Statistics – JEE Main MathematicsPractice Questions & PYQs

Calculating the mean and variance of the combined set requires determining the total sum of the new elements and the sum of their squares. Given the initial sets $A$ and $B$ have 5 elements each with means $\bar{A} = 5$ and $\bar{B} = 8$, the sums are $\sum a_i = 25$ and $\sum b_i = 40$. Using the formula $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$, the sum of the squares for each set is $\sum a_i^2 = 5(12 + 5^2) = 185$ and $\sum b_i^2 = 5(20 + 8^2) = 420$.

The new set $C$ contains 10 elements: five elements from $A$ shifted by $-3$ and five elements from $B$ shifted by $+2$. The mean of the combined set is $\bar{C} = \frac{5(5-3) + 5(8+2)}{10} = \frac{10 + 50}{10} = 6$. To find the variance, we calculate the sum of the squares of the elements in $C$: $\sum (a_i - 3)^2 + \sum (b_i + 2)^2$. Expanding these expressions yields $\sum a_i^2 - 6\sum a_i + 5(3^2) + \sum b_i^2 + 4\sum b_i + 5(2^2)$. Substituting the known sums gives $185 - 6(25) + 45 + 420 + 4(40) + 20 = 680$. The variance of $C$ is therefore $\frac{680}{10} - (6)^2 = 68 - 36 = 32$. Adding the mean and the variance of set $C$ results in $6 + 32 = 38$.

Recognizing that the expression $\mu^2 + \sigma^2$ is equivalent to $E(X^2)$ makes this problem straightforward, as the variance of a distribution is defined by the relationship $\sigma^2 = E(X^2) - \mu^2$. This simplification eliminates the need to calculate the mean and variance individually, allowing for the direct computation of the expected value of the squared random variable.

The value of $k$ is found by setting the sum of the frequencies to 62: $(k+2) + 2k + (k^{2-}1) + (k^{2-}1) + (k^{2+}1) + (k-3) = 62$. Simplifying this polynomial results in $3k^2 + 4k - 64 = 0$. Factoring the quadratic gives $(3k+16)(k-4) = 0$, and because frequency must be positive, $k=4$ is the only valid solution.

With $k=4$, the frequencies corresponding to the values 0, 1, 2, 3, 4, 5 become 6, 8, 15, 15, 17, 1. Calculating the sum of squares $E(X^2)$ involves summing the product of each squared value and its corresponding frequency: $0^2(6) + 1^2(8) + 2^2(15) + 3^2(15) + 4^2(17) + 5^2(1) = 500$. Dividing this by the total sum of frequencies, 62, results in $\frac{500}{62}$, which is approximately 8.06. The greatest integer $\leq 8.06$ is 8.

Mean is calculated as the sum of observations divided by the total count, while variance is defined by the difference between the mean of the squares and the square of the mean. The initial set of 7 observations with a mean of 8 implies a total sum of $7 \times 8 = 56$. Removing the observation 14 reduces this total to $56 - 14 = 42$, which makes the new mean $a = \frac{42}{6} = 7$.

The sum of the squares of the original observations follows from the variance formula $16 = \frac{\sum x_i^2}{7} - 8^2$, resulting in a sum of $7 \times 80 = 560$. Excluding the value 14 decreases this sum of squares by $14^2 = 196$, leaving $364$ for the remaining 6 entries. The new variance $b$ is computed as $\frac{364}{6} - 7^2$, which simplifies to $\frac{364}{6} - 49 = \frac{70}{6}$.

Substituting these values into the expression $a + 3b - 5$ provides $7 + 3(\frac{70}{6}) - 5$. The multiplication $3 \times \frac{70}{6}$ reduces to $35$, resulting in $7 + 35 - 5$, which equals 37.

The mean of a frequency distribution is defined as the weighted average of the class midpoints, which allows us to solve for any missing frequency by equating the given mean to the ratio of the total sum of products to the total count. For this distribution, the midpoints are 5, 15, 25, 35, and 45. Using the mean formula $\bar{x} = \frac{\sum f_i x_i}{N}$, we set up the equation $28 = \frac{(2 \times 5) + (3 \times 15) + (x \times 25) + (5 \times 35) + (4 \times 45)}{2 + 3 + x + 5 + 4}$. This simplifies to $28 = \frac{410 + 25x}{14 + x}$, and solving for $x$ yields $28(14 + x) = 410 + 25x$, resulting in $392 + 28x = 410 + 25x$, so $x = 6$.

With the frequency determined, the variance is calculated using the formula $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$. Calculating the sum of the products of each frequency and the square of its corresponding midpoint, we find $\sum f_i x_i^2 = (2 \times 25) + (3 \times 225) + (6 \times 625) + (5 \times 1225) + (4 \times 2025) = 18700$. Dividing this sum by the total frequency of 20 gives 935, and subtracting the square of the mean, 784, results in a final variance of 151.

The mean of 10 observations is 20 and the standard deviation is 8, implying the initial sum is 200 and the variance is 64. Using the relationship $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$, we find the original sum of squares is $10 \times (64 + 400) = 4640$. Correcting the erroneous entry by removing 50 and adding 40 results in a new mean of 19 and a sum of squares of $4640 - 50^2 + 40^2 = 3740$. The corrected variance is then calculated as
$\frac{3740}{10} - 19^2 = 374 - 361 = 13$

The mean deviation of a data set provides a measure of spread that depends only on the relative distances of values from the mean, not their absolute position on the number line. For a sequence of 100 consecutive positive integers starting at $a_1$, the terms are $a_1, a_1+1, \dots, a_1+99$. The mean $\bar{x}$ of this arithmetic progression is the average of the first and last terms, calculated as $\bar{x} = \frac{a_1 + (a_1+99)}{2} = a_1 + 49.5$.

The mean deviation is defined as $\frac{1}{100} \sum_{i=1}^{100} |x_i - \bar{x}|$. Substituting the terms, the absolute differences become $|a_1 + k - (a_1 + 49.5)| = |k - 49.5|$ for $k$ ranging from $0$ to $99$. These differences form the set $\{49.5, 48.5, \dots, 0.5, 0.5, \dots, 48.5, 49.5\}$. Summing these 100 absolute deviations, we calculate $2(0.5 + 1.5 + \dots + 49.5) = 1 + 3 + \dots + 99$.

This sum of the first 50 odd integers is $50^2 = 2500$. Dividing this total by the number of terms, we get $\frac{2500}{100} = 25$. Because the starting term $a_1$ cancels out entirely during the subtraction from the mean, the mean deviation is always 25 regardless of the specific positive integer chosen for $a_1$, meaning the set of all valid values for $a_1$ is the set of all natural numbers.

Variance essentially measures the spread of data points from their mean, calculated via the relationship $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left( \frac{\sum f_i x_i}{\sum f_i} \right)^2$. A much faster approach involves taking deviations from an assumed mean of 5, which centers the data and simplifies the arithmetic significantly. With this shift, the sum of deviations becomes zero, reducing the variance formula to the simple ratio of the sum of weighted squared deviations to the total frequency.

Summing the frequencies results in a total of $45 + \alpha$, while calculating the squared deviations weighted by their frequencies yields a sum of 150. Equating the ratio of these values to the provided variance of 3 gives the expression $3 = \frac{150}{45 + \alpha}$. Multiplying both sides by the denominator results \in $135 + 3\alpha = 150$, which simplifies to $3\alpha = 15$, confirming that $\alpha = 5$.

The mean and variance of a frequency distribution are determined by the formulas $\overline{x} = \frac{\sum f_i x_i}{N}$ and $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\overline{x})^2$, where $N = \sum f_i$ is the total frequency. Summing the given frequencies provides $N = 40 + \alpha + \beta$, and the sum of products $\sum f_i x_i$ is $360 + 6\alpha + 12\beta$. Equating the mean to 9 leads to $\frac{360 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$, which simplifies to $360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$, indicating that $\alpha = \beta$.

Calculating $\sum f_i x_i^2$ results in $3904 + 36\alpha + 144\beta$. Using the variance formula with $\sigma^2 = 15.08$ and $\overline{x}^2 = 81$, we have $\frac{3904 + 36\alpha + 144\beta}{40 + \alpha + \beta} - 81 = 15.08$. Substituting $\alpha$ for $\beta$ into this expression gives $\frac{3904 + 180\alpha}{40 + 2\alpha} = 96.08$. Solving for $\alpha$ yields $3904 + 180\alpha = 3843.2 + 192.16\alpha$, which simplifies to $12.16\alpha = 60.8$, confirming $\alpha = 5$. With $\beta = 5$, the required value $\alpha^2 + \beta^2 - \alpha\beta$ becomes $5^2 + 5^2 - 5^2$, which equals 25.

Variance is defined by the relationship $\sigma^2 = \frac{\sum x^2}{n} - (\frac{\sum x}{n})^2$. With 12 observations having a mean of $\frac{9}{2}$ and a variance of 4, the initial sum is $\sum x = 12 \times \frac{9}{2} = 54$, while the sum of squares is found by $4 = \frac{\sum x^2}{12} - (\frac{9}{2})^2$, which simplifies to $\frac{\sum x^2}{12} = 4 + \frac{81}{4} = \frac{97}{4}$, yielding $\sum x^2 = 291$.

Adjusting these figures to replace the incorrect observations of 9 and 10 with the correct values of 7 and 14 updates the sum to $\sum x_{\text{new}} = 54 - (9 + 10) + (7 + 14) = 56$ and the sum of squares to $\sum x^2_{\text{new}} = 291 - (81 + 100) + (49 + 196) = 355$. Applying these updated totals to the variance formula gives $\sigma_{\text{new}}^2 = \frac{355}{12} - (\frac{56}{12})^2 = \frac{355}{12} - \frac{196}{9}$. Calculating this yields a common denominator of 36, resulting in $\frac{1065 - 784}{36} = \frac{281}{36}$. Since 281 and 36 are coprime, the sum $m+n$ equals 317.

Representing the five positive numbers in a geometric progression as $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$ allows for a symmetric approach to the calculations. Given that the mean is $\frac{31}{10}$, the sum of these terms is $5 \times \frac{31}{10} = \frac{31}{2}$, while the mean of their reciprocals, $\frac{31}{40}$, implies the sum of reciprocals is $5 \times \frac{31}{40} = \frac{31}{8}$. Dividing the sum of the original terms by the sum of their reciprocals results in the relation:

$\frac{a(\frac{1}{r^2} + \frac{1}{r} + 1 + r + r^2)}{\frac{1}{a}(r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2})} = \frac{31/2}{31/8} = 4$

This simplifies to $a^2 = 4$, which means $a=2$ because the terms must be positive. Applying the condition $a_3 + a_4 + a_5 = 14$, we substitute $a=2$ to obtain $2 + 2r + 2r^2 = 14$, which reduces to the quadratic equation $r^2 + r - 6 = 0$. Factoring this as $(r+3)(r-2) = 0$ yields $r=2$ for positive terms, identifying the sequence as $0.5, 1, 2, 4, 8$. The variance is then calculated using the formula $\sigma^2 = \frac{\sum x_i^2}{N} - (\frac{\sum x_i}{N})^2$, which gives:

$\sigma^2 = \frac{0.25 + 1 + 4 + 16 + 64}{5} - \left(\frac{31}{10}\right)^2 = \frac{85.25}{5} - 9.61 = 17.05 - 9.61 = 7.44$

Expressing $7.44$ as a fraction gives $\frac{744}{100}$, which simplifies to $\frac{186}{25}$. With $m=186$ and $n=25$ being co-prime, the sum $m+n$ equals $211$.

Variance is fundamentally determined by the sum of observations and the sum of their squares, meaning that when data is recorded incorrectly, the error must be rectified by adjusting these initial aggregates before recalculating the measure of dispersion. Given a mean of 50 for 10 students, the original sum of marks is $10 \times 50 = 500$, and using the relationship $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, we can find the original sum of squares by substituting $144 = \frac{\sum x_i^2}{10} - 2500$, which yields $\sum x_i^2 = 26440$.

To correct the data, subtract the wrongly read values and add the true values to obtain the updated sums. The corrected sum of marks is $500 - 45 - 50 + 20 + 25 = 450$, and the corrected sum of squares is $26440 - (45)^2 - (50)^2 + (20)^2 + (25)^2$, which simplifies to $26440 - 2025 - 2500 + 400 + 625 = 22940$. Applying the variance formula with these updated values, the final result is $\frac{22940}{10} - (\frac{450}{10})^2$, resulting in $2294 - 2025 = 269$.

The combined mean of two datasets is the weighted average of their individual means. Given that 100 students have a mean mark of 40 and $n$ students have a mean of 55, the combined mean of 50 for the $100+n$ students is expressed by the equation $\frac{100(40) + n(55)}{100+n} = 50$. Solving this yields $4000 + 55n = 5000 + 50n$, which simplifies to $5n = 1000$, confirming that $n = 200$.

The combined variance is determined by the formula $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1+n_2}$, where $d_1$ and $d_2$ represent the deviations of the class means from the combined mean. Here, $d_1 = 40 - 50 = -10$ and $d_2 = 55 - 50 = 5$. With $\sigma_1 = \alpha$ and $\sigma_2 = 30 - \alpha$, we substitute the known values into the equation: $350 = \frac{100(\alpha^2 + (-10)^2) + 200((30-\alpha)^2 + 5^2)}{100 + 200}$.

Simplifying this expression results in $350 \times 300 = 100(\alpha^2 + 100) + 200((30-\alpha)^2 + 25)$, which becomes $105000 = 100\alpha^2 + 10000 + 200(30-\alpha)^2 + 5000$. Subtracting 15000 from both sides leaves $90000 = 100\alpha^2 + 200(30-\alpha)^2$, and dividing by 100 gives $\alpha^2 + 2(30-\alpha)^2 = 900$. Expanding the quadratic term, we obtain $\alpha^2 + 2(900 - 60\alpha + \alpha^2) = 900$, which reduces to $3\alpha^2 - 120\alpha + 900 = 0$ or $\alpha^2 - 40\alpha + 300 = 0$.

Factoring the resulting quadratic equation yields $(\alpha - 10)(\alpha - 30) = 0$, so $\alpha$ is either 10 or 30. The sum of the variances is $\sigma_1^2 + \sigma_2^2 = \alpha^2 + (30-\alpha)^2$. For the solution $\alpha = 30$, this sum equals $30^2 + 0^2 = 900$, which matches the correct answer.

Since the selection of apples occurs without replacement from a finite population containing three rotten and seven good apples, the random variable $X$ representing the number of rotten apples follows a hypergeometric distribution. The expression $10(\mu^2 + \sigma^2)$ is simplified by recognizing that the sum of the square of the mean and the variance is equivalent to the second raw moment of the distribution, denoted as E\[X^2]$ = \sum x^2 P(x)$.

The probability distribution for $X$ involves drawing four apples from ten, where the probabilities $P(x)$ for each possible number of rotten apples $x \in \{0, 1, 2, 3\}$ are $\frac{1}{6}$, $\frac{1}{2}$, $\frac{3}{10}$, and $\frac{1}{30}$ respectively. Calculating the second raw moment results in the following equation:
$E\$[X^2]\$ = 0^2 \cdot \frac{1}{6} + 1^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{3}{10} + 3^2 \cdot \frac{1}{30} = 0 + \frac{1}{2} + \frac{12}{10} + \frac{9}{30} = 2$
Using this result, the value of the required expression is $10(\mu^2 + \sigma^2) = 10 \cdot 2 = 20$.

The mean and variance of a dataset allow us to define the sum and sum of squares of unknown observations through a system of algebraic equations. Given that the mean of the 5 observations is 5, their total sum is 25, meaning the two unknown values $a$ and $b$ satisfy $a + b = 16$. Substituting these into the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ gives $8 = \frac{1^2 + 3^2 + 5^2 + a^2 + b^2}{5} - 25$, which simplifies to $165 = 35 + a^2 + b^2$, resulting in $a^2 + b^2 = 130$. We determine the product $ab$ using the identity $a^2 + b^2 + 2ab = (a + b)^2$, yielding $130 + 2ab = 16^2$, so $2ab = 126$ and $ab = 63$. Finally, evaluating the identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$ with our derived values provides $16 \times (130 - 63)$, which equals 1072.

When combining two datasets, the total variance is not simply a direct average of the individual variances. Instead, it must account for both the internal spread of each set and the distance between their respective means, as the separation between group averages introduces additional dispersion into the combined pool.

The combined variance formula incorporates the weighted average of individual variances and adds a term representing the difference between the group means. With $n_1 = 15$ and $n_2 = 15$, means $\bar{x}_1 = 12$ and $\bar{x}_2 = 14$, and variances $\sigma^2_1 = 14$ and $\sigma^2_2 = \sigma^2$, the relationship is expressed as
$13 = \frac{15(14) + 15(\sigma^2)}{15+15} + \frac{15 \cdot 15(12 - 14)^2}{(15+15)^2}$
Simplifying the first part of the expression yields $\frac{14 + \sigma^2}{2}$, while the second term accounts for the mean difference as $\frac{225(4)}{900}$. Since the second fraction resolves to $1$, the equation simplifies to $13 = \frac{14 + \sigma^2}{2} + 1$. Subtracting $1$ from both sides results in $12 = \frac{14 + \sigma^2}{2}$, leading to $24 = 14 + \sigma^2$, which yields $\sigma^2 = 10$.

The variance of any arithmetic progression with $n$ terms and common difference $d$ follows the relationship $\sigma^2 = \frac{d^2(n^{2-}1)}{12}$. For this seven-term progression, the variance formula results in $\frac{d^2(49-1)}{12}$, which simplifies to $4d^2$. Since the given standard deviation is 4, its square is 16, allowing us to establish $4d^2 = 16$, which confirms the common difference $d = 2$.

With $d$ determined and the first term $x_1$ equal to 9, the sequence consists of 9, 11, 13, 15, 17, 19, 21. The mean of this symmetric progression is the middle term, 15, and the sixth term $x_6$ is 19. Adding the mean and the sixth term together gives $15 + 19 = 34$.

The terms of the arithmetic progression are represented as $a_1, a_1+d, a_1+2d, a_1+3d, a_1+4d, a_1+5d$. Since $a_1+a_3=10$, it follows that $a_1 + (a_1+2d) = 2a_1+2d=10$, which simplifies to $a_1+d=5$. Because the mean of these six numbers is $\frac{19}{2}$, their total sum is 57. Using the formula for the sum of an arithmetic progression, the following equation holds:
$3(2a_1+5d)=57 \implies 2a_1+5d=19$
Solving the system of equations $a_1+d=5$ and $2a_1+5d=19$ yields $d=3$ and $a_1=2$, resulting in the sequence 2, 5, 8, 11, 14, 17.

The variance $\sigma^2$ is defined as the mean of the squares of the numbers minus the square of the mean. Calculating the sum of the squares:
$\sum_{i=1}^6 a_i^2 = 2^2 + 5^2 + 8^2 + 11^2 + 14^2 + 17^2 = 699$
The mean of these squares is $\frac{699}{6} = \frac{233}{2}$. The variance is then calculated as:
$\sigma^2 = \frac{233}{2} - \left(\frac{19}{2}\right)^2 = \frac{466-361}{4} = \frac{105}{4}$
Multiplying by 8 leads to the final result:
$8 \sigma^2 = 8 \times \frac{105}{4} = 210$

Variance is fundamentally defined by the difference between the mean of squares and the square of the mean, expressed as $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. Because changing a single data point shifts both the total sum and the sum of the squares, we must first determine the number of students by observing how the mean fluctuates. Since the sum of marks increases by 4 while the mean rises from 10 to 10.2, we can state that the new mean is $\frac{\sum x_i + 4}{n} = 10.2$. Knowing the initial sum is $\sum x_i = 10n$, this leads directly to $10n + 4 = 10.2n$, which confirms there are $n = 20$ students.

With $n = 20$ determined, we derive the initial sum of squares using the given variance: $4 = \frac{\sum x_i^2}{20} - 10^2$, meaning the original sum of squares must be $\sum x_i^2 = 20(104) = 2080$. When the mark of 8 is updated to 12, the new sum of squares becomes $2080 - 8^2 + 12^2$, which simplifies to 2160. Finally, applying the variance formula with these updated values, the new variance is calculated as $\sigma_{new}^2 = \frac{2160}{20} - 10.2^2 = 108 - 104.04 = 3.96$.

The mean deviation about the mean is the average of the absolute differences between each data point and the mean of 5. We first determine the values of the missing observations $x$ and $y$ using the given statistical properties. Since the mean is 5 for 6 observations, their sum must be 30, leading to $1 + 2 + 4 + 5 + x + y = 30$, or $x + y = 18$. The variance formula, $\frac{\sum x_i^2}{n} - (\bar{x})^2 = 10$, provides the equation:
$\frac{1^2 + 2^2 + 4^2 + 5^2 + x^2 + y^2}{6} - 25 = 10$
Multiplying by 6 results in $46 + x^2 + y^2 = 210$, or $x^2 + y^2 = 164$. Solving the system $x + y = 18$ and $x^2 + y^2 = 164$ identifies the missing numbers as 8 and 10.

With the complete dataset identified as $1, 2, 4, 5, 8, 10$, the mean deviation is calculated by finding the absolute difference of each observation from 5. These differences are $|1-5|=4$, $|2-5|=3$, $|4-5|=1$, $|5-5|=0$, $|8-5|=3$, and $|10-5|=5$. Summing these results yields $4 + 3 + 1 + 0 + 3 + 5 = 16$. Dividing by the total of 6 observations gives $\frac{16}{6}$, which simplifies to $\frac{8}{3}$.

The mean of a dataset is defined as the sum of all observations divided by the total number of observations. For the eight numbers $x, y, 10, 12, 6, 12, 4,$ and $8$, the sum is $x+y+52$. Setting the mean to 9, we have the equation $\frac{x+y+52}{8} = 9$, which simplifies to $x+y=20$.

Variance can be calculated as the mean of the squares of the observations minus the square of the mean. Applying this definition, we can express the variance as $\frac{x^{2+}y^{2+}10^{2+}12^{2+}6^{2+}12^{2+}4^{2+}8^2}{8} - 9^2 = 9.25$. Simplifying the sum of squares, we get $\frac{x^{2+}y^{2+}504}{8} - 81 = 9.25$, which means $\frac{x^{2+}y^{2+}504}{8} = 90.25$. Multiplying by 8 results in $x^{2+}y^{2+}504 = 722$, giving the equation $x^{2+}y^2=218$.

We now have a system of equations with $x+y=20$ and $x^{2+}y^2=218$. Substituting $y = 20-x$ into the second equation yields $x^2 + (20-x)^2 = 218$. Expanding this gives $x^2 + 400 - 40x + x^2 = 218$, which simplifies to $2x^2 - 40x + 182 = 0$. Dividing by 2, we obtain $x^2 - 20x + 91 = 0$, which factors as $(x-13)(x-7) = 0$.

Since the condition $x > y$ must be satisfied, we determine that $x=13$ and $y=7$. Calculating the final expression $3x-2y$ gives $3(13) - 2(7)$, which equals $39 - 14$, resulting in 25.