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31 January, 08:17

One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 35 individuals who consider themselves to be avid Internet users results in a mean time of 2.01 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.01 hours or less from a population whose mean is presumed to be 2.45 hours.

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  1. 31 January, 09:19
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    sample size, n = 55

    /sigma = s//sqrt{n} = 2//sqrt{55}

    = 0.26968

    likelihood of obtaining a sample mean of 2.00 hours or less

    P (X<=2) = P (Z < = (2-2.35) / 0.26968) = P (Z < = - 1.292835)

    = 0.0972

    i. e likelihood percentage is 9.72%
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