1. **State the problem:** We have a frequency distribution with scores and frequencies involving $x$, and the total score sum is 300. We need to find the variance and standard deviation of the scores. 2. **Set up the frequency table:** | Score ($x_i$) | Frequency ($f_i$) | |---|---| | 2 | $x-1$ | | 4 | 9 | | 5 | 11 | | 6 | $x+2$ | | 7 | 9 | | 8 | $x-5$ | 3. **Find $x$ using total frequency:** Total frequency $= (x-1) + 9 + 11 + (x+2) + 9 + (x-5) = 3x + 25$ 4. **Find $x$ using total score sum:** Total score sum $= 2(x-1) + 4(9) + 5(11) + 6(x+2) + 7(9) + 8(x-5) = 300$ Simplify: $$2x - 2 + 36 + 55 + 6x + 12 + 63 + 8x - 40 = 300$$ $$ (2x + 6x + 8x) + (-2 + 36 + 55 + 12 + 63 - 40) = 300$$ $$16x + 124 = 300$$ $$16x = 176$$ $$x = 11$$ 5. **Calculate frequencies with $x=11$:** - $f_1 = 11 - 1 = 10$ - $f_4 = 11 + 2 = 13$ - $f_6 = 11 - 5 = 6$ 6. **Calculate mean ($\bar{x}$):** Total frequency $= 10 + 9 + 11 + 13 + 9 + 6 = 58$ Sum of scores $= 2(10) + 4(9) + 5(11) + 6(13) + 7(9) + 8(6) = 20 + 36 + 55 + 78 + 63 + 48 = 300$ Mean: $$\bar{x} = \frac{300}{58} \approx 5.17$$ 7. **Calculate variance:** Use formula: $$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$$ Calculate each term: - $(2 - 5.17)^2 = 10.03$, contribution $= 10 \times 10.03 = 100.3$ - $(4 - 5.17)^2 = 1.37$, contribution $= 9 \times 1.37 = 12.33$ - $(5 - 5.17)^2 = 0.03$, contribution $= 11 \times 0.03 = 0.33$ - $(6 - 5.17)^2 = 0.69$, contribution $= 13 \times 0.69 = 8.97$ - $(7 - 5.17)^2 = 3.34$, contribution $= 9 \times 3.34 = 30.06$ - $(8 - 5.17)^2 = 7.98$, contribution $= 6 \times 7.98 = 47.88$ Sum contributions: $$100.3 + 12.33 + 0.33 + 8.97 + 30.06 + 47.88 = 199.87$$ Variance: $$\sigma^2 = \frac{199.87}{58} \approx 3.45$$ 8. **Calculate standard deviation:** $$\sigma = \sqrt{3.45} \approx 1.86$$ **Final answers:** - Variance $\approx 3.45$ - Standard deviation $\approx 1.86$ (to 2 decimal places)