Suppose 52% of the doctors in America are dentists. If a random sample of size 775 is selected, what is the probability that the proportion of doctors who are dentists will differ from the population proportion by less than 5%? Round your answer to four decimal places.

0.134%

Step-by-step explanation:

We have to

sd * P = [p * (1 - p) / n] ^ (1/2)

where p = 0.52 and n = 775, replacing:

sd * P = [0.52 * (1 - 0.52) / 775] ^ (1/2) = 0.0179

Now the probability would be:

P = 1 - P [(- 0.05) /0.0179 <z <(0.05) / 0.0179]

P = 1 - P (-2.79 <z <2.79)

P = 1 - P (z <2.79) - P (z <-2.79)

P = 1 - 0.9974 - 0.00126

P = 1 - 0.99614

P = 0.00134

Therefore the probability is 0.134%

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