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It is estimated that 1 in 9 Canadian men will develop prostate cancer during their lifetime. (1) Approximate the probability that at most 20 of the 180 Canadian men will eventually develop prostate cancer. (2) And consider a random sample of three Canadian men, and let Pˆ represent the proportion of these men who will eventually develop prostate cancer. Derive the sampling distribution of Pˆ .

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Final answer:

To approximate the probability that at most 20 of the 180 Canadian men will eventually develop prostate cancer, we can use the binomial distribution. For the second part of the question, we are considering a random sample of three Canadian men and we want to derive the sampling distribution of the proportion who will eventually develop prostate cancer (P-hat).

Step-by-step explanation:

To approximate the probability that at most 20 of the 180 Canadian men will eventually develop prostate cancer, we can use the binomial distribution. The probability of success (developing prostate cancer) for each individual is estimated to be 1/9. The formula for the binomial distribution is P(X <= k) = sum from i=0 to k of (n choose i) * p^i * (1-p)^(n-i), where n is the number of trials, k is the number of successes we are interested in, p is the probability of success, and (n choose i) is the number of ways to choose i successes from n trials. Plugging in the values, the approximate probability is:

P(X <= 20) = sum from i=0 to 20 of (180 choose i) * (1/9)^i * (8/9)^(180-i)

For the second part of the question, we are considering a random sample of three Canadian men and we want to derive the sampling distribution of the proportion who will eventually develop prostate cancer (P-hat). Since each individual can either develop prostate cancer or not, and the sample size is small, we can assume that the sampling distribution follows a binomial distribution. So the sampling distribution of P-hat will be a binomial distribution with parameters n (sample size) and p (probability of success).

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