Answer to Question #5183 in Statistics and Probability for SASHA
computers. In the survey, the number of hours was normally distributed, with a mean of 8
hours and a standard deviation of 1 hour. A survey participant is randomly selected. Find the
probability that the hours spent on the home computer by the participant are between 5.5 and
9.5 hours per week.
and 0.62%, so the answer is P=94.41%-0.62%=93.79%.
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