Question

One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 2.00 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.00 hours or less from a population whose mean is presumed to be 2.35 hours.

Answer 1

Answer:

9.72%

Step-by-step explanation:

Our values here are:

Sample Size, n= 55

Standard Deviation is 2.

$\sigma = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{n}}$

$\sigma = 0.26968$

Our probability of obtaining a sample equal or less to 2 hours is

$P(X \leq 2) = P ( Z \leq \frac{2-2.35}{0.26968})$

$P=(Z \leq -1.292835) = 0.0972$

We can conclude that our probability is 9.72% from a population whose mean is presumed to be 2.35hours