The A.C. Nielsen Company collected data on the weekly TV viewing times, in hours, of 200 people. Suppose that the sample mean is 30.25, the

Question

The A.C. Nielsen Company collected data on the weekly TV viewing times, in hours, of 200 people. Suppose that the sample mean is 30.25, the sample standard deviation is 12.60, and that the histogram of the viewing times is bell shaped. Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85

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Piper 4 months 2021-10-08T13:16:27+00:00 1 Answer 0 views 0

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    2021-10-08T13:18:10+00:00

    Answer:

    By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85

    Step-by-step explanation:

    The Empirical Rule states that, for a normally distributed(bell-shaped) random variable:

    68% of the measures are within 1 standard deviation of the mean.

    95% of the measures are within 2 standard deviation of the mean.

    99.7% of the measures are within 3 standard deviations of the mean.

    In this problem, we have that:

    Mean = 30.25

    Standard deviation = 12.60

    Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85

    17.65 = 30.25 – 1*12.60

    So 17.65 is one standard deviation below the mean.

    42.85 = 30.25 + 1*12.60

    So 42.85 is one standard deviation above the mean

    By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85

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