Answer:
4.72 hours/day
Step-by-step explanation:
Mean time spent watching TV (μ) = 2.8 hours a day
Standard deviation (σ) = 1.5 hours a day
The 90th percentile (upper 10%) of a normal distribution has an equivalent z-score of roughly z = 1.282. The minimum time spent watching TV, X, at the 90th percentile is:
[tex]z=\frac{X-\mu}{\sigma} \\1.282=\frac{X-2.8}{1.5}\\ X=4.72\ hours/day[/tex]
On a typical day, you must watch at least 4.72 hours of TV to be in the upper 10%.
