1. **State the problem:** We are given the mean and standard deviation for Grade 11 males' gaming hours and a frequency distribution for Grade 11 females. We need to calculate the mean and standard deviation for females and then compare both groups. 2. **Recall formulas:** - Mean for grouped data: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is frequency and $x_i$ is the midpoint of each interval. - Standard deviation for grouped data: $$\sigma = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}}$$ 3. **Calculate midpoints for female intervals:** - 3-5: midpoint $= \frac{3+5}{2} = 4$ - 5-7: midpoint $= 6$ - 7-9: midpoint $= 8$ - 9-11: midpoint $= 10$ - 11-13: midpoint $= 12$ - 13-15: midpoint $= 14$ 4. **Calculate total frequency:** $$N = 7 + 11 + 16 + 19 + 12 + 5 = 70$$ 5. **Calculate weighted sum of midpoints:** $$\sum f_i x_i = 7\times4 + 11\times6 + 16\times8 + 19\times10 + 12\times12 + 5\times14$$ $$= 28 + 66 + 128 + 190 + 144 + 70 = 626$$ 6. **Calculate mean for females:** $$\bar{x} = \frac{626}{70} \approx 8.94$$ 7. **Calculate squared deviations weighted by frequency:** Calculate each $(x_i - \bar{x})^2$: - $(4 - 8.94)^2 = 24.40$ - $(6 - 8.94)^2 = 8.64$ - $(8 - 8.94)^2 = 0.88$ - $(10 - 8.94)^2 = 1.12$ - $(12 - 8.94)^2 = 9.36$ - $(14 - 8.94)^2 = 25.62$ Multiply by frequencies: $$7\times24.40 + 11\times8.64 + 16\times0.88 + 19\times1.12 + 12\times9.36 + 5\times25.62$$ $$= 170.8 + 95.04 + 14.08 + 21.28 + 112.32 + 128.1 = 541.62$$ 8. **Calculate standard deviation for females:** $$\sigma = \sqrt{\frac{541.62}{70}} = \sqrt{7.74} \approx 2.78$$ 9. **Summary:** - Males: $\bar{x} = 12.84$, $\sigma = 2.16$ - Females: $\bar{x} \approx 8.94$, $\sigma \approx 2.78$ 10. **Interpretation:** Males play more hours on average (12.84 vs. 8.94). Females have a slightly higher variability in hours played (2.78 vs. 2.16). The distribution for females is more spread out, while males tend to cluster closer to their mean.