To find the amount of time a person must spend on their phone to be in the top 15% of cell phone users, we will use the properties of the normal distribution.

Steps to solve:

1. Identify the known values:

   • Mean (µ) = 223 minutes
   • Standard deviation (σ) = 31 minutes
   • Top 15% means that the cutoff time is at the 85th percentile (since 100% - 15% = 85%).

2. Find the z-score corresponding to the 85th percentile:

   • From standard normal distribution tables, the z-score for the 85th percentile is approximately 1.036.

3. Use the z-score formula to solve for the corresponding time (X):

$z = \frac{X - \mu}{\sigma}$

Where:

• $z$ is the z-score (1.036),
• $\mu$ is the mean (223 minutes),
• $\sigma$ is the standard deviation (31 minutes),
• $X$ is the value we want to find.

Rearranging the formula to solve for $X$:

$X = z \cdot \sigma + \mu$

$X = 1.036 \cdot 31 + 223$

$X \approx 32.116 + 223 = 255.116$

Conclusion:

To be in the top 15% of cell phone users, you would need to spend approximately 255 minutes (or 4 hours and 15 minutes) on your phone.

Would you like further details or clarifications?

Here are 5 related questions:

1. How do we interpret z-scores in a normal distribution?
2. What is the significance of standard deviation in this context?
3. How would the result change if the standard deviation were larger?
4. What is the probability of spending less than 200 minutes on the phone?
5. How do percentiles relate to the normal distribution?

Tip: The area under a normal curve represents probabilities, and z-scores help us find how much of the data falls below a certain point!