1st question: One way that healthcare providers evaluate and monitor the quality
ID: 3050859 • Letter: 1
Question
1st question: One way that healthcare providers evaluate and monitor the quality of their services is by sending “secret shoppers” posing as patients.
In a recent study, a team of physicians was used to evaluate the quality of care provided at pharmacies serving a major HMO network in Southern California. Since budget restrictions kept the team from evaluating every pharmacy in the network, they instead picked three major Southern California cities at random and focus their energy on evaluating the network pharmacies in those cities.
3a. Does this appear to be a random or a non-random sampling design? Why?
2nd question: Perform the following calculations. Please show all work.
You roll a six-sided dice four times. What is the probability of rolling 1, 2, 3, and 4?
3rd question: Answer the following questions using the information below:
Blood Type Rh Factor Worldwide HIV/AIDS Prevalence
O: 46% Positive: 84% HIV positive: 0.9%
A: 40% Negative: 16% HIV negative: 99.1%
B: 10%
AB: 4%
a)How many different outcomes are possible combining the above three data?
b)What is the probability of having blood type O or blood type B?
c)What is the probability of being HIV positive with blood type AB, Rh negative?
4th question: You have a bag of 10 red, 10 blue, and 30 white marbles.
You draw five marbles without returning any to the bag.
What is the probability of drawing 1 red, 1 red, 1 red, 1 blue, and then 1 white?
4a. Same question as above – but after each drawing you put the marble back into the bag
Explanation / Answer
Answer to 3a : Although the three cities are picked randomly, there appears to be a problem in the random sampling because of this phrase "they instead picked three major Southern California cities at random". It is important that the group of pharmacies selected be representative of the population of pharmacies, and not be biased in a systematic manner (only pharmacies from big cities). For example, a group comprised of the wealthiest individuals in a given area probably would not accurately reflect the opinions of the entire population in that area. The physicians should have instead used multi stage random sampling (first randomly sample distiricts, then randomly sample counties and then randomly sample pharmacies in those counties).
Answer to 2nd question : Rolling a dice four times has a sample space of 6*6*6*6 = 1296 possible outcomes. For the event in question, it can only happen in 1 way for the first throw (since there's only one way for digit 1 to show up), 1 way for the second throw (since there's only one way for digit 2 to show up) and so on. Therefore, since all these sub events must happen together, there is only 1 * 1 * 1 * 1 = 1 desirable outcome. Therefore, the total probability = 1 / 1296 .
Answer to 3rd question : a) There are 3 different variables given (Blood type, Rh factor, Hiv prevalence). Blood type has 4 different levels, wheras Rh factor and Hiv prevalence each have 2 levels. By multiplication rule of counting, the number of different outcomes are 4 * 2 * 2 = 16 different outcomes.
Answer to 3rd question : b) By addition rule of counting, the probability of being blood type O or being blood type B (assuming one can't be both O and B), is the sum of the probabilities of blood type O and blood type B. So the answer is 46% + 10% = 56%
Answer to 3rd question : c) This is a trick question. The probability of being HIV positive does not change with the blood type or the rhesus factor. Therefore, the answer is still 0.9% or probability wise, 0.009.
Answer to 4th question : Remember that without returning balls back, sample space after each picking changes. Therefore, firstly the probability of drawing a red ball first = 10/(10+10+30) = 1 / 5. After that, probability of drawing another red = 9(Since 1 red was removed previously) / 49 (1 ball already removed) = 9 / 49. After that, probability of drawing another red = 8 / 48 = 1 / 6 . After that, probability of drawing a blue ball = 10 / 47 . After that, probability of drawing a white ball = 30 / 46 = 15 / 23. Finally, since all these events necessarily happen together, the total probability = (1 / 5) * (9 / 49) * (1 / 6) * (10 / 47) * (15 / 23) = 0.00084955351.
Answer to 4th question 4a : Here, since balls are returned back, the sample space is always the same (i.e. 50). Therefore, the total probability is calculated with similar procedure as above i.e. = (10 / 50) * (10 / 50) * (10 / 50) * (10 / 50) * (30 / 50) = 0.00096.
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