The mean and variance of a data of 10$10$ observations are 10$10$ and 2$2$, respectively. If an observation alpha$\alpha$ in this data is replaced by beta$\beta$, then the mean and variance become 10.1$10.1$ and 1.99$1.99$, respectively. Then alpha + beta$\alpha + \beta$ equals.

### 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 Given Mean = 8$= 8$ for 7 observations: 2, 4, 10, x, 12, 14, y$2, 4, 10, x, 12, 14, y$. frac2 + 4 + 10 + x + 12 + 14 + y7 = 8$$\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$$ Rightarrow x + y + 42 = 56 Rightarrow x + y = 14 dots(1)$$\Rightarrow x + y + 42 = 56 \Rightarrow x + y = 14 \dots(1)$$ Given Variance = 16$= 16$: 16 = frac2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^27 - 8^2$$16 = \frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2$$ 16 + 64 = frac4 + 16 + 100 + x^2 + 144 + 196 + y^27$$16 + 64 = \frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7}$$ 80 times 7 = 460 + x^2 + y^2 Rightarrow 560 = 460 + x^2 + y^2 Rightarrow x^2 + y^2 = 100 dots(2)$$80 \times 7 = 460 + x^2 + y^2 \Rightarrow 560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100 \dots(2)$$ ### Step 1: Solve for x and y Using algebraic identity (x+y)^2 = x^2 + y^2 + 2xy$(x+y)^2 = x^2 + y^2 + 2xy$: 14^2 = 100 + 2xy Rightarrow 196 - 100 = 2xy Rightarrow 2xy = 96 Rightarrow xy = 48$$14^2 = 100 + 2xy \Rightarrow 196 - 100 = 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48$$ Since x+y=14$x+y=14$ and xy=48$xy=48$, roots of quadratic t^2 - 14t + 48 = 0$t^2 - 14t + 48 = 0$ are 8, 6$8, 6$. Given x > y$x > y$, we must select x = 8$x = 8$ and y = 6$y = 6$. ### Step 2: Construct the set and evaluate Probability The new set X$X$ is formed by \1, 2, 3, x-4, y, 5\$\{1, 2, 3, x-4, y, 5\}$. Substituting x=8$x=8$ and y=6$y=6$, we get \1, 2, 3, 4, 6, 5\$\{1, 2, 3, 4, 6, 5\}$. There are 6 distinct elements: \1, 2, 3, 4, 5, 6\$\{1, 2, 3, 4, 5, 6\}$. We choose two numbers without replacement. Total outcomes = 6 times 5 = 30$6 \times 5 = 30$ permutations (or binom62 = 15$\binom{6}{2} = 15$ combinations). Let's use combinations. Total ways to choose 2 numbers = binom62 = 15$\binom{6}{2} = 15$. We need the probability that the *smaller* number is less than 4$4$. P(textsmaller < 4) = 1 - P(textsmaller geq 4)$P(\text{smaller} < 4) = 1 - P(\text{smaller} \geq 4)$. ### Step 3: Final Calculation via Complement For the smaller number to be geq 4$\geq 4$, BOTH chosen numbers must be geq 4$\geq 4$. The available numbers geq 4$\geq 4$ in the set are \4, 5, 6\$\{4, 5, 6\}$ (Total 3 numbers). Ways to choose two numbers from these 3 is binom32 = 3$\binom{3}{2} = 3$. P(textsmaller geq 4) = frac315 = frac15$$P(\text{smaller} \geq 4) = \frac{3}{15} = \frac{1}{5}$$ P(textsmaller < 4) = 1 - frac15 = frac45$$P(\text{smaller} < 4) = 1 - \frac{1}{5} = \frac{4}{5}$$ ### Pattern Recognition When statistical problems ask for "at least one" or "minimum bounding", calculating the complement probability (e.g., both elements strictly exceeding the threshold) cuts combinatorial checks from 3+$3+$ cases down to exactly 1$1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics Class 12 Maths: Probability

### Related Formula Mean deviation about median is given by: textM.D. = fracsum_i=1^n |X_i - X_M|n$$\text{M.D.} = \frac{\sum_{i=1}^{n} |X_i - X_M|}{n}$$ ### Core Logic For 1000$1000$ numbers in A.P., median X_M = frac1001k2$X_M = \frac{1001k}{2}$. textM.D. = frac2 left(frack2 + frac3k2 + frac5k2 + dots + 500 text termsright)1000$$\text{M.D.} = \frac{2 \left(\frac{k}{2} + \frac{3k}{2} + \frac{5k}{2} + \dots + 500 \text{ terms}\right)}{1000}$$ textM.D. = frac2 cdot frack2 (500)^21000 = frac500k2$$\text{M.D.} = \frac{2 \cdot \frac{k}{2} (500)^2}{1000} = \frac{500k}{2}$$ ### Step 1: Evaluation of k and k^2 Given mean deviation is 500$500$: frac500k2 = 500 implies k = 2$$\frac{500k}{2} = 500 \implies k = 2$$ Therefore, k^2 = 4$k^2 = 4$. ### Pattern Recognition Use symmetry of A.P. around median to quickly evaluate absolute deviation sum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics

### Related Formula mu = fracsum x_iN$$\mu = \frac{\sum x_i}{N}$$ sigma^2 = fracsum x_i^2N - mu^2$$\sigma^2 = \frac{\sum x_i^2}{N} - \mu^2$$ ### Core Logic Calculate x+y$x+y$ using the mean formula: frac-10 - 7 - 1 + x + y + 9 + 2 + 168 = frac72$$\frac{-10 - 7 - 1 + x + y + 9 + 2 + 16}{8} = \frac{7}{2}$$ frac9 + x + y8 = frac72 Rightarrow 9 + x + y = 28 Rightarrow x + y = 19 dots (1)$$\frac{9 + x + y}{8} = \frac{7}{2} \Rightarrow 9 + x + y = 28 \Rightarrow x + y = 19 \dots (1)$$ ### Step 1: Calculate sum of squares using Variance Using the variance formula: frac(-10)^2 + (-7)^2 + (-1)^2 + x^2 + y^2 + 9^2 + 2^2 + 16^28 - left(frac72right)^2 = frac2934$$\frac{(-10)^2 + (-7)^2 + (-1)^2 + x^2 + y^2 + 9^2 + 2^2 + 16^2}{8} - \left(\frac{7}{2}\right)^2 = \frac{293}{4}$$ frac100 + 49 + 1 + x^2 + y^2 + 81 + 4 + 2568 - frac494 = frac2934$$\frac{100 + 49 + 1 + x^2 + y^2 + 81 + 4 + 256}{8} - \frac{49}{4} = \frac{293}{4}$$ frac491 + x^2 + y^28 = frac2934 + frac494 = frac3424 = frac6848$$\frac{491 + x^2 + y^2}{8} = \frac{293}{4} + \frac{49}{4} = \frac{342}{4} = \frac{684}{8}$$ 491 + x^2 + y^2 = 684 Rightarrow x^2 + y^2 = 193 dots (2)$$491 + x^2 + y^2 = 684 \Rightarrow x^2 + y^2 = 193 \dots (2)$$ ### Step 2: Solve for x and y From (1), y = 19 - x$y = 19 - x$. Substitute into (2): x^2 + (19 - x)^2 = 193$$x^2 + (19 - x)^2 = 193$$ x^2 + 361 - 38x + x^2 = 193$$x^2 + 361 - 38x + x^2 = 193$$ 2x^2 - 38x + 168 = 0 Rightarrow x^2 - 19x + 84 = 0$$2x^2 - 38x + 168 = 0 \Rightarrow x^2 - 19x + 84 = 0$$ (x - 12)(x - 7) = 0$$(x - 12)(x - 7) = 0$$ Thus, x = 12$x = 12$ and y = 7$y = 7$ (or vice-versa, which doesn't affect absolute differences). ### Step 3: Calculate the New Mean We need the mean of 4$4$ numbers: x, y, x + y + 1, |x - y|$x, y, x + y + 1, |x - y|$. Substitute x = 12, y = 7$x = 12, y = 7$: The numbers are 12, 7, (12+7+1), |12-7| = 12, 7, 20, 5$12, 7, (12+7+1), |12-7| = 12, 7, 20, 5$. New Mean = frac12 + 7 + 20 + 54 = frac444 = 11$\frac{12 + 7 + 20 + 5}{4} = \frac{44}{4} = 11$. ### Pattern Recognition Statistics questions mapping x+y$x+y$ and x^2+y^2$x^2+y^2$ inherently hide a quadratic symmetric system (x+y)^2 - 2xy = x^2+y^2$(x+y)^2 - 2xy = x^2+y^2$. Solving for the roots immediately grants the discrete terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics

### Related Formula mu = fracsum f_i x_isum f_i$$\mu = \frac{\sum f_i x_i}{\sum f_i}$$ sigma^2 = fracsum f_i x_i^2sum f_i - mu^2$$\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \mu^2$$ ### Core Logic Midpoints (x_i$x_i$): 6, 10, 14, 18$6, 10, 14, 18$. Frequencies (f_i$f_i$): 3, lambda, 4, 7$3, \lambda, 4, 7$. Sum of frequencies sum f_i = 3 + lambda + 4 + 7 = 14 + lambda$\sum f_i = 3 + \lambda + 4 + 7 = 14 + \lambda$. Calculate Mean (mu$\mu$): mu = frac6(3) + 10(lambda) + 14(4) + 18(7)14 + lambda$$\mu = \frac{6(3) + 10(\lambda) + 14(4) + 18(7)}{14 + \lambda}$$ mu = frac18 + 10lambda + 56 + 12614 + lambda = frac10lambda + 200lambda + 14 = 10 + frac60lambda + 14$$\mu = \frac{18 + 10\lambda + 56 + 126}{14 + \lambda} = \frac{10\lambda + 200}{\lambda + 14} = 10 + \frac{60}{\lambda + 14}$$ Since data is standard, mu$\mu$ and lambda$\lambda$ typically hold integer values, hinting lambda + 14$\lambda + 14$ must be a factor of 60. Therefore lambda$\lambda$ could be 1, 6, 16$1, 6, 16$. ### Step 1: Applying Variance Variance sigma^2 = 19$\sigma^2 = 19$. sigma^2 = fracsum f_i x_i^214 + lambda - mu^2$$\sigma^2 = \frac{\sum f_i x_i^2}{14 + \lambda} - \mu^2$$ sum f_i x_i^2 = 3(36) + lambda(100) + 4(196) + 7(324) = 108 + 100lambda + 784 + 2268 = 100lambda + 3160$$\sum f_i x_i^2 = 3(36) + \lambda(100) + 4(196) + 7(324) = 108 + 100\lambda + 784 + 2268 = 100\lambda + 3160$$ Substitute into variance: 19 = frac100lambda + 3160lambda + 14 - left(frac10lambda + 200lambda + 14right)^2$$19 = \frac{100\lambda + 3160}{\lambda + 14} - \left(\frac{10\lambda + 200}{\lambda + 14}\right)^2$$ ### Step 2: Testing Candidates Let's test the potential integer lambda = 6$\lambda = 6$ (a factor of 60 is 14+6=20$14+6=20$): If lambda = 6$\lambda = 6$: mu = 10 + frac6020 = 10 + 3 = 13$$\mu = 10 + \frac{60}{20} = 10 + 3 = 13$$ Let's verify variance for lambda = 6$\lambda = 6$: sigma^2 = frac100(6) + 316020 - 13^2 = frac376020 - 169 = 188 - 169 = 19$$\sigma^2 = \frac{100(6) + 3160}{20} - 13^2 = \frac{3760}{20} - 169 = 188 - 169 = 19$$ This perfectly matches the given variance. Thus, lambda = 6$\lambda = 6$ and mu = 13$\mu = 13$. lambda + mu = 6 + 13 = 19$$\lambda + \mu = 6 + 13 = 19$$ ### Pattern Recognition Instead of solving a horrible cubic/quadratic equation for the variance parameter, use the integer division trait of the mean to test valid candidates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics