Question

A director of reservations believes that 6% of the ticketed passengers are no-shows. If the director is right, what is the probability that the proportion of no-shows in a sample of 531 ticketed passengers would be greater than 4%? Round your answer to four decimal places.

Answer 1

Solution

Given that,

p = 0.06

1 - p = 1-0.06=0.94

n = 531

= p =0.06

=  [p( 1 - p ) / n] = [(0.06*0.94) / 531 ] = 0.0103

P( > 0.04) = 1 - P( <0.04 )

= 1 - P(( - ) / < (0.04-0.06) / 0.0103)

= 1 - P(z <-1.94 )

Using z table

= 1 -0.0262

=0.9738

probability=0.9738