Question

The amount of time that a truck spends in waiting at the distribution center for unloading is a random variable with a mean of 30 minutes and a standard deviation of 20 minutes. Suppose that a random sample of 9 trucks is observed. Find the probability that the average waiting time of trucks is:

(1) Less than 20 minutes

(2) Between 20 and 40 minutes

Answer 1

Solution :

Given that ,

mean = = 30

standard deviation = = 20

n = 9

=    = 30

= / n = 20 / 9 = 6.67

1) P( < 20) = P(( - ) / < (20 - 30) / 6.67)

= P(z < - 1.50 )

Using z table

= 0.0668

2) P(20< < 40 )  

= P[(20 - 30) / 6.67 < ( - ) / < (40 - 30) / 6.67)]

= P(- 1.50 < Z < 1.50 )

= P(Z < 1.50) - P(Z < - 1.50)

Using z table,  

= 0.9332 - 0.0668  

= 0.8664