### Related Formula textMean (barx) = fracsum x_in$\text{Mean } (\bar{x}) = \frac{\sum x_i}{n}$ textVariance (sigma^2) = fracsum x_i^2n - (barx)^2$\text{Variance } (\sigma^2) = \frac{\sum x_i^2}{n} - (\bar{x})^2$ ### Core Logic We are given 10 observations: 65, 68, 58, 44, 48, 45, 60, alpha, beta, 60$65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$. Total n=10$n=10$. Sum of known observations: S = 65 + 68 + 58 + 44 + 48 + 45 + 60 + 60 = 448$S = 65 + 68 + 58 + 44 + 48 + 45 + 60 + 60 = 448$ The mean barx = 56$\bar{x} = 56$: frac448 + alpha + beta10 = 56$\frac{448 + \alpha + \beta}{10} = 56$ 448 + alpha + beta = 560$448 + \alpha + \beta = 560$ alpha + beta = 112$\alpha + \beta = 112$ ### Step 1: Use Variance Equation The variance sigma^2 = 66.2$\sigma^2 = 66.2$. Using the computational formula for variance: fracsum x_i^210 - (56)^2 = 66.2$\frac{\sum x_i^2}{10} - (56)^2 = 66.2$ Calculate the sum of squares of known observations: sum x_known^2 = 65^2 + 68^2 + 58^2 + 44^2 + 48^2 + 45^2 + 60^2 + 60^2$\sum x_{known}^2 = 65^2 + 68^2 + 58^2 + 44^2 + 48^2 + 45^2 + 60^2 + 60^2$ = 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600$= 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600$ = 25678$= 25678$ Insert this into the variance equation: frac25678 + alpha^2 + beta^210 - 3136 = 66.2$\frac{25678 + \alpha^2 + \beta^2}{10} - 3136 = 66.2$ ### Step 2: Solve for Squares Sum Isolate alpha^2 + beta^2$\alpha^2 + \beta^2$: frac25678 + alpha^2 + beta^210 = 3136 + 66.2$\frac{25678 + \alpha^2 + \beta^2}{10} = 3136 + 66.2$ frac25678 + alpha^2 + beta^210 = 3202.2$\frac{25678 + \alpha^2 + \beta^2}{10} = 3202.2$ 25678 + alpha^2 + beta^2 = 32022$25678 + \alpha^2 + \beta^2 = 32022$ alpha^2 + beta^2 = 32022 - 25678$\alpha^2 + \beta^2 = 32022 - 25678$ alpha^2 + beta^2 = 6344$\alpha^2 + \beta^2 = 6344$ ### Pattern Recognition If a question asks solely for alpha^2+beta^2$\alpha^2+\beta^2$ given mean and variance, you do not need to solve the complex polynomial system to find the individual values of alpha$\alpha$ and beta$\beta$. The variance equation isolates alpha^2+beta^2$\alpha^2+\beta^2$ automatically as a single chunk. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

### Related Formula M = l + left( fracfracN2 - cff right) times h$M = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ ### Core Logic Constructing the Cumulative Frequency (CF) table: Total frequency N = 36$N = 36$. Therefore, fracN2 = 18$\frac{N}{2} = 18$. ### Step 1: Identifying median class Since 18$18$ lies in the cumulative frequency interval > 12$> 12$ and leq 22$\leq 22$, the median class is 8-12$8-12$. Here, l = 8$l = 8$ (lower limit), cf = 12$cf = 12$ (CF of previous class), f = 10$f = 10$ (frequency of current class), h = 4$h = 4$ (class size). ### Step 2: Calculating median M = 8 + left( frac18 - 1210 right) times 4$M = 8 + \left( \frac{18 - 12}{10} \right) \times 4$ M = 8 + frac610 times 4$M = 8 + \frac{6}{10} \times 4$ M = 8 + 2.4 = 10.4$M = 8 + 2.4 = 10.4$ We need to find 20M$20M$: 20M = 20 times 10.4 = 208$20M = 20 \times 10.4 = 208$ ### Pattern Recognition Finding the cumulative frequency sequence safely identifies the median class. Plugging into the standard linear interpolation formula yields the exact median. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics

### Related Formula textMean barx = fracsum x_in$\text{Mean } \bar{x} = \frac{\sum x_i}{n}$ textVariance sigma^2 = fracsum (x_i - barx)^2n$\text{Variance } \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$ ### Core Logic Mean is 55: fraca + b + 68 + 44 + 48 + 606 = 55$\frac{a + b + 68 + 44 + 48 + 60}{6} = 55$ 220 + a + b = 330 implies a + b = 110$220 + a + b = 330 \implies a + b = 110$ Variance is 194: frac(a-55)^2 + (b-55)^2 + (68-55)^2 + (44-55)^2 + (48-55)^2 + (60-55)^26 = 194$\frac{(a-55)^2 + (b-55)^2 + (68-55)^2 + (44-55)^2 + (48-55)^2 + (60-55)^2}{6} = 194$ (a-55)^2 + (b-55)^2 + 13^2 + (-11)^2 + (-7)^2 + 5^2 = 1164$(a-55)^2 + (b-55)^2 + 13^2 + (-11)^2 + (-7)^2 + 5^2 = 1164$ (a-55)^2 + (b-55)^2 + 169 + 121 + 49 + 25 = 1164$(a-55)^2 + (b-55)^2 + 169 + 121 + 49 + 25 = 1164$ (a-55)^2 + (b-55)^2 = 800$(a-55)^2 + (b-55)^2 = 800$ Expand the squares using a+b=110$a+b=110$: a^2 + b^2 - 110(a+b) + 2(3025) = 800$a^2 + b^2 - 110(a+b) + 2(3025) = 800$ a^2 + b^2 - 110(110) + 6050 = 800 implies a^2 + b^2 = 6850$a^2 + b^2 - 110(110) + 6050 = 800 \implies a^2 + b^2 = 6850$ Using (a+b)^2 = 12100 implies a^2+b^2+2ab = 12100 implies 2ab = 12100 - 6850 = 5250$(a+b)^2 = 12100 \implies a^2+b^2+2ab = 12100 \implies 2ab = 12100 - 6850 = 5250$. (a-b)^2 = a^2+b^2 - 2ab = 6850 - 5250 = 1600 implies a-b = 40$(a-b)^2 = a^2+b^2 - 2ab = 6850 - 5250 = 1600 \implies a-b = 40$ (since a>b$a>b$). Solving a+b=110$a+b=110$ and a-b=40$a-b=40$: a = 75, quad b = 35$a = 75, \quad b = 35$ Finally, evaluate a + 3b$a + 3b$: a + 3b = 75 + 3(35) = 180$a + 3b = 75 + 3(35) = 180$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics