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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.

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Answer:

9.72%

Step-by-step explanation:

Our values here are:

Sample Size, n= 55

Standard Deviation is 2.

[tex]\sigma = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{n}}[/tex]

[tex]\sigma = 0.26968[/tex]

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

[tex]P(X \leq 2) = P ( Z \leq \frac{2-2.35}{0.26968})[/tex]

[tex]P=(Z \leq -1.292835) = 0.0972[/tex]

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

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