Question

Suppose 41%41% of the doctors in America are dentists.

If a random sample of size 826826 is selected, what is the probability that the proportion of doctors who are dentists will be less than 40%40%? Round your answer to four decimal places.

Answer 1

Solution

Given that,

p = 0.41

1 - p = 1 - 0.41 = 0.59

n = 826

$\mu$_{$\hat p$} = p =0.41

$\sigma$_{$\hat p$} =  $\sqrt$[p ( 1 - p ) / n] =  $\sqrt$ [(0.41*0.59) /826 ] = 0.0171

P( $\hat p$ < 0.40) =

= P[( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (0.40- 0.41 ) / 0.0171 ]

= P(z <-0.58 )

Using z table,

= 0.2810