SOLUTION: A claim was recently made on national television that two of every three doctors
recommend a particular pain killer. Suppose a random sample of n = 300 doctors revealed that
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-> SOLUTION: A claim was recently made on national television that two of every three doctors
recommend a particular pain killer. Suppose a random sample of n = 300 doctors revealed that
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Question 884000: A claim was recently made on national television that two of every three doctors
recommend a particular pain killer. Suppose a random sample of n = 300 doctors revealed that
180 said that they would recommend the painkiller. If the TV claim is correct
a) Describe the sampling distribution of ̂.
b) What is the probability of 180 or fewer in the sample agreeing?
You can put this solution on YOUR website! p = 2/3 = proportion of doctors claimed to be recommending a particular pain killer.
q = 1 - p = 1/3 = proportion of doctors not claimed to be recommending a particular pain killer.
n = 300 = number of doctors in the sample.
population mean is assumed to be n * p = 300 * 2/3 = 200.
the population standard deviation is assumed to be sqrt (p * q * n) which is equal to sqrt (300 * 2/3 * 1/3) = sqrt(600/9).
z score is equal to (180 - 200) / sqrt(600/9) which is equal to -2.45.
look up in z score table for z score of -2.45 and you'll find that the area to the left of that z score is equal to .0071.
the probability that someone will have a score less than 180 is equal to .0071.
