Astudy of one thousand teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 20 hours. the population standard deviation is 2 hours. what is the 95% confidence interval for the mean? 19.88−20.12 hours 19.76−10.24 hours 19.88−20 hours 20−20.12 hours

The z-value that corresponds to a two-tailed 95% confidence interval is z = +/- 1.96. Then the bounds of the confidence interval can be determined as:
Lower bound = mean - z*SD/sqrt(n) = 20 - 1.96*2/sqrt(100) = 20 - 0.12 = 19.88 hours
Upper bound = mean + z*SD/sqrt(n) = 20 + 1.96*2/sqrt(100) = 20 + 0.12 = 20.12 hours
So the first choice is the correct 19.88-20.12 hours