### 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 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

### Related Formula The standard statistical variance equation for a sample size n$n$ is defined as: sigma^2 = fracsum x_i^2n - (barx)^2$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ ### Core Logic Calculate the initial incorrect \sum of observations using the given incorrect mean: barx_textold = 5.5 = fracsum x_textold10 implies sum x_textold = 55$\bar{x}_{\text{old}} = 5.5 = \frac{\sum x_{\text{old}}}{10} \implies \sum x_{\text{old}} = 55$ The given incorrect \sum of squares is: sum x_textold^2 = 371$\sum x_{\text{old}}^2 = 371$ ### Step 1: Compute Corrected Sum of Observations Adjust the linear \sum by subtracting the incorrect inputs and adding the true values: sum x_textnew = 55 - (4 + 5) + (6 + 8) = 55 - 9 + 14 = 60$\sum x_{\text{new}} = 55 - (4 + 5) + (6 + 8) = 55 - 9 + 14 = 60$ Calculate the new corrected mean value: barx_textnew = frac6010 = 6$\bar{x}_{\text{new}} = \frac{60}{10} = 6$ ### Step 2: Compute Corrected Sum of Squares Adjust the \sum of squares by swapping the squared entries: sum x_textnew^2 = 371 - (4^2 + 5^2) + (6^2 + 8^2)$\sum x_{\text{new}}^2 = 371 - (4^2 + 5^2) + (6^2 + 8^2)$ sum x_textnew^2 = 371 - (16 + 25) + (36 + 64)$\sum x_{\text{new}}^2 = 371 - (16 + 25) + (36 + 64)$ sum x_textnew^2 = 371 - 41 + 100 = 430$\sum x_{\text{new}}^2 = 371 - 41 + 100 = 430$ ### Step 3: Calculate Corrected Variance Substitute the corrected values into the standard variance formula: sigma_textnew^2 = fracsum x_textnew^210 - (barx_textnew)^2$\sigma_{\text{new}}^2 = \frac{\sum x_{\text{new}}^2}{10} - (\bar{x}_{\text{new}})^2$ sigma_textnew^2 = frac43010 - (6)^2 = 43 - 36 = 7$\sigma_{\text{new}}^2 = \frac{430}{10} - (6)^2 = 43 - 36 = 7$ ### Pattern Recognition When updating statistical aggregates like mean and variance after data correction, always compute the corrected linear \sum and \sum of squares separately before recombining them into the variance formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics