probability.
Part (d)
When considering a random sample of three Canadian men, ˉPˆ can assume four values: 0, 1/3, 2/3, or 1, representing the proportion of men in the sample who will develop prostate cancer. The sampling distribution of ˉPˆ can then be derived by assessing the probability of each outcome.