Astudy of three hundred teenagers found that the number of hours they spend on instant messaging sites each week is normally distributed with a mean of 13 hours. the population standard deviation is 5 hours. what is the margin of error for a 99% confidence interval?
0.743
0.334
0.320
0.413

OPtion A

Step-by-step explanation:

Given that a study of three hundred teenagers found that the number of hours they spend on instant messaging sites each week is normally distributed with a mean of 13 hours. The population standard deviation is 5 hours.

Std deviation = 5 hours

Sample size = 300

Hence std error of sample = 5/sqrt 300

=0.289

Margin of error = z Critical x std error

Since confidence level = 99%

Z critical = 2.58

Margin of error = 2.58 (0.289)

=0.743

Option A is right

im not quite sure if this is right but i got 5.6 oisutuve

step-by-step explanation:

x/3-10=8

x/3=18

x=48