Question

the amount of daily time that teenagers spend on a brand a cell phone is normally distributed with a given mean $mu = 2.5$ hr and standard deviation $sigma = 0.6$ hr. what percentage of the teenagers spend more than 3.1 hr?○ 5%○ 10%○ 16%○ 32%

Explanation:

Step1: Calculate z-score

The z-score formula is $z = \frac{x - \mu}{\sigma}$. Substitute $x=3.1$, $\mu=2.5$, $\sigma=0.6$:
$z = \frac{3.1 - 2.5}{0.6} = 1$

Step2: Use empirical rule

For normal distribution, ~68% of data lies within $\mu\pm\sigma$ (z=-1 to z=1). So data above z=1 is $\frac{100\% - 68\%}{2} = 16\%$.

Answer:

16% (Option C)