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Suppose 49% of the doctors in America are dentists. If a random sample of size 689 is selected, what is the probability that the proportion of doctors who are dentists will be less than 47%

Sagot :

okpalawalter8

Answer:

[tex]P(<47\%)=0.1468[/tex]

Step-by-step explanation:

From the question we are told that:

Percentage of Dentist Doctors P(D)=49\%

Sample size n=689

Generally the equation for  probability that the proportion of doctors who are dentists will be less than [tex]P(<47\%)[/tex] is mathematically given by

 [tex]P(<47\%)=Z>(\frac{\=x-P(D)}{\sqrt{\frac{P(D)*1-P(D)}{n}}})[/tex]

 [tex]P(<47\%)=Z>(\frac{0.47-0.49}{\sqrt{\frac{0.49*0.51}{689}}})[/tex]

 [tex]P(<47\%)=Z>(1.05)[/tex]

Therefore from table

 [tex]P(<47\%)=0.1468[/tex]

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