anonymous
  • anonymous
Approximately 15% of Americans have a blood type of B. The local blood bank has 80 donors stop by on a given day (which can be considered a random sample from the population). What is the probability that between 15% and 20% of the donors on a given day have a blood type of B? Be sure to state and justify your assumptions.
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So I can justify the fact that this is normal, since there is a large sample size. That's all I got, though.
WolframWizard
  • WolframWizard
Use the binomcdf function on a calculator. 15% of 80 is 12 and 20% of 80 is 16. Plug in binomcdf(80,0.15,11) to find the P(X<12) which is .45219. Then plug in binomcdf(80,0.15,16) to find the P(X≤16) which is .91629. Subtract P(X<12) from P(X≤16) to get P(12≤x≤16) which is .46410. So the probability that between 15% and 20% (12 people and 16 people) have blood type B is about 0.46410.
anonymous
  • anonymous
Thank you. I had one more question. If you could help me with that, that'd be great!! An opinion poll asks an SRS of 1500 adults, "Do you happen to jog?" Suppose that the population proportion who jog (a parameter) is p = 0.15. To estimate p, we use the proportion p-hat in the sample who answer "Yes." Justify the use of a normal approximation and find the following probabilities. a) P(p > 0.16) b) P(0.14 < p < 0.16)

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anonymous
  • anonymous
Again, I can justify the normal approximation, but I have trouble with the actual probabilities.
WolframWizard
  • WolframWizard
Use z distributions for answering the question. z=(O - E)/SEp O is the observed proportion for p-hat E is the expected proportion or p SEp is the standard error of the sample proportion =√(p*(1-p)/n) a) z=(0.16 - 0.15)/0.0092=+1.09 so P(p>0.16) is P(z>1.09) This means you have to find the probability a number is 1.09 standard deviations above the mean. In a normal distribution, this means how much of the area is above 1.09 standard deviations from the mean. This is 0.138 b) Using the same steps as part a, convert the probability into P(-1.09
anonymous
  • anonymous
I see!! Thank you!!!!

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