Question

Suppose that the population proportion of people that enter a hospital lobby that have a contagious disease is 0.18. Further suppose the hospital gets a random sample of size 1894. What is the probability that the sample proportion is within ±.02 of the population proportion?

p=0.18

n=1894

value=0.02

Answer 1

Solution

Given that,

p = 0.18

1 - p = 1 - 0.18 = 0.82

n = 1894

= p = 0.18

=  [p ( 1 - p ) / n] = [(0.18 * 0.82) / 1894 ] = 0.0088

P(0.16 < < 0.20 )

= P[(0.16 - 0.18) / 0.0088 < ( - ) / < (0.20 - 0.18) /0.0088 ]

= P(-2.27 < z < 2.27)

= P(z < 2.27) - P(z < -2.27)

Using z table,   

= 0.9884 - 0.0116

= 0.9768