Wayne Merritt drives from Cleveland to Chicago frequently and has noticed that traffic and weather make a big difference in the time it takes to make the trip. As a result, he has a hard time planning activities around his arrival time. To better plan his business, Wayne wants to calculate his average driving time as well as a measure of how much an actual trip is likely to vary from that average. To do that, he clocked 10 trips with the results in table.

Driving Time	Number of trips
6 hrs, 0 min	1
6 hrs, 15 min	1
6 hrs, 25 min	2
6 hrs, 45 min	3
7 hrs, 11 min	1
7 hrs, 30 min	1
9 hrs, 20 min	1

1. Calculate the mean, standard deviation, and coefficient of variation of Wayne's driving time to Chicago. (Hint: Treat the 10 trips as the 10 possible outcomes of a discrete probability distribution, each of which has a probability of 0.1.) Do not round intermediate calculations. Round the answers to two decimal places.

   Mean	fill in the blank 1 minutes
   Standard deviation	fill in the blank 2 minutes
   Coefficient of variation	fill in the blank 3

2. Calculate the average variation in driving time. Round the answer to two decimal places. (Hint: the average variation is calculated using the formula .) fill in the blank 4 minutes The average variation is

   significantly higher thansignificantly lower thanabout the same assignificantly lower than

   the standard deviation. An average deviation would probably be

   morelessmore

   meaningful to Wayne.