If the mean deviation about the median of the numbers k, 2k, 3k, dots, 1000k$k, 2k, 3k, \dots, 1000k$ is 500$500$, then k^2$k^2$ is equal to:

### 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 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 set up algebraic equations using the definitions of mean and variance to determine the values of a+b$a+b$ and ab$ab$. ### Step 1: Apply the Mean condition Given mean barx = 9$\bar{x} = 9$ for N=8$N=8$ observations: frac6 + 4 + a + 8 + b + 12 + 10 + 138 = 9$$\frac{6 + 4 + a + 8 + b + 12 + 10 + 13}{8} = 9$$ 53 + a + b = 72 implies a + b = 19 quad text--- (1)$$53 + a + b = 72 \implies a + b = 19 \quad \text{--- (1)}$$ ### Step 2: Apply the Variance condition Given variance sigma^2 = 9.25 = frac374$\sigma^2 = 9.25 = \frac{37}{4}$: frac36 + 16 + a^2 + 64 + b^2 + 144 + 100 + 1698 - 81 = frac374$$\frac{36 + 16 + a^2 + 64 + b^2 + 144 + 100 + 169}{8} - 81 = \frac{37}{4}$$ frac529 + a^2 + b^28 = 81 + 9.25 = 90.25 = frac3614$$\frac{529 + a^2 + b^2}{8} = 81 + 9.25 = 90.25 = \frac{361}{4}$$ 529 + a^2 + b^2 = 722 implies a^2 + b^2 = 193 quad text--- (2)$$529 + a^2 + b^2 = 722 \implies a^2 + b^2 = 193 \quad \text{--- (2)}$$ ### Step 3: Solve for ab and calculate the target expression We know (a+b)^2 = a^2 + b^2 + 2ab$(a+b)^2 = a^2 + b^2 + 2ab$. Substitute equations (1) and (2): 19^2 = 193 + 2ab$$19^2 = 193 + 2ab$$ 361 = 193 + 2ab implies 2ab = 168 implies ab = 84$$361 = 193 + 2ab \implies 2ab = 168 \implies ab = 84$$ Now, calculate the target value: a + b + ab = 19 + 84 = 103$$a + b + ab = 19 + 84 = 103$$ ### Pattern Recognition Direct symmetric evaluation: Statistics problems in JEE with missing observations often ask for symmetric combinations of variables like a+b+ab$a+b+ab$ or a^2+b^2$a^2+b^2$. These can be calculated using quadratic expansions without solving for a$a$ and b$b$ individually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

### Related Formula Mean of N$N$ observations: barx = fracsum x_iN$$\bar{x} = \frac{\sum x_i}{N}$$ Variance of N$N$ observations: sigma^2 = fracsum x_i^2N - (barx)^2$$\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$$ ### Core Logic Given mean is 5$5$ for 5$5$ observations: frac1 + 3 + a + 7 + b5 = 5 implies a + b + 11 = 25 implies a + b = 14 quad text--- (1)$$\frac{1 + 3 + a + 7 + b}{5} = 5 \implies a + b + 11 = 25 \implies a + b = 14 \quad \text{--- (1)}$$ Given variance is 10$10$: frac1^2 + 3^2 + a^2 + 7^2 + b^25 - 5^2 = 10 implies frac59 + a^2 + b^25 = 35$$\frac{1^2 + 3^2 + a^2 + 7^2 + b^2}{5} - 5^2 = 10 \implies \frac{59 + a^2 + b^2}{5} = 35$$ a^2 + b^2 = 175 - 59 = 116 quad text--- (2)$$a^2 + b^2 = 175 - 59 = 116 \quad \text{--- (2)}$$ ### Step 1: Finding a$a$ and b$b$ Using standard algebraic identity (a+b)^2 = a^2 + b^2 + 2ab$(a+b)^2 = a^2 + b^2 + 2ab$: 14^2 = 116 + 2ab implies 196 = 116 + 2ab implies ab = 40$$14^2 = 116 + 2ab \implies 196 = 116 + 2ab \implies ab = 40$$ Solving a+b=14$a+b=14$ and ab=40$ab=40$: a(14-a) = 40 implies a^2 - 14a + 40 = 0 implies (a-10)(a-4) = 0$$a(14-a) = 40 \implies a^2 - 14a + 40 = 0 \implies (a-10)(a-4) = 0$$ Since a > b$a > b$, we obtain a = 10$a = 10$ and b = 4$b = 4$. ### Step 2: Constructing new set and finding variance The original set is x_1 = 1, x_2 = 3, x_3 = 10, x_4 = 7, x_5 = 4$x_1 = 1, x_2 = 3, x_3 = 10, x_4 = 7, x_5 = 4$. We construct the new set y_n = n + x_n$y_n = n + x_n$: - y_1 = 1 + 1 = 2$y_1 = 1 + 1 = 2$ - y_2 = 2 + 3 = 5$y_2 = 2 + 3 = 5$ - y_3 = 3 + 10 = 13$y_3 = 3 + 10 = 13$ - y_4 = 4 + 7 = 11$y_4 = 4 + 7 = 11$ - y_5 = 5 + 4 = 9$y_5 = 5 + 4 = 9$ Mean of new set: bary = frac2 + 5 + 13 + 11 + 95 = frac405 = 8$$\bar{y} = \frac{2 + 5 + 13 + 11 + 9}{5} = \frac{40}{5} = 8$$ Variance of new set: sigma_new^2 = frac2^2 + 5^2 + 13^2 + 11^2 + 9^25 - 8^2$$\sigma_{new}^2 = \frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5} - 8^2$$ sigma_new^2 = frac4 + 25 + 169 + 121 + 815 - 64 = frac4005 - 64 = 80 - 64 = 16$$\sigma_{new}^2 = \frac{4 + 25 + 169 + 121 + 81}{5} - 64 = \frac{400}{5} - 64 = 80 - 64 = 16$$ ### Pattern Recognition Note that adding a changing factor like +n$+n$ is different from adding a constant C$C$ to each observation (which leaves variance unchanged). In this case, calculate individual x_n$x_n$ variables directly first before applying transformations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics and Probability

### Related Formula Sum of all probabilities in a distribution: sum_x=0^infty P(X = x) = 1$$\sum_{x=0}^{\infty} P(X = x) = 1$$ Complementary probability: P(X ge 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$P(X \ge 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$ ### Core Logic Let's first determine the constant k$k$: k sum_x=0^infty (x+1) 3^-x = 1$$k \sum_{x=0}^{\infty} (x+1) 3^{-x} = 1$$ This is an Arithmetico-Geometric Progression (AGP). Let S = sum_x=0^infty (x+1)left(frac13right)^x$S = \sum_{x=0}^{\infty} (x+1)\left(\frac{1}{3}\right)^x$: S = 1 + frac23 + frac39 + frac427 + dots quad text--- (1)$$S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots \quad \text{--- (1)}$$ frac13S = frac13 + frac29 + frac327 + dots quad text--- (2)$$\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots \quad \text{--- (2)}$$ ### Step 1: Finding k$k$ Subtracting (2) from (1): Sleft(1 - frac13right) = 1 + frac13 + frac19 + frac127 + dots$$S\left(1 - \frac{1}{3}\right) = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$$ frac23S = frac11 - 1/3 = frac32 implies S = frac94$$\frac{2}{3}S = \frac{1}{1 - 1/3} = \frac{3}{2} \implies S = \frac{9}{4}$$ Substitute back: k cdot left(frac94right) = 1 implies k = frac49$$k \cdot \left(\frac{9}{4}\right) = 1 \implies k = \frac{4}{9}$$ ### Step 2: Calculating P(X ge 3)$P(X \ge 3)$ Calculate initial probability values: - P(X=0) = k(1)(1) = frac49$P(X=0) = k(1)(1) = \frac{4}{9}$ - P(X=1) = k(2)left(frac13right) = frac23 cdot frac49 = frac827$P(X=1) = k(2)\left(\frac{1}{3}\right) = \frac{2}{3} \cdot \frac{4}{9} = \frac{8}{27}$ - P(X=2) = k(3)left(frac19right) = frac13 cdot frac49 = frac427$P(X=2) = k(3)\left(\frac{1}{9}\right) = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27}$ textSum P(X < 3) = frac1227 + frac827 + frac427 = frac2427 = frac89$$\text{Sum } P(X < 3) = \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{24}{27} = \frac{8}{9}$$ P(X ge 3) = 1 - frac89 = frac19$$P(X \ge 3) = 1 - \frac{8}{9} = \frac{1}{9}$$ ### Pattern Recognition The infinite AGP sum with factor (x+1)r^x$(x+1)r^x$ always converges to frac1(1-r)^2$\frac{1}{(1-r)^2}$. Here r = 1/3$r = 1/3$, so the sum is frac1(2/3)^2 = 9/4$\frac{1}{(2/3)^2} = 9/4$. This mental check saves doing the full subtraction sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics and Probability Class 11 Mathematics: Sequences and Series