Question

In a restaurant, 90% of customers order coffee with their dinner. A simple random sample of 144 patrons of the restaurant is taken. What is the probability that the proportion of people (sample proportion) who will order coffee with their dinner is more than 94.5%?

Answer 1

Solution

Given that,

p = 0.90

1 - p = 1-090=0.10

n = 144

= p =0.90

=  [p( 1 - p ) / n] = [(0.90*0.10) /144 ] = 0.025

P( > 0.945) = 1 - P( <0.945 )

= 1 - P(( - ) / < (0.945 -0.90) / 0.025)

= 1 - P(z <1.8 )

Using z table

= 1 -0.9641

=0.0359

probability=0.0359