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 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.

Answer 1

likelihood percentage is 9.34%

Explanation:

given,

sample size (n) = 55

mean time of the result (μ)= 1.91 hours

`\sigma = (s)/(√(n))`

`\sigma = (1.91)/(√(55))`

`\sigma = 0.25754`

likelihood of obtaining a sample mean of 1.91 hours or less

`P(x\leq 1.91) = P(Z\leq (x - \mu)/(\sigma))`

`P(x\leq 1.91) = P(Z\leq (1.91 - 2.25)/(0.25754))`

`P(x\leq 1.91) = P(Z\leq (-1.3201))`

`P(x\leq 1.91) = 0.0934`

likelihood percentage is 9.34%