In january 2003, the american worker spent an average of 77 hours logged on to the internet while at work (CNBC, march 15,2003). Assume the times are normally distribute and that the standard deviation is 20 hours.
a) What is the probability a randomly selected worker spent fewer than 50 hours logged on to the internet?
b) What percentage of workers spent more than 100 hours logged on to hte internet?
c) A person is classified as a heavy user if he/she is in the upper 20% of usage how many hours must a worker have logged on to the internet to be considered a heavy user?

Step-by-step explanation:

Given mean = 77hours, standard deviation = 20hours

a) the probability a randomly selected worker spent fewer than 50 hours logged on to the internet? = P(X < 50)

z = X - mean/SD, where SD = standard deviation

= 50 - 77/20

z = -1.35, P(X < 50) = P(z < -1.35) = 0.0885 from z -table

b) percentage of workers spent more than 100 hours logged on to the internet

= P(X > 100)

z = 100 - 70/20

z = 1.15

P(z> 1.15) = 1 - P(z < 1.15)

= 0.1251, hence percentage of workers spent more than 100 hours logged on to the internet 12.51%

c) from the appendix in the z table

P(z < 0.84) = 1 - 0.20 = 0.80

from x - mean /SD = z

x = mean + zSD

x = 77 + 20 x 0.84

x = 94, how many hours must a worker have logged on to the internet to be considered a heavy user = 94hrs

two data sets with the same median will not have the same varaibility

step-by-step explanation:

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?