Question

A total of 350 households were survey in an O-D survey. The average trip length 15 minutes, with a standard deviation of 7 minutes. If 30 households were chosen at random, what is the probability they have an average trip length greater than 12 minutes and less than 30 minutes?.

Answer 1

Solution :

Given that,

mean = = 15

standard deviation = = 7

n = 30

= = 15

= / n = 7/ 30 = 1.278

P(12 < < 30 )  

= P[(12 - 15) / 1.278 < ( - ) / < (30 - 15) / 1.278)]

= P(-2.35 < Z < 11.74)

= P(Z < 11.74) - P(Z < -2.35)

Using z table,  

= 1 - 0.0094

= 0.9906