Television viewing reached a new high when the global information and measurement company reported a mean daily viewing time of 8.35 hours per household. 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 4 and 10 hours a day? (to 4 decimals)
b. how many hours of television viewing must a household have in order to be in the top 7% of all television viewing household? (to 2 decimals)
c. what is the probability that a household views television more than 4 hours a day? (to 4 decimals)
Let X denote the daily viewing time
We are given :
a.
Required probability = Probability that household views television between 4 and 10 hours a day = P(4<X<10)
We use Excel function "NORMSDIST()" as :
Hence,
Probability that household views television between 4 and 10 hours a day = 0.7044
b.
We need to find the 93rd Percentile
Let P93 denote the 93rd Percentile
We have :
We use the Excel function "NORMSINV()" as :
Number of hours of television viewing must a household have in order to be in the top 7% of all television viewing household = 12.04
c.
Required probability = Probability that a household views television more than 4 hours a day = P(X>4)
We use Excel function "NORMSDIST()" as:
Hence,
Probability that a household views television more than 4 hours a day = 0.9591
Television viewing reached a new high when the global information and measurement company reported a mean...
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...
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