A consumer products research website ranks the smartphones currently available according to the longest battery life. The website conducts a test of active use and records the length of the smartphones’ battery life. In order for a smartphone to make the

To determine the mean battery life required for a smartphone to be in the top 15% of battery lives tested, we can use the properties of the normal distribution. We know that the battery life follows a normal distribution, and the cutoff for the top 15% is at least 45 hours. 1. **Identify the Z-score corresponding to the top 15%**: The top 15% corresponds to the 85th percentile of the distribution (since the top 15% means that 85% of values are below this point). We can use a standard normal distribution table or calculator to find the Z-score that corresponds to the 85th percentile. The Z-score for the 85th percentile is approximately **1.036**. 2. **Use the Z-score formula**: The Z-score is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where: - \(Z\) is the Z-score, - \(X\) is the value in the distribution (in this case, 45 hours), - \(\mu\) is the mean we want to find, and - \(\sigma\) is the standard deviation (given as 4.9 hours). 3. **Plug in the values we have**: \[ 1.036 = \frac{45 - \mu}{4.9} \] 4. **Solve for \(\mu\)**: Rearranging the equation gives: \[ 45 - \mu = 1.036 \times 4.9 \] Calculating \(1.036 \times 4.9\): \[ 1.036 \times 4.9 \approx 5.0834 \] Now, substituting this value back in: \[ 45 - \mu = 5.0834 \] Rearranging gives: \[ \mu = 45 - 5.0834 \approx 39.9166 \] Therefore, the approximate mean battery life during their test should be **about 39.92 hours** for a smartphone to make the top 15% list.