Answer
$P(X \leq 35) = 0.0793$. This implies that about 8 out of the 100 samples of size 35 will have a mean time less than or equal to 1.89 hours. Since the probability is not unusual, we can not conclude that avid internet users watch less television.
Work Step by Step
Here n = 35
$σ_{x̅} = \frac{σ}{n} = \frac{1.93 }{\sqrt 35} = 0.319$
$z = \frac{x̅ - μ}{σ_{x̅}} = \frac{1.89 - 2.35}{0.319} = -1.41$
$P(X \leq 35) = P(z < -1.41) = 0.0793$
