Media researchers report the average daily TV viewing time for U.S. adult males to be 4.28 hours. Assume a normal distribution with a standard deviation of sigma 1.30 hours. Note: This problem has nothing to do with sampling distributions. It is just a normal distribution problem based on the random variable X = amount of viewing time for U.S. adult males. (Report all answers out to 4 decimal places.)

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Chapter1: Starting With Matlab
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Media researchers report the average daily TV viewing time for U.S. adult males to
be 4.28 hours. Assume a normal distribution with a standard deviation of sigma =
1.30 hours.
Note: This problem has nothing to do with sampling distributions. It is just a
normal distribution problem based on the random variable X = amount of
viewing time for U.S. adult males.
(Report all answers out to 4 decimal places.)
What is the probability that a randomly selected U.S. adult male watches TV less
than 2 hours per day?
A
How much TV would a U.S. adult male have to watch in order to be at the 99th
percentile (i.e., only 1% of his counterparts are more "TV intensive" than he is)?
95% of adult males typically watch between
A and
hours of TV in a day.
(Make sure values are equidistant from the mean.)
Transcribed Image Text:Media researchers report the average daily TV viewing time for U.S. adult males to be 4.28 hours. Assume a normal distribution with a standard deviation of sigma = 1.30 hours. Note: This problem has nothing to do with sampling distributions. It is just a normal distribution problem based on the random variable X = amount of viewing time for U.S. adult males. (Report all answers out to 4 decimal places.) What is the probability that a randomly selected U.S. adult male watches TV less than 2 hours per day? A How much TV would a U.S. adult male have to watch in order to be at the 99th percentile (i.e., only 1% of his counterparts are more "TV intensive" than he is)? 95% of adult males typically watch between A and hours of TV in a day. (Make sure values are equidistant from the mean.)
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