Question

Part 1. Case 6-2b: Internet Usage 2003
In January 2003, the American worker spent an average of 77 hours logged on to the Internet while at work. Assume the times are normally distributed and that the standard deviation is 20 hours.

Refer to Case 6-2b: Internet Usage 2003
Which of the following statements about Internet usage by American workers is most probably FALSE?

	1.	A worker must have logged on to the Internet at least 123.60 hours to be in the upper 1% in terms of hours of Internet usage.
	2.	The percentage of workers who spent 10 hours or less logged on to the Internet is less than or equal to 0.04%
	3.	The percentage of workers who spent 20 hours or less logged on to the Internet is less than or equal to 22%
	4.	The percentage of workers who spent less than 100 hours logged on to the Internet is 87.49%
	5.	A worker must have logged on to the Internet at most 44.10 hours to be in the bottom 5% in terms of hours of Internet usage.

Part 2.

Case 6-2: Internet Usage 1997
In January 1997, the American worker spent an average of 60 hours logged on to the Internet while at work. Assume the times are normally distributed and that the standard deviation is 25 hours.

Refer to Case 6-2: Internet Usage 1997
What is the probability a randomly selected worker spent fewer than 50 hours logged on to the Internet?

	1.	.1554
	2.	.6554
	3.	.7119
	4.	-.4000
	5.	.3446

Answer 1

Part 1:

The statement 1,2,4 and 5 are correct

Statement 3 is FALSE

The percentage of workers who spent 20 hours or less who logged on to the internet = P(X ≤ 20)

= P{Z ≤ (20 - 77)/20}

= P(Z ≤ -2.85)

= 0.0022 = 0.22%

Part 2:

The probability that a randomly selected worker spent fewer than 50 hours logged on to the internet

= P(X < 50)

= P{Z < (50 - 60)/25}

= P(Z < -0.4)

= 0.3446