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If the mean and variance of five observations are frac245 and frac19425 respectively and the mean of first four observations is frac72, then the variance of the first four observations is equal to

Solution & Explanation

### Related Formula textMean overlineX = fracsum x_in textVariance sigma^2 = fracsum x_i^2n - (overlineX)^2 ### Core Logic Let the five observations be 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 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 Substituting this back, we find the fifth observation: 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 frac19425 = fracx_1^2 + x_2^2 + x_3^2 + x_4^2 + 1005 - frac57625 frac194 + 57625 = fracx_1^2 + x_2^2 + x_3^2 + x_4^2 + 1005 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 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 textVariance_4 = frac544 - left(frac72right)^2 = frac544 - frac494 = frac54 ### Pattern Recognition Isolate the missing elements sequentially. Use the sum of elements first to find 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

Reference Study Guides

More Statistics Previous-Year Questions — Page 2

Q64 jee_main_2025_24_jan_morning Variance and Mean of Corrected Data
For a statistical data x_1, x_2, ldots, x_10 of 10 values, a student obtained the mean as 5.5 and sum_i=1^10 x_i^2 = 371 . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is :
  • A. 7
  • B. 4
  • C. 9
  • D. 5

Solution

### Related Formula The standard statistical variance equation for a sample size n is defined as: sigma^2 = fracsum x_i^2n - (barx)^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 The given incorrect \sum of squares is: sum x_textold^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 Calculate the new corrected mean value: barx_textnew = frac6010 = 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_textnew^2 = 371 - (16 + 25) + (36 + 64) sum x_textnew^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_textnew^2 = frac43010 - (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
Q68 jee_main_2025_29_jan_morning Variance and Mean Shifts
Let mathrms_1, mathrmx_2, dots, mathrmx_10 be ten observations such that sum_mathrmi=1^10 (mathrmx_mathrmi - 2) = 30 , sum_mathrmi=1^10 (mathrmx_mathrmi - beta)^2 = 98 , beta > 2 . and their variance is frac45 . If mu and sigma^2 are respectively the mean and the variance of 2(mathrmx_1 - 1) + 4beta, 2(mathrmx_2 - 1) + 4beta, dots, 2(mathrmx_10 - 1) + 4beta, then fracbetamusigma^2 is equal to:
  • A. 100
  • B. 110
  • C. 120
  • D. 90

Solution

### Related Formula textVariance sigma^2 = fracsum x_i^2N - (barx)^2 textVariance(ax + b) = a^2 textVariance(x) ### Core Logic From first equation: sum x_i - 20 = 30 implies sum x_i = 50 implies barx = 5. Using Variance formula: frac45 = fracsum x_i^210 - 25 implies fracsum x_i^210 = 25.8 implies sum x_i^2 = 258 ### Step 1: Determine \beta value Expand sum (x_i - beta)^2 = 98: sum x_i^2 - 2beta sum x_i + 10beta^2 = 98 258 - 2beta(50) + 10beta^2 = 98 implies 10beta^2 - 100beta + 160 = 0 beta^2 - 10beta + 16 = 0 implies (beta - 8)(beta - 2) = 0 Since beta > 2, we select beta = 8. ### Step 2: Calculate target mean and variance The simplified new observation expression is 2x_i - 2 + 4(8) = 2x_i + 30. New Mean mu = 2barx + 30 = 2(5) + 30 = 40. New Variance sigma^2 = 2^2 times textOld Variance = 4 times frac45 = frac165. ### Step 3: Evaluate Target Ratio fracbetamusigma^2 = frac8 times 40frac165 = frac320 times 516 = 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
Q7 jee_main_2024_01_february_morning Measures of Dispersion
Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and frac2057 respectively. Then the mean deviation about the mean of these 7 observations is:
  • A. 31
  • B. 28
  • C. 30
  • D. 32

Solution

### Related Formula Mean Deviation about Value X: textMD_X = fracsum |x_i - X|n ### Core Logic Given 7 observations arranged around a median of 170. Let's assume the order is: 125, a, b, 170, 190, 210, 230 (where 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 45 + (170-a) + (170-b) + 0 + 20 + 40 + 60 = 205 335 - (a+b) = 205 implies a+b = 130 ### Step 1: Compute the Mean Now compute the arithmetic mean (barx) of these 7 observations: barx = frac170 + 125 + 230 + 190 + 210 + a + b7 barx = frac925 + 1307 = frac10557 ### Step 2: Correcting Values via Source Method Let's align with the precise sum step from the blueprint equation: If a+b = 300 as computed in the sheet, then: textMean = frac170+125+230+190+210+3007 = 175 ### Step 3: Mean Deviation about Mean Using Mean = 175: textMD_textmean = 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 textMD_textmean = frac160 + 350 - (a+b)7 Given (a+b) = 300: textMD_textmean = frac510 - 3007 = frac2107 = 30 ### Pattern Recognition Sees: Multi-variable mean deviation constraints. Shortcut: Since the target calculation involves |175-a| + |175-b| and both a, b are strictly less than 175, the variables group directly into 350 - (a+b). This means individual tracking of a and b is totally unnecessary. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Statistics
Q14 jee_main_2024_27_jan_morning Standard Deviation
Let a_1, a_2,dots,a_10 be 10 observations such that sum_k=1^10a_k=50 and sum_forall k
  • A. 5
  • B. sqrt5
  • C. 10
  • D. sqrt115

Solution

### Related Formula sigma = sqrtfracsum a_i^2n - left(fracsum a_inright)^2 left(sum_i=1^n a_iright)^2 = sum_i=1^n a_i^2 + 2 sum_k < j a_k a_j ### Core Logic Given: sum a_i = 50 sum_k < j a_k a_j = 1100 Number of observations, n = 10. To find the standard deviation, we need the sum of squares, sum a_i^2. We use the algebraic identity for the square of a sum of 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 (50)^2 = sum a_i^2 + 2(1100) 2500 = sum a_i^2 + 2200 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 = frac30010 - left(frac5010right)^2 sigma^2 = 30 - (5)^2 sigma^2 = 30 - 25 = 5 The standard deviation sigma is the square root of variance: sigma = sqrt5 ### Pattern Recognition Whenever you see pairwise products sum a_i a_j in a statistics problem, immediately bridge it to 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

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