Forty-two percent of primary care doctors think their patients receive unnecessa

Question

Forty-two percent of primary care doctors think their patients receive unnecessary medical care.

a).  Suppose a sample of (100) primary care doctors was taken. What is the probability the proportion of primary care doctors who think their patients receive unnecessary medical care is between 45% and 50%?

b).  Suppose a sample of (100) primary care doctors was taken. What is the probability the proportion of primary care doctors who think their patients receive unnecessary medical care is between 38% and 48%?

c).  Within what symmetrical limits of the population proportion will 88% of the sample proportions fall?

Explanation / Answer

Normal Distribution
Proportion ( P ) =0.42
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.42*0.58/100)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.45) = (0.45-0.42)/0.0494
= 0.03/0.0494 = 0.6073
= P ( Z <0.6073) From Standard Normal Table
= 0.72817
P(X < 0.5) = (0.5-0.42)/0.0494
= 0.08/0.0494 = 1.6194
= P ( Z <1.6194) From Standard Normal Table
= 0.94732
P(0.45 < X < 0.5) = 0.94732-0.72817 = 0.2192                  
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.38) = (0.38-0.42)/0.0494
= -0.04/0.0494 = -0.8097
= P ( Z <-0.8097) From Standard Normal Table
= 0.20905
P(X < 0.4) = (0.4-0.42)/0.0494
= -0.02/0.0494 = -0.4049
= P ( Z <-0.4049) From Standard Normal Table
= 0.34279
P(0.38 < X < 0.4) = 0.34279-0.20905 = 0.1337

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