Solution & Explanation
### 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
Given Mean = 8$= 8$ for 7 observations: 2, 4, 10, x, 12, 14, y$2, 4, 10, x, 12, 14, y$.
frac2 + 4 + 10 + x + 12 + 14 + y7 = 8$$\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$$
Rightarrow x + y + 42 = 56 Rightarrow x + y = 14 dots(1)$$\Rightarrow x + y + 42 = 56 \Rightarrow x + y = 14 \dots(1)$$
Given Variance = 16$= 16$:
16 = frac2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^27 - 8^2$$16 = \frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2$$
16 + 64 = frac4 + 16 + 100 + x^2 + 144 + 196 + y^27$$16 + 64 = \frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7}$$
80 times 7 = 460 + x^2 + y^2 Rightarrow 560 = 460 + x^2 + y^2 Rightarrow x^2 + y^2 = 100 dots(2)$$80 \times 7 = 460 + x^2 + y^2 \Rightarrow 560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100 \dots(2)$$
### Step 1: Solve for x and y
Using algebraic identity (x+y)^2 = x^2 + y^2 + 2xy$(x+y)^2 = x^2 + y^2 + 2xy$:
14^2 = 100 + 2xy Rightarrow 196 - 100 = 2xy Rightarrow 2xy = 96 Rightarrow xy = 48$$14^2 = 100 + 2xy \Rightarrow 196 - 100 = 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48$$
Since x+y=14$x+y=14$ and xy=48$xy=48$, roots of quadratic t^2 - 14t + 48 = 0$t^2 - 14t + 48 = 0$ are 8, 6$8, 6$.
Given x > y$x > y$, we must select x = 8$x = 8$ and y = 6$y = 6$.
### Step 2: Construct the set and evaluate Probability
The new set X$X$ is formed by \1, 2, 3, x-4, y, 5\$\{1, 2, 3, x-4, y, 5\}$.
Substituting x=8$x=8$ and y=6$y=6$, we get \1, 2, 3, 4, 6, 5\$\{1, 2, 3, 4, 6, 5\}$.
There are 6 distinct elements: \1, 2, 3, 4, 5, 6\$\{1, 2, 3, 4, 5, 6\}$.
We choose two numbers without replacement. Total outcomes = 6 times 5 = 30$6 \times 5 = 30$ permutations (or binom62 = 15$\binom{6}{2} = 15$ combinations).
Let's use combinations. Total ways to choose 2 numbers = binom62 = 15$\binom{6}{2} = 15$.
We need the probability that the *smaller* number is less than 4$4$.
P(textsmaller < 4) = 1 - P(textsmaller geq 4)$P(\text{smaller} < 4) = 1 - P(\text{smaller} \geq 4)$.
### Step 3: Final Calculation via Complement
For the smaller number to be geq 4$\geq 4$, BOTH chosen numbers must be geq 4$\geq 4$.
The available numbers geq 4$\geq 4$ in the set are \4, 5, 6\$\{4, 5, 6\}$ (Total 3 numbers).
Ways to choose two numbers from these 3 is binom32 = 3$\binom{3}{2} = 3$.
P(textsmaller geq 4) = frac315 = frac15$$P(\text{smaller} \geq 4) = \frac{3}{15} = \frac{1}{5}$$
P(textsmaller < 4) = 1 - frac15 = frac45$$P(\text{smaller} < 4) = 1 - \frac{1}{5} = \frac{4}{5}$$
### Pattern Recognition
When statistical problems ask for "at least one" or "minimum bounding", calculating the complement probability (e.g., both elements strictly exceeding the threshold) cuts combinatorial checks from 3+$3+$ cases down to exactly 1$1$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Statistics
Class 12 Maths: Probability
More Statistics and Probability Previous-Year Questions — Page 2
Q70
jee_main_2025_04_april_evening
Measures of Dispersion
Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and sqrt2$\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is :
- A. 1$1$
- B. frac34$\frac{3}{4}$
- C. 2$2$
- D. frac12$\frac{1}{2}$
Solution
### 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
Q64
jee_main_2025_24_jan_morning
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
jee_main_2025_29_jan_morning
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
Q7
jee_main_2024_01_february_morning
Measures of Dispersion
Let the median and the mean deviation about the median of 7$7$ observations 170, 125, 230, 190, 210, a, b$170, 125, 230, 190, 210, a, b$ be 170$170$ and frac2057$\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7$7$ observations is:
- A. 31$31$
- B. 28$28$
- C. 30$30$
- D. 32$32$
Solution
### 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
Q4
jee_main_2024_29_january_evening
Mean and Variance
If the mean and variance of five observations are frac245$\frac{24}{5}$ and frac19425$\frac{194}{25}$ respectively and the mean of first four observations is frac72$\frac{7}{2}$, then the variance of the first four observations is equal to
- A. frac45$\frac{4}{5}$
- B. frac7712$\frac{77}{12}$
- C. frac54$\frac{5}{4}$
- D. frac1054$\frac{105}{4}$
Solution
### 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
