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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 1.91 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.91 hours or less from a population whose mean is presumed to be 2.25 hours.

Respuesta :

Answer:

likelihood percentage is 9.34%

Step-by-step explanation:

given,

sample size (n) = 55

mean time of the result (μ)= 1.91 hours

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

[tex]\sigma = \dfrac{1.91}{\sqrt{55}}[/tex]

[tex]\sigma = 0.25754[/tex]

likelihood of obtaining a sample mean of 1.91 hours or less

[tex]P(x\leq 1.91) = P(Z\leq \dfrac{x - \mu}{\sigma})[/tex]

[tex]P(x\leq 1.91) = P(Z\leq \dfrac{1.91 - 2.25}{0.25754})[/tex]

[tex]P(x\leq 1.91) = P(Z\leq (-1.3201))[/tex]

[tex]P(x\leq 1.91) = 0.0934[/tex]

likelihood percentage is 9.34%

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