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
Q61
jee_main_2025_02_april_evening
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
Q54
jee_main_2025_03_april_evening
Measures of Dispersion
Let the
Mean and Variance of five observations x_1 = 1, x_2 = 3, x_3 = a, x_4 = 7$x_1 = 1, x_2 = 3, x_3 = a, x_4 = 7$ and
x_5 = b, a > b,$x_5 = b, a > b,$ be
5$5$ and
10$10$ respectively. Then the Variance of the observations
n + x_n, n = 1, 2, dots dots 5$n + x_n, n = 1, 2, \dots \dots 5$ is
- A. 17$17$
- B. 16.4$16.4$
- C. 17.4$17.4$
- D. 16$16$
Solution
### Related Formula
Mean of N$N$ observations:
barx = fracsum x_iN$$\bar{x} = \frac{\sum x_i}{N}$$
Variance of N$N$ observations:
sigma^2 = fracsum x_i^2N - (barx)^2$$\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$$
### Core Logic
Given mean is 5$5$ for 5$5$ observations:
frac1 + 3 + a + 7 + b5 = 5 implies a + b + 11 = 25 implies a + b = 14 quad text--- (1)$$\frac{1 + 3 + a + 7 + b}{5} = 5 \implies a + b + 11 = 25 \implies a + b = 14 \quad \text{--- (1)}$$
Given variance is 10$10$:
frac1^2 + 3^2 + a^2 + 7^2 + b^25 - 5^2 = 10 implies frac59 + a^2 + b^25 = 35$$\frac{1^2 + 3^2 + a^2 + 7^2 + b^2}{5} - 5^2 = 10 \implies \frac{59 + a^2 + b^2}{5} = 35$$
a^2 + b^2 = 175 - 59 = 116 quad text--- (2)$$a^2 + b^2 = 175 - 59 = 116 \quad \text{--- (2)}$$
### Step 1: Finding a$a$ and b$b$
Using standard algebraic identity (a+b)^2 = a^2 + b^2 + 2ab$(a+b)^2 = a^2 + b^2 + 2ab$:
14^2 = 116 + 2ab implies 196 = 116 + 2ab implies ab = 40$$14^2 = 116 + 2ab \implies 196 = 116 + 2ab \implies ab = 40$$
Solving a+b=14$a+b=14$ and ab=40$ab=40$:
a(14-a) = 40 implies a^2 - 14a + 40 = 0 implies (a-10)(a-4) = 0$$a(14-a) = 40 \implies a^2 - 14a + 40 = 0 \implies (a-10)(a-4) = 0$$
Since a > b$a > b$, we obtain a = 10$a = 10$ and b = 4$b = 4$.
### Step 2: Constructing new set and finding variance
The original set is x_1 = 1, x_2 = 3, x_3 = 10, x_4 = 7, x_5 = 4$x_1 = 1, x_2 = 3, x_3 = 10, x_4 = 7, x_5 = 4$.
We construct the new set y_n = n + x_n$y_n = n + x_n$:
- y_1 = 1 + 1 = 2$y_1 = 1 + 1 = 2$
- y_2 = 2 + 3 = 5$y_2 = 2 + 3 = 5$
- y_3 = 3 + 10 = 13$y_3 = 3 + 10 = 13$
- y_4 = 4 + 7 = 11$y_4 = 4 + 7 = 11$
- y_5 = 5 + 4 = 9$y_5 = 5 + 4 = 9$
Mean of new set:
bary = frac2 + 5 + 13 + 11 + 95 = frac405 = 8$$\bar{y} = \frac{2 + 5 + 13 + 11 + 9}{5} = \frac{40}{5} = 8$$
Variance of new set:
sigma_new^2 = frac2^2 + 5^2 + 13^2 + 11^2 + 9^25 - 8^2$$\sigma_{new}^2 = \frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5} - 8^2$$
sigma_new^2 = frac4 + 25 + 169 + 121 + 815 - 64 = frac4005 - 64 = 80 - 64 = 16$$\sigma_{new}^2 = \frac{4 + 25 + 169 + 121 + 81}{5} - 64 = \frac{400}{5} - 64 = 80 - 64 = 16$$
### Pattern Recognition
Note that adding a changing factor like +n$+n$ is different from adding a constant C$C$ to each observation (which leaves variance unchanged). In this case, calculate individual x_n$x_n$ variables directly first before applying transformations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Statistics and Probability
Q63
jee_main_2025_03_april_evening
Probability Distributions
If the probability that the random variable X$X$ takes the value x$x$ is given by P(X = x) = k(x + 1) 3^-x$P(X = x) = k(x + 1) 3^{-x}$, x = 0, 1, 2, 3 dots$x = 0, 1, 2, 3 \dots$, where k$k$ is a constant, then P(X ge 3)$P(X \ge 3)$ is equal to
- A. frac727$\frac{7}{27}$
- B. frac49$\frac{4}{9}$
- C. frac827$\frac{8}{27}$
- D. frac19$\frac{1}{9}$
Solution
### Related Formula
Sum of all probabilities in a distribution:
sum_x=0^infty P(X = x) = 1$$\sum_{x=0}^{\infty} P(X = x) = 1$$
Complementary probability:
P(X ge 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$P(X \ge 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$
### Core Logic
Let's first determine the constant k$k$:
k sum_x=0^infty (x+1) 3^-x = 1$$k \sum_{x=0}^{\infty} (x+1) 3^{-x} = 1$$
This is an Arithmetico-Geometric Progression (AGP). Let S = sum_x=0^infty (x+1)left(frac13right)^x$S = \sum_{x=0}^{\infty} (x+1)\left(\frac{1}{3}\right)^x$:
S = 1 + frac23 + frac39 + frac427 + dots quad text--- (1)$$S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots \quad \text{--- (1)}$$
frac13S = frac13 + frac29 + frac327 + dots quad text--- (2)$$\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots \quad \text{--- (2)}$$
### Step 1: Finding k$k$
Subtracting (2) from (1):
Sleft(1 - frac13right) = 1 + frac13 + frac19 + frac127 + dots$$S\left(1 - \frac{1}{3}\right) = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$$
frac23S = frac11 - 1/3 = frac32 implies S = frac94$$\frac{2}{3}S = \frac{1}{1 - 1/3} = \frac{3}{2} \implies S = \frac{9}{4}$$
Substitute back:
k cdot left(frac94right) = 1 implies k = frac49$$k \cdot \left(\frac{9}{4}\right) = 1 \implies k = \frac{4}{9}$$
### Step 2: Calculating P(X ge 3)$P(X \ge 3)$
Calculate initial probability values:
- P(X=0) = k(1)(1) = frac49$P(X=0) = k(1)(1) = \frac{4}{9}$
- P(X=1) = k(2)left(frac13right) = frac23 cdot frac49 = frac827$P(X=1) = k(2)\left(\frac{1}{3}\right) = \frac{2}{3} \cdot \frac{4}{9} = \frac{8}{27}$
- P(X=2) = k(3)left(frac19right) = frac13 cdot frac49 = frac427$P(X=2) = k(3)\left(\frac{1}{9}\right) = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27}$
textSum P(X < 3) = frac1227 + frac827 + frac427 = frac2427 = frac89$$\text{Sum } P(X < 3) = \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{24}{27} = \frac{8}{9}$$
P(X ge 3) = 1 - frac89 = frac19$$P(X \ge 3) = 1 - \frac{8}{9} = \frac{1}{9}$$
### Pattern Recognition
The infinite AGP sum with factor (x+1)r^x$(x+1)r^x$ always converges to frac1(1-r)^2$\frac{1}{(1-r)^2}$. Here r = 1/3$r = 1/3$, so the sum is frac1(2/3)^2 = 9/4$\frac{1}{(2/3)^2} = 9/4$. This mental check saves doing the full subtraction sequence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Statistics and Probability
Class 11 Mathematics: Sequences and Series
Q62
jee_main_2025_07_april_morning
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
