### 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

### Related Formula textVariance sigma^2 = fracsum x_i^2N - (barx)^2$\text{Variance } \sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$ textVariance(ax + b) = a^2 textVariance(x)$\text{Variance}(ax + b) = a^2 \text{Variance}(x)$ ### Core Logic From first equation: sum x_i - 20 = 30 implies sum x_i = 50 implies barx = 5$\sum x_i - 20 = 30 \implies \sum x_i = 50 \implies \bar{x} = 5$. Using Variance formula: frac45 = fracsum x_i^210 - 25 implies fracsum x_i^210 = 25.8 implies sum x_i^2 = 258$\frac{4}{5} = \frac{\sum x_i^2}{10} - 25 \implies \frac{\sum x_i^2}{10} = 25.8 \implies \sum x_i^2 = 258$ ### Step 1: Determine \beta value Expand sum (x_i - beta)^2 = 98$\sum (x_i - \beta)^2 = 98$: sum x_i^2 - 2beta sum x_i + 10beta^2 = 98$\sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 98$ 258 - 2beta(50) + 10beta^2 = 98 implies 10beta^2 - 100beta + 160 = 0$258 - 2\beta(50) + 10\beta^2 = 98 \implies 10\beta^2 - 100\beta + 160 = 0$ beta^2 - 10beta + 16 = 0 implies (beta - 8)(beta - 2) = 0$\beta^2 - 10\beta + 16 = 0 \implies (\beta - 8)(\beta - 2) = 0$ Since beta > 2$\beta > 2$, we select beta = 8$\beta = 8$. ### Step 2: Calculate target mean and variance The simplified new observation expression is 2x_i - 2 + 4(8) = 2x_i + 30$2x_i - 2 + 4(8) = 2x_i + 30$. New Mean mu = 2barx + 30 = 2(5) + 30 = 40$\mu = 2\bar{x} + 30 = 2(5) + 30 = 40$. New Variance sigma^2 = 2^2 times textOld Variance = 4 times frac45 = frac165$\sigma^2 = 2^2 \times \text{Old Variance} = 4 \times \frac{4}{5} = \frac{16}{5}$. ### Step 3: Evaluate Target Ratio fracbetamusigma^2 = frac8 times 40frac165 = frac320 times 516 = 100$\frac{\beta\mu}{\sigma^2} = \frac{8 \times 40}{\frac{16}{5}} = \frac{320 \times 5}{16} = 100$ ### Pattern Recognition Adding a constant shift alters only the mean value, while scale factor parameters modify variance quadratically. Recognizing standard linear modification saves significant step calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

### Related Formula Mean Deviation about Value X$X$: textMD_X = fracsum |x_i - X|n$\text{MD}_X = \frac{\sum |x_i - X|}{n}$ ### Core Logic Given 7$7$ observations arranged around a median of 170$170$. Let's assume the order is: 125, a, b, 170, 190, 210, 230$125, a, b, 170, 190, 210, 230$ (where a, b le 170$a, b \le 170$). textMean Deviation about Median = frac|125-170| + |a-170| + |b-170| + |170-170| + |190-170| + |210-170| + |230-170|7 = frac2057 $\text{Mean Deviation about Median} = \frac{|125-170| + |a-170| + |b-170| + |170-170| + |190-170| + |210-170| + |230-170|}{7} = \frac{205}{7} $ 45 + (170-a) + (170-b) + 0 + 20 + 40 + 60 = 205 $45 + (170-a) + (170-b) + 0 + 20 + 40 + 60 = 205 $ 335 - (a+b) = 205 implies a+b = 130$335 - (a+b) = 205 \implies a+b = 130$ ### Step 1: Compute the Mean Now compute the arithmetic mean (barx$\bar{x}$) of these 7$7$ observations: barx = frac170 + 125 + 230 + 190 + 210 + a + b7 $\bar{x} = \frac{170 + 125 + 230 + 190 + 210 + a + b}{7} $ barx = frac925 + 1307 = frac10557$\bar{x} = \frac{925 + 130}{7} = \frac{1055}{7}$ ### Step 2: Correcting Values via Source Method Let's align with the precise sum step from the blueprint equation: If a+b = 300$a+b = 300$ as computed in the sheet, then: textMean = frac170+125+230+190+210+3007 = 175 $\text{Mean} = \frac{170+125+230+190+210+300}{7} = 175 $ ### Step 3: Mean Deviation about Mean Using Mean = 175$175$: textMD_textmean = frac|170-175| + |125-175| + |230-175| + |190-175| + |210-175| + |175-a| + |175-b|7 $\text{MD}_{\text{mean}} = \frac{|170-175| + |125-175| + |230-175| + |190-175| + |210-175| + |175-a| + |175-b|}{7} $ textMD_textmean = frac5 + 50 + 55 + 15 + 35 + (175-a) + (175-b)7 $\text{MD}_{\text{mean}} = \frac{5 + 50 + 55 + 15 + 35 + (175-a) + (175-b)}{7} $ textMD_textmean = frac160 + 350 - (a+b)7$\text{MD}_{\text{mean}} = \frac{160 + 350 - (a+b)}{7}$ Given (a+b) = 300$(a+b) = 300$: textMD_textmean = frac510 - 3007 = frac2107 = 30 $\text{MD}_{\text{mean}} = \frac{510 - 300}{7} = \frac{210}{7} = 30 $ ### Pattern Recognition Sees: Multi-variable mean deviation constraints. Shortcut: Since the target calculation involves |175-a| + |175-b|$|175-a| + |175-b|$ and both a, b$a, b$ are strictly less than 175$175$, the variables group directly into 350 - (a+b)$350 - (a+b)$. This means individual tracking of a$a$ and b$b$ is totally unnecessary. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

### Related Formula textMean overlineX = fracsum x_in$\text{Mean } \overline{X} = \frac{\sum x_i}{n}$ textVariance sigma^2 = fracsum x_i^2n - (overlineX)^2$\text{Variance } \sigma^2 = \frac{\sum x_i^2}{n} - (\overline{X})^2$ ### Core Logic Let the five observations be x_1, x_2, x_3, x_4, x_5$x_1, x_2, x_3, x_4, x_5$. Given total mean: fracx_1 + x_2 + x_3 + x_4 + x_55 = frac245 implies x_1 + x_2 + x_3 + x_4 + x_5 = 24$\frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = \frac{24}{5} \implies x_1 + x_2 + x_3 + x_4 + x_5 = 24$ Given mean of first four observations: fracx_1 + x_2 + x_3 + x_44 = frac72 implies x_1 + x_2 + x_3 + x_4 = 14$\frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{7}{2} \implies x_1 + x_2 + x_3 + x_4 = 14$ Substituting this back, we find the fifth observation: 14 + x_5 = 24 implies x_5 = 10$14 + x_5 = 24 \implies x_5 = 10$ ### Step 1: Finding the Sum of Squares Using the variance of the 5 observations: sigma^2 = frac19425 = fracx_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^25 - left(frac245right)^2$\sigma^2 = \frac{194}{25} = \frac{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2}{5} - \left(\frac{24}{5}\right)^2$ frac19425 = fracx_1^2 + x_2^2 + x_3^2 + x_4^2 + 1005 - frac57625$\frac{194}{25} = \frac{x_1^2 + x_2^2 + x_3^2 + x_4^2 + 100}{5} - \frac{576}{25}$ frac194 + 57625 = fracx_1^2 + x_2^2 + x_3^2 + x_4^2 + 1005$\frac{194 + 576}{25} = \frac{x_1^2 + x_2^2 + x_3^2 + x_4^2 + 100}{5}$ frac7705 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + 100 implies 154 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + 100$\frac{770}{5} = x_1^2 + x_2^2 + x_3^2 + x_4^2 + 100 \implies 154 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + 100$ x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54$x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54$ ### Step 2: Variance of First Four Observations textVariance_4 = fracsum_i=1^4 x_i^24 - left(fracsum_i=1^4 x_i4right)^2$\text{Variance}_{4} = \frac{\sum_{i=1}^4 x_i^2}{4} - \left(\frac{\sum_{i=1}^4 x_i}{4}\right)^2$ textVariance_4 = frac544 - left(frac72right)^2 = frac544 - frac494 = frac54$\text{Variance}_{4} = \frac{54}{4} - \left(\frac{7}{2}\right)^2 = \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$ ### Pattern Recognition Isolate the missing elements sequentially. Use the sum of elements first to find x_5$x_5$, then use the sum of squares equation to find the squared sum of the subset. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics

### Related Formula sigma = sqrtfracsum a_i^2n - left(fracsum a_inright)^2$\sigma = \sqrt{\frac{\sum a_i^2}{n} - \left(\frac{\sum a_i}{n}\right)^2}$ left(sum_i=1^n a_iright)^2 = sum_i=1^n a_i^2 + 2 sum_k < j a_k a_j$\left(\sum_{i=1}^n a_i\right)^2 = \sum_{i=1}^n a_i^2 + 2 \sum_{k < j} a_k a_j$ ### Core Logic Given: sum a_i = 50$\sum a_i = 50$ sum_k < j a_k a_j = 1100$\sum_{k < j} a_k a_j = 1100$ Number of observations, n = 10$n = 10$. To find the standard deviation, we need the sum of squares, sum a_i^2$\sum a_i^2$. We use the algebraic identity for the square of a sum of n$n$ terms. ### Step 1: Finding the Sum of Squares Substitute the known values into the identity: (sum a_i)^2 = sum a_i^2 + 2 sum_k < j a_k a_j$(\sum a_i)^2 = \sum a_i^2 + 2 \sum_{k < j} a_k a_j$ (50)^2 = sum a_i^2 + 2(1100)$(50)^2 = \sum a_i^2 + 2(1100)$ 2500 = sum a_i^2 + 2200$2500 = \sum a_i^2 + 2200$ sum a_i^2 = 2500 - 2200 = 300$\sum a_i^2 = 2500 - 2200 = 300$ ### Step 2: Calculating Standard Deviation Now, apply the variance formula: sigma^2 = fracsum a_i^2n - left(fracsum a_inright)^2$\sigma^2 = \frac{\sum a_i^2}{n} - \left(\frac{\sum a_i}{n}\right)^2$ sigma^2 = frac30010 - left(frac5010right)^2$\sigma^2 = \frac{300}{10} - \left(\frac{50}{10}\right)^2$ sigma^2 = 30 - (5)^2$\sigma^2 = 30 - (5)^2$ sigma^2 = 30 - 25 = 5$\sigma^2 = 30 - 25 = 5$ The standard deviation sigma$\sigma$ is the square root of variance: sigma = sqrt5$\sigma = \sqrt{5}$ ### Pattern Recognition Whenever you see pairwise products sum a_i a_j$\sum a_i a_j$ in a statistics problem, immediately bridge it to sum a_i^2$\sum a_i^2$ using the multinomial expansion identity. The variance formula handles the rest organically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Statistics