Question

The National Coffee Association reported that 63% of U.S. adults drink coffee daily. A random sample of 300 U.S. adults is selected. Round your answers to at least four decimal places as needed.

Find probability that less then 58% of sampled adults drink coffee daily

Answer 1

Solution

Given that,

p = 0.63

1 - p = 1-0.63=0.37

n = 300

= p =0.63

=  [p ( 1 - p ) / n] =   [(0.63*37) / 300 ] = 0.02787

P( <0.58 ) =

= P[( - ) / < (0.58-0.63) / 0.02787]

= P(z <-1.79 )

Using z table,   

probability =0.0367