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19 May, 03:38

Television viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 8.35 hours per household (USA Today, November 11, 2009). Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household.

A) What is the probability that a household views television between 5 and 10 hours a day?

B) How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households?

C) What is the probability that a household views television more than 3 hours a day?

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  1. 19 May, 03:59
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    Step-by-step explanation:

    Using normal distribution,

    z = (x - u) / s

    Where u = mean = 8.35

    s = standard deviation = 2.5

    x = viewing rate of households

    A) the probability that a household views television between 5 and 10 hours a day

    P (5 lesser than or equal to x lesser than or equal to 10)

    For P (5)

    z = (5-8.35) / 2.5 = - 3.35/2.5 = - 1.34

    Looking at the normal distribution table,

    P (5) = 0.0901

    For P (10)

    z = (10 - 8.35) / 2.5 = 1.65/2.5 = 0.66

    Looking at the normal distribution table,

    P (10) = 0.7454

    P (5 lesser than or equal to x lesser than or equal to 10)

    = P (10) - P (5) = 0.7454-0.0901 = 0.6553

    B) They must be viewing up to 97% = 0.97

    Looking at the table, z value for 0.97=1.88

    z = 1.88 = (x - 8.35) / 2.5

    x = 4.7+8.35 = 13.05 hours

    C) P (x greater than 3) = 1-P (0 lesser than or equal to 3)

    z = (3-8.35) / 2.5

    = - 2.14

    Looking at the table

    P (0 lesser than or equal to 3) = 0.0162

    P (x greater than 3) = 1-0.0162 = 0.9838
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