To determine what percentage of teenagers spend more than 3.1 hours on their brand A cell phone, we need to follow these steps:
1. Identify the given parameters:
- Mean ([tex]\(\mu\)[/tex]) = 2.5 hours
- Standard deviation ([tex]\(\sigma\)[/tex]) = 0.6 hours
- Time of interest = 3.1 hours
2. Calculate the z-score for the time of interest:
The z-score is a measure of how many standard deviations a particular value is from the mean. It can be calculated using the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
Substituting the given values:
[tex]\[
z = \frac{3.1 - 2.5}{0.6} = \frac{0.6}{0.6} = 1
\][/tex]
3. Find the cumulative probability associated with this z-score:
To find the percentage of teenagers who spend more than 3.1 hours on their cell phone, we need to look up the cumulative probability for a z-score of 1. This cumulative probability represents the area under the normal distribution curve to the left of [tex]\(z = 1\)[/tex].
From standard normal distribution tables or using statistical software, we can find the cumulative distribution function (CDF) value for [tex]\(z = 1\)[/tex]:
[tex]\[
\text{CDF}(1) \approx 0.8413
\][/tex]
4. Calculate the percentage of teenagers who spend more than 3.1 hours:
The cumulative probability we found (0.8413) represents the proportion of teenagers who spend up to 3.1 hours on their phone. To find the proportion who spend more than 3.1 hours, we subtract this value from 1:
[tex]\[
P(X > 3.1) = 1 - \text{CDF}(1) = 1 - 0.8413 = 0.1587
\][/tex]
5. Convert this probability to a percentage:
[tex]\[
0.1587 \times 100 \approx 15.87\%
\][/tex]
Therefore, about 16% of the teenagers spend more than 3.1 hours on their brand A cell phone. The correct choice from the given options is:
[tex]\[
\boxed{16\%}
\][/tex]
