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

Answer:

$z =\frac{0.39-0.43}{0.0226}= -1.766$

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

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

Step-by-step 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:

$\hat p sim N( p, \sqrt{\frac{p (1-p)}{n}})$

With the following parameters:

$\mu_{\hat p} = 0.43$

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

And we want to find this probability:

$P(\hat p >0.39)$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_{\hat p}}{\sigma_{\hat p}}$

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

$z =\frac{0.39-0.43}{0.0226}= -1.766$

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

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