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=1.2780

P(12<    <30 ) = P[(12 -15) / 1.2780 < ( - ) / < (30 -15) /1.2780 )]

= P(-2.35 < Z <11.74 )

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

Using z table

=1 -0.0094

=0.9906

probability= 0.9906