Question

According to the Department of Transportation, the population proportion of late flights equals 0.204. A random sample of 140 domestic flights was selected. What z-score would you use to test the probability that 35 or fewer flights from this sample were late?

Answer 1

Solution :

Given that,

p = 0.204

q = 1 - p =1-0.204=0.796

n = 140

Using binomial distribution,

Mean = = n * p = 140*0.204=28.56

Standard deviation = = n * p * q =  104*0.204*0.796=4.7680

Using continuity correction ,

P(x< 35 ) = P((x - ) < (35.5-28.56) /4.7680 )

= P(z < 1.46)

Using z table

Probability =0.9279

z-score=1.46