A lot of research is conducted on the sleeping habits of U.S. adults. One study reported that X, the amount of sleep per night of U.S. adult

Question

A lot of research is conducted on the sleeping habits of U.S. adults. One study reported that X, the amount of sleep per night of U.S. adults follows a normal distribution with mean μ=7.5 hours and standard deviation σ=1.2 hours.

Using the Standard Deviation Rule, what is the probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night?

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Margaret 19 hours 2021-09-10T14:36:58+00:00 1 Answer 0

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    2021-09-10T14:38:55+00:00

    Answer:

    16% probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night

    Step-by-step explanation:

    The Empirical Rule(Standard Deviation) states that, for a normally distributed 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 = 7.5

    Standard deviation = 1.2

    Using the Standard Deviation Rule, what is the probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night?

    8.7 = 7.5 + 1.2

    So 8.7 is one standard deviation above the mean.

    By the Empirical Rule, 68% of the measures are within 1 standard deviation of the mean. The other 100-68 = 32% are more than one standard deviation from the mean. Since the normal probability distribution is symmetric, 16% are more than one standard deviation below the mean and 16% are more than one standard deviation above the mean(above 8.7 hours)

    So, 16% probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night

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45:7+7-4:2-5:5*4+35:2 =? ( )