In 2010 42% primary care doctors in the Unites States believe their patients rec
ID: 3397860 • Letter: I
Question
In 2010 42% primary care doctors in the Unites States believe their patients receive unnecessary medical care. The health care industry in the state of New Jersey wants to determine if the proportion of primary care doctors in New Jersey believe their patients review unnecessary medical care. To study this, the sample of 300 primary care doctors were randomly selected. What is the probability the result of the study will be between 39% and 44%? Use proportion probability Formula. In 2010 42% primary care doctors in the Unites States believe their patients receive unnecessary medical care. The health care industry in the state of New Jersey wants to determine if the proportion of primary care doctors in New Jersey believe their patients review unnecessary medical care. To study this, the sample of 300 primary care doctors were randomly selected. What is the probability the result of the study will be between 39% and 44%? Use proportion probability Formula. In 2010 42% primary care doctors in the Unites States believe their patients receive unnecessary medical care. The health care industry in the state of New Jersey wants to determine if the proportion of primary care doctors in New Jersey believe their patients review unnecessary medical care. To study this, the sample of 300 primary care doctors were randomly selected. What is the probability the result of the study will be between 39% and 44%? Use proportion probability Formula.Explanation / Answer
Proportion ( P ) =0.42
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.42*0.58/300)
Normal Distribution = Z= X- u / sd ~ N(0,1)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.39) = (0.39-0.42)/0.0285
= -0.03/0.0285 = -1.0526
= P ( Z <-1.0526) From Standard Normal Table
= 0.14625
P(X < 0.44) = (0.44-0.42)/0.0285
= 0.02/0.0285 = 0.7018
= P ( Z <0.7018) From Standard Normal Table
= 0.75858
P(0.39 < X < 0.44) = 0.75858-0.14625 = 0.6123
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