1. The problem asks to find the 75th percentile ($P_{75}$) of the given radiation levels in $\frac{W}{kg}$. The 75th percentile is the value below which 75% of the data falls. 2. The data set has 50 values sorted in ascending order: $$0.20, 0.29, 0.30, 0.50, 0.56, 0.58, 0.62, 0.89, 0.90, 0.92, 0.93, 0.94, 0.98, 1.02, 1.11, 1.11, 1.14, 1.15, 1.18, 1.19, 1.19, 1.23, 1.23, 1.26, 1.28, 1.28, 1.29, 1.30, 1.34, 1.36, 1.36, 1.39, 1.41, 1.41, 1.44$$ 3. To find the position of the 75th percentile, use the formula: $$P_{75} = \text{value at position } k = \frac{75}{100} \times (n + 1)$$ where $n=50$ is the number of data points. 4. Calculate $k$: $$k = 0.75 \times (50 + 1) = 0.75 \times 51 = 38.25$$ 5. Since $k=38.25$ is not an integer, the 75th percentile lies between the 38th and 39th values in the ordered list. 6. Identify the 38th and 39th values: - 38th value = 1.41 - 39th value = 1.41 7. Interpolate to find $P_{75}$: $$P_{75} = 1.41 + 0.25 \times (1.41 - 1.41) = 1.41$$ 8. Therefore, the 75th percentile $P_{75}$ is: $$\boxed{1.41} \frac{W}{kg}$$ This means 75% of the cell phones have radiation levels at or below 1.41 $\frac{W}{kg}$.