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 30 hours. the population standard deviation is two hours.
what is the 95% confidence interval for the mean?

Conf interval = (29.876, 30.124)

Step-by-step explanation:

Given that A study 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 30 hours.

Also population std deviation sigma = 2 hours

For 95% confidence interval z critical value is

±1.96

Margin of error = Std error x Z critical

Std error = std dev/sq rt of n where n = sample size = 1000

Hence std error = 2/√1000=0.0632

Margin of error = ±1.96(0.0632)=±0.1240

Confidence interval lower bound = 30-0.124=29.876

Upper bound = 30+0.124 = 30.124

Conf interval = (29.876, 30.124)

answer: procent 54 wynosi 33,33%

1/8 3/16 3/4

step-by-step explanation:

you put them as all the same denominator and multiply