### 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

### Related Formula Mean command: barx = fracsum x_in$\bar{x} = \frac{\sum x_i}{n}$ Variance command: sigma^2 = fracsum x_i^2n - (barx)^2$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ ### Core Logic Given values: n = 100, quad barx_textold = 40, quad sigma_textold = 5.1$n = 100, \quad \bar{x}_{\text{old}} = 40, \quad \sigma_{\text{old}} = 5.1$ Incorrect item = 50$50$, Correct replacement = 40$40$. Find the correct mean mu$\mu$: sum x_i,textold = 100 times 40 = 4000$\sum x_{i,\text{old}} = 100 \times 40 = 4000$ sum x_i,textcorrect = 4000 - 50 + 40 = 3990$\sum x_{i,\text{correct}} = 4000 - 50 + 40 = 3990$ mu = frac3990100 = 39.9$\mu = \frac{3990}{100} = 39.9$ ### Step 1: Compute Correct Variance From the incorrect variance formulation: sigma_textold^2 = (5.1)^2 = 26.01$\sigma_{\text{old}}^2 = (5.1)^2 = 26.01$ 26.01 = fracsum x_i,textold^2100 - (40)^2 = fracsum x_i,textold^2100 - 1600$26.01 = \frac{\sum x_{i,\text{old}}^2}{100} - (40)^2 = \frac{\sum x_{i,\text{old}}^2}{100} - 1600$ fracsum x_i,textold^2100 = 1626.01 implies sum x_i,textold^2 = 162601$\frac{\sum x_{i,\text{old}}^2}{100} = 1626.01 \implies \sum x_{i,\text{old}}^2 = 162601$ Adjust the squared summation block: sum x_i,textcorrect^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701$\sum x_{i,\text{correct}}^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701$ Now compute the updated variance sigma^2$\sigma^2$: sigma^2 = frac161701100 - (39.9)^2$\sigma^2 = \frac{161701}{100} - (39.9)^2$ sigma^2 = 1617.01 - 1592.01 = 25$\sigma^2 = 1617.01 - 1592.01 = 25$ sigma = sqrt25 = 5$\sigma = \sqrt{25} = 5$ ### Step 2: Final Calculation Substitute the evaluated correct parameters: 10(mu + sigma) = 10(39.9 + 5) = 10(44.9) = 449$10(\mu + \sigma) = 10(39.9 + 5) = 10(44.9) = 449$ ### Pattern Recognition Notice how the subtraction of 50^2$50^2$ and subsequent addition of 40^2$40^2$ directly reduces the total squared variance sum by an exact round value of 900$900$, landing beautifully on a perfect square root target value of 25$25$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

### Core Logic We have 8 observations: 2, 3, 3, 4, 5, 7, a, b$2, 3, 3, 4, 5, 7, a, b$. Given mean barx = 4$\bar{x} = 4$: frac2 + 3 + 3 + 4 + 5 + 7 + a + b8 = 4 implies 24 + a + b = 32 implies a + b = 8$\frac{2 + 3 + 3 + 4 + 5 + 7 + a + b}{8} = 4 \implies 24 + a + b = 32 \implies a + b = 8$ ### Step 1: Using the Variance Property Given standard deviation sigma = sqrt2 implies textVariance sigma^2 = 2$\sigma = \sqrt{2} \implies \text{Variance } \sigma^2 = 2$. The variance formula is sigma^2 = fracsum x_i^2n - (barx)^2$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$: 2 = frac2^2 + 3^2 + 3^2 + 4^2 + 5^2 + 7^2 + a^2 + b^28 - 4^2$2 = \frac{2^2 + 3^2 + 3^2 + 4^2 + 5^2 + 7^2 + a^2 + b^2}{8} - 4^2$ 2 = frac4 + 9 + 9 + 16 + 25 + 49 + a^2 + b^28 - 16 = frac112 + a^2 + b^28 - 16$2 = \frac{4 + 9 + 9 + 16 + 25 + 49 + a^2 + b^2}{8} - 16 = \frac{112 + a^2 + b^2}{8} - 16$ 2 + 16 = frac112 + a^2 + b^28 implies 18 times 8 = 112 + a^2 + b^2$2 + 16 = \frac{112 + a^2 + b^2}{8} \implies 18 \times 8 = 112 + a^2 + b^2$ 144 = 112 + a^2 + b^2 implies a^2 + b^2 = 32$144 = 112 + a^2 + b^2 \implies a^2 + b^2 = 32$ ### Step 2: Solving for a and b and finding the Mode We know (a+b)^2 = a^2 + b^2 + 2ab implies 8^2 = 32 + 2ab implies 64 = 32 + 2ab implies 2ab = 32 implies ab = 16$(a+b)^2 = a^2 + b^2 + 2ab \implies 8^2 = 32 + 2ab \implies 64 = 32 + 2ab \implies 2ab = 32 \implies ab = 16$. Solving a+b=8$a+b=8$ and ab=16$ab=16$ gives a=4$a=4$ and b=4$b=4$. The complete data set is \2, 3, 3, 4, 4, 4, 5, 7\$\{2, 3, 3, 4, 4, 4, 5, 7\}$. The value appearing with highest frequency is 4$4$ (occurs 3 times), so textMode = 4$\text{Mode} = 4$. ### Step 3: Calculating Mean Deviation about Mode Mean Deviation about Mode is: textM.D. = fracsum |x_i - textMode|n$\text{M.D.} = \frac{\sum |x_i - \text{Mode}|}{n}$ = frac|2-4| + |3-4| + |3-4| + |4-4| + |4-4| + |4-4| + |5-4| + |7-4|8$= \frac{|2-4| + |3-4| + |3-4| + |4-4| + |4-4| + |4-4| + |5-4| + |7-4|}{8}$ = frac2 + 1 + 1 + 0 + 0 + 0 + 1 + 38 = frac88 = 1$= \frac{2 + 1 + 1 + 0 + 0 + 0 + 1 + 3}{8} = \frac{8}{8} = 1$ ### Pattern Recognition When a+b=2sqrtab$a+b=2\sqrt{ab}$ (here 8 = 2sqrt16$8 = 2\sqrt{16}$), the roots are guaranteed to be equal, meaning a=b$a=b$. Recognizing this condition instantly avoids full quadratic polynomial substitution steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics