SOLUTION: A newspaper article reported that people spend a mean of
7.5
hours per day watching TV, with a standard deviation of
2.1
hours. A psychologist would like to conduct inter
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-> SOLUTION: A newspaper article reported that people spend a mean of
7.5
hours per day watching TV, with a standard deviation of
2.1
hours. A psychologist would like to conduct inter
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Question 1081409: A newspaper article reported that people spend a mean of
7.5
hours per day watching TV, with a standard deviation of
2.1
hours. A psychologist would like to conduct interviews with the
10
%
of the population who spend the most time watching TV. She assumes that the daily time people spend watching TV is normally distributed. At least how many hours of daily TV watching are necessary for a person to be eligible for the interview? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The 90th percentile for the normal distribution is z=1.282
z=(x-mean)/sd
1.282=(x-7.5)/2.1
x-7.5=2.6922
x=10.1922=10.2 hours
