Question

In 2009, the American Time Use Survey found that the average American spends 2.8 hours a day watching TV. Suppose the standard deviation is 1.5 hours and that TV time is normally distributed. If you are in the upper 10% of all TV watchers, at the minimum, how many hours do you watch on a typical day

Answer 1

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:

$z=\frac{X-\mu}{\sigma} \\1.282=\frac{X-2.8}{1.5}\\ X=4.72\ hours/day$  

On a typical day, you must watch at least 4.72 hours of TV to be in the upper 10%.