1. **State the problem:** We are given a frequency distribution table with scores and frequencies involving a variable $x$: | Score | 2 | 4 | 5 | 6 | 7 | 8 | |-------|---|---|---|---|---|---| | Frequency | $x-1$ | 9 | 11 | $x+2$ | 9 | $x-5$ | The total score (sum of frequency times score) is 300. We need to find: (i) The variance of the scores. (ii) The standard deviation of the scores, correct to 2 decimal places. 2. **Find $x$ using the total score:** The total score is the sum of (score $\times$ frequency): $$ 2(x-1) + 4(9) + 5(11) + 6(x+2) + 7(9) + 8(x-5) = 300 $$ Simplify: $$ 2x - 2 + 36 + 55 + 6x + 12 + 63 + 8x - 40 = 300 $$ Combine like terms: $$ (2x + 6x + 8x) + (-2 + 36 + 55 + 12 + 63 - 40) = 300 $$ $$ 16x + 124 = 300 $$ Solve for $x$: $$ 16x = 300 - 124 = 176 $$ $$ x = \frac{176}{16} = 11 $$ 3. **Calculate frequencies with $x=11$:** - $x-1 = 11 - 1 = 10$ - $x+2 = 11 + 2 = 13$ - $x-5 = 11 - 5 = 6$ Frequencies: 10, 9, 11, 13, 9, 6 4. **Calculate mean $\bar{x}$:** $$ \bar{x} = \frac{\sum (score \times frequency)}{\sum frequency} $$ Sum of frequencies: $$ 10 + 9 + 11 + 13 + 9 + 6 = 58 $$ Sum of (score $\times$ frequency): $$ 2(10) + 4(9) + 5(11) + 6(13) + 7(9) + 8(6) = 20 + 36 + 55 + 78 + 63 + 48 = 300 $$ Mean: $$ \bar{x} = \frac{300}{58} \approx 5.1724 $$ 5. **Calculate variance $\sigma^2$:** Variance formula for grouped data: $$ \sigma^2 = \frac{\sum f (x_i - \bar{x})^2}{\sum f} $$ Calculate each $(x_i - \bar{x})^2$: - $(2 - 5.1724)^2 = 10.06$ - $(4 - 5.1724)^2 = 1.37$ - $(5 - 5.1724)^2 = 0.03$ - $(6 - 5.1724)^2 = 0.68$ - $(7 - 5.1724)^2 = 3.34$ - $(8 - 5.1724)^2 = 7.98$ Multiply by frequencies: - $10.06 \times 10 = 100.6$ - $1.37 \times 9 = 12.33$ - $0.03 \times 11 = 0.33$ - $0.68 \times 13 = 8.84$ - $3.34 \times 9 = 30.06$ - $7.98 \times 6 = 47.88$ Sum: $$ 100.6 + 12.33 + 0.33 + 8.84 + 30.06 + 47.88 = 199.04 $$ Variance: $$ \sigma^2 = \frac{199.04}{58} \approx 3.43 $$ 6. **Calculate standard deviation $\sigma$:** $$ \sigma = \sqrt{3.43} \approx 1.85 $$ **Final answers:** (i) Variance $\approx 3.43$ (ii) Standard deviation $\approx 1.85$