Answer:
A) To find the probability that less than 2 people out of a random sample of 300 will be diagnosed with the rare virus, we can use the binomial distribution.
Let X be the number of people in the sample diagnosed with the virus. Then, X follows a binomial distribution with n = 300 and p = 0.005, since each person in the sample has a 0.5% chance of being diagnosed with the virus. We want to find P(X < 2).
Using a binomial probability table or calculator, we find:
P(X < 2) = P(X = 0) + P(X = 1) = 0.9722
Therefore, the probability that less than 2 people out of a random sample of 300 will be diagnosed with the rare virus this year is approximately 0.9722 or 97.22%.
B) To determine whether 15 positive cases out of a sample of 300 residents in a small town located beside a medical military research facility is rare or normal, we need to compare this result to what we would expect by chance alone.
In a random sample of 300 Americans, we would expect 300 x 0.005 = 1.5 people to be diagnosed with the virus on average. The standard deviation of a binomial distribution is sqrt(np(1-p)), which for this case is sqrt(300 x 0.005 x 0.995) = 1.09.
We can use a normal approximation to the binomial distribution to calculate the probability of observing 15 or more positive cases in the sample:
Z = (15 - 1.5) / 1.09 ≈ 12.39
Using a standard normal distribution table or calculator, we find that P(Z > 12.39) is extremely small, close to zero.
Therefore, observing 15 or more positive cases in a sample of 300 is extremely rare and unlikely to occur by chance alone. It is possible that the medical military research facility located beside the town is the source of the virus, or that there is another explanation for the high number of cases in the town. Further investigation would be needed to determine the cause.