Question

Suppose 43% of the doctors in a hospital are surgeons. If a sample of 478 doctors is selected, what is the probability that the sample proportion of surgeons will be greater than 39%? Round your answer to four decimal places.

Answer 1

`z =(0.39-0.43)/(0.0226)= -1.766`

And we can find this probability using the complement rule and the normal standard table or excel and we got:

`P(\hat p >0.39) =P(Z>-1.766) =1-P(Z<-1.766) = 1-0.0387=0.9613`

Explanation:

For this case we can define the population proportion p as "true proportion of surgeons" and we can check if we can use the normal approximation for the distribution of
`\hat p`

1)
`np =478*0.43 =205.54 >10`

2)
`n(1-p) =478*(1-0.43) =272.46 >10`

3) Random sample: We assume that the data comes from a random sample

Since we can use the normal approximation the distribution for
`\hat p` is given by:

`\hat p sim N( p, \sqrt{(p (1-p))/(n)})`

With the following parameters:

`\mu_(\hat p) = 0.43`

`\sigma_(\hat p) =\sqrt{(0.43*(1-0.43))/(478)}= 0.0226`

And we want to find this probability:

`P(\hat p >0.39)`

And we can use the z score formula given by:

`z = (\hat p -\mu_(\hat p))/(\sigma_(\hat p))`

And if we calculate the z score for
`\hat p = 0.39` we got:

`z =(0.39-0.43)/(0.0226)= -1.766`

And we can find this probability using the complement rule and the normal standard table or excel and we got:

`P(\hat p >0.39) =P(Z>-1.766) =1-P(Z<-1.766) = 1-0.0387=0.9613`