Solution & Explanation
### 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
More Statistics Previous-Year Questions
Q61
2025
Mean and Variance
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13$6, 4, a, 8, b, 12, 10, 13$ are 9$9$ and 9.25$9.25$ respectively, then a + b + ab$a + b + ab$ is equal to :
- A. 105
- B. 103
- C. 100
- D. 106
Solution
### 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
Q62
2025
Mean and Variance
The mean and standard deviation of 100 observations are 40 and 5.1, respectively, By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are mu$\mu$ and sigma$\sigma$ respectively, then 10(mu + sigma)$10(\mu + \sigma)$ is equal to
- A. 445$445$
- B. 451$451$
- C. 447$447$
- D. 449$449$
Solution
### 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
Q64
2025
Variance and Mean of Corrected Data
For a statistical data x_1, x_2, ldots, x_10$x_1, x_2, \ldots, x_{10}$ of 10$10$ values, a student obtained the mean as 5.5$5.5$ and sum_i=1^10 x_i^2 = 371$\sum_{i=1}^{10} x_i^2 = 371$ . He later found that he had noted two values in the data incorrectly as 4$4$ and 5$5$, instead of the correct values 6$6$ and 8$8$, respectively. The variance of the corrected data is :
- A. 7$7$
- B. 4$4$
- C. 9$9$
- D. 5$5$
Solution
### 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
Q68
2025
Variance and Mean Shifts
Let mathrms_1, mathrmx_2, dots, mathrmx_10$\mathrm{s}_1, \mathrm{x}_2, \dots, \mathrm{x}_{10}$ be ten observations such that sum_mathrmi=1^10 (mathrmx_mathrmi - 2) = 30$\sum_{\mathrm{i}=1}^{10} (\mathrm{x}_\mathrm{i} - 2) = 30$ , sum_mathrmi=1^10 (mathrmx_mathrmi - beta)^2 = 98$\sum_{\mathrm{i}=1}^{10} (\mathrm{x}_\mathrm{i} - \beta)^2 = 98$ , beta > 2$\beta > 2$ . and their variance is frac45$\frac{4}{5}$ . If mu$\mu$ and sigma^2$\sigma^2$ are respectively the mean and the variance of 2(mathrmx_1 - 1) + 4beta, 2(mathrmx_2 - 1) + 4beta, dots, 2(mathrmx_10 - 1) + 4beta,$2(\mathrm{x}_1 - 1) + 4\beta, 2(\mathrm{x}_2 - 1) + 4\beta, \dots, 2(\mathrm{x}_{10} - 1) + 4\beta,$ then fracbetamusigma^2$\frac{\beta\mu}{\sigma^2}$ is equal to:
- A. 100
- B. 110
- C. 120
- D. 90
Solution
### 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