Statistics – JEE Main MathematicsPractice Questions & PYQs

Variance is defined by the difference between the mean of the squares of the observations and the square of their mean, expressed as $\sigma^2 = \frac{1}{N} \sum x_i^2 - \bar{x}^2$. This relationship allows one to determine the sum of the squares of a set when the variance and mean are already known.

For the initial set of $3n$ numbers, the overall mean is determined by the weighted average of the two subsets: $\bar{x} = \frac{2n(6) + n(3)}{3n} = 5$. Since the variance is 4, the mean of the squares must satisfy $\frac{\sum x_i^2}{3n} - 5^2 = 4$, which implies $\frac{\sum x_i^2}{3n} = 29$, or a total sum of squares $\sum x_i^2 = 87n$.

Transforming the set by adding 1 to the first $2n$ numbers and subtracting 1 from the remaining $n$ numbers creates a new mean of $\bar{x}' = \frac{2n(7) + n(2)}{3n} = \frac{16}{3}$. Expanding the sum of the squares of the new set, $\sum (x_i \pm 1)^2$, yields the original sum of squares plus $2(2n(6)) - 2(n(3)) + 3n$, resulting in a new sum of $87n + 24n - 6n + 3n = 108n$. Applying the variance formula, $k = \frac{108n}{3n} - (\frac{16}{3})^2 = 36 - \frac{256}{9}$, and multiplying by 9 results in $324 - 256$, which equals 68.

The variance of the first $n$ natural numbers, $1, 2, \ldots, n$, is calculated using the formula $\frac{n^2 - 1}{12}$. Since the variance is 14, we set up the equation $14 = \frac{n^2 - 1}{12}$ to solve for $n$. Multiplying both sides by 12 results in $n^2 - 1 = 168$, which simplifies to $n^2 = 169$. Taking the square root of 169, we find that $n = 13$, confirming the result is an odd natural number as required.

The median represents the value at the center of a data set, and for a continuous distribution, it occupies the position equivalent to half the total frequency. Summing the given frequencies $\alpha + 110 + 54 + 30 + \beta = 584$ yields the relationship $\alpha + \beta = 390$. As the median value 45 falls within the 40-50 class, this range functions as the median class for the formula calculation:

$45 = 40 + \left( \frac{292 - (\alpha + 110 + 54)}{30} \right) \times 10$

Solving the expression $5 = \left( \frac{292 - (\alpha + 164)}{30} \right) \times 10$ reduces to $15 = 128 - \alpha$, identifying $\alpha = 113$. With $\alpha$ determined, the remaining variable $\beta$ is found by subtracting 113 from 390, resulting \in $\beta = 277$. The absolute difference $|\alpha - \beta|$ is then computed as $|113 - 277| = 164$.

The mean age of a group is determined by the total sum of all ages divided by the number of individuals. With 25 teachers averaging 40 years each, their total combined age is $25 \times 40 = 1000$. Replacing the 60-year-old teacher with a new teacher of age $x$ changes the total sum to $1000 - 60 + x$, or $940 + x$. Since the group size remains 25 and the resulting mean is 39, the relationship is expressed as $\frac{940 + x}{25} = 39$. Calculating the product of the new mean and the group size gives $25 \times 39 = 975$, and setting $940 + x = 975$ confirms that the age of the new teacher is 35.

Standard deviation is a measure of dispersion that remains invariant when a constant is added to every observation. For the dataset $a+2, b+2, c+2$, the arithmetic mean $\mu$ is calculated as the sum divided by three, resulting in $\frac{a+b+c+6}{3}$. Given the condition $b = a+c$, we can simplify this expression for the mean to $\frac{2b+6}{3}$, which is equal to $\frac{2b}{3} + 2$.

To determine the variance $d^2$, we compute the squared deviations of each term from the mean, expressed as $(a+2 - \mu)^2$, $(b+2 - \mu)^2$, and $(c+2 - \mu)^2$. Substituting $\mu = \frac{2b}{3} + 2$ leads to the differences $(a - \frac{2b}{3})^2$, $(\frac{b}{3})^2$, and $(c - \frac{2b}{3})^2$. Expanding these squares yields $a^2 - \frac{4ab}{3} + \frac{4b^2}{9}$, $\frac{b^2}{9}$, and $c^2 - \frac{4bc}{3} + \frac{4b^2}{9}$. Summing these values gives $a^2 + c^2 + b^2 - \frac{4b}{3}(a+c)$, which reduces to $a^2 + c^2 + b^2 - \frac{4b^2}{3}$ or $a^2 + c^2 - \frac{b^2}{3}$. Since $d^2 = \frac{1}{3}(a^2 + c^2 - \frac{b^2}{3})$, multiplying both sides by 3 results in $3d^2 = a^2 + c^2 - \frac{b^2}{3}$. Multiplying the entire expression by 3 again yields $9d^2 = 3(a^2 + c^2) - b^2$, which rearranges to $b^2 = 3(a^2 + c^2) - 9d^2$.

The mean and variance of a dataset provide a system of equations that allow us to determine unknown values within a group. Given the mean of 6 observations is 6.5, the total sum of all observations is 39; subtracting the sum of the four known observations (2, 4, 5, and 7, which total 18) reveals that the remaining two must sum to 21. By representing these as $a$ and $21 - a$ and applying the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, we can substitute the known values into the equation $10.25 = \frac{2^2 + 4^2 + 5^2 + 7^2 + a^2 + (21 - a)^2}{6} - (6.5)^2$. This simplifies to $10.25 + 42.25 = \frac{94 + a^2 + (21 - a)^2}{6}$, which further resolves to $315 = 94 + a^2 + (21 - a)^2$, or $221 = a^2 + (21 - a)^2$. Solving the resulting quadratic equation $2a^2 - 42a + 220 = 0$ yields $(a-10)(a-11) = 0$, confirming that the missing observations are 10 and 11.

The properties of mean and standard deviation dictate that adding a constant to every observation shifts the mean by that constant but leaves the standard deviation entirely unaffected. In this specific series, the observations consist of equal counts of $a$ and $-a$, which forces the original mean to be $\overline{x} = 0$ since these values perfectly balance each other. Because the mean is zero, the variance is the average of the squared values, simplifying to $\sigma^2 = a^2$, which means the original standard deviation is $\sigma = a$.

When a constant $b$ is added to each observation, the new mean becomes the original mean plus $b$. Given that the new mean is $\overline{y} = 5$, the relationship $\overline{x} + b = 5$ confirms $b = 5$. Because the standard deviation remains unchanged after adding a constant, the original standard deviation $\sigma = a$ must equal the new standard deviation $\sigma_y = 20$, establishing that $a = 20$. Combining these results leads to the final calculation $a^2 + b^2 = 20^2 + 5^2 = 425$.

The terms in the sequence are generated by multiplying corresponding elements from two distinct arithmetic progressions. The first sequence, $7, 10, 13, \ldots$, yields the general term $a_n = 3n + 4$, and the second sequence, $8, 10, 12, \ldots$, provides the term $b_n = 2n + 6$. By finding the product of these two expressions, we obtain the general term for the series as $T_n = (3n + 4)(2n + 6) = 6n^2 + 26n + 24$.

To calculate the mean of the first $10$ terms, we find the sum of this expression from $n=1$ to $10$ and divide by $10$. Applying summation formulas, the sum of $6n^2$ over $10$ terms is $6 \times \frac{10 \times 11 \times 21}{6} = 2310$, while the sum of $26n$ is $26 \times \frac{10 \times 11}{2} = 1430$, and the constant $24$ summed $10$ times is $240$. Adding these components gives a total sum of $3980$, and dividing by $10$ yields the final mean of $398$.

Determining the standard deviation of the second sample relies on the principle of combined variance, which accounts for both the internal variance of each sample and the variance arising from the difference between the sample means. The entire group consists of 250 items with 100 in the first sample, meaning the second sample contains 150 items. The mean of the second sample follows from the weighted average formula, where the total mean of 15.6 equates to the sum of the products of the means and their respective counts divided by 250, resulting in a mean of 16 for the second sample.

Applying the combined variance formula:
$\sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{x}_1 - \overline{x}_2)^2$
Substituting the known values where the total variance $\sigma^2$ is 13.44, the variance of the first sample $\sigma_1^2$ is 9, and the squared difference of the means $(\overline{x}_1 - \overline{x}_2)^2$ is 1, the equation becomes $13.44 = \frac{100 \times 9 + 150 \times \sigma_2^2}{250} + \frac{100 \times 150}{(250)^2} \times 1$. Simplifying this expression leads to $13.44 = 3.6 + 0.6 \sigma_2^2 + 0.24$, which further reduces to $0.6 \sigma_2^2 = 9.6$. Dividing both sides by 0.6 results in $\sigma_2^2 = 16$, which confirms the standard deviation of the second sample is 4.

The variance of a set of observations is linked to the mean $\bar{X}$ and any arbitrary constant $\beta$ by the formula $\sigma^2 = \frac{1}{18} \sum_{i=1}^{18} (X_i - \beta)^2 - (\bar{X} - \beta)^2$. With 18 observations and a standard deviation of 1, the variance is 1. The first equation $\sum_{i=1}^{18} (X_i - \alpha) = 36$ implies that the mean $\bar{X}$ satisfies $\bar{X} - \alpha = \frac{36}{18}$, or $\bar{X} - \alpha = 2$. Substituting the second sum $\sum_{i=1}^{18} (X_i - \beta)^2 = 90$ into the variance formula gives $1 = \frac{90}{18} - (\bar{X} - \beta)^2$, which simplifies to $1 = 5 - (\bar{X} - \beta)^2$, or $(\bar{X} - \beta)^2 = 4$. Substituting $\bar{X} = \alpha + 2$ into this expression yields $(\alpha + 2 - \beta)^2 = 4$. This leads to the two possibilities $\alpha - \beta + 2 = 2$ or $\alpha - \beta + 2 = -2$. Since $\alpha$ and $\beta$ are distinct, $\alpha - \beta$ cannot be 0, leaving only $\alpha - \beta = -4$. Calculating the absolute value gives $|\alpha - \beta| = 4$.

The mean of a dataset is defined as the sum of its observations divided by the total count, which for these six observations gives:
$10 = \frac{7+10+11+15+a+b}{6}$
This equation simplifies to $a+b=17$. The variance is calculated as the average of the squared observations minus the square of the mean, represented by the formula $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{Mean})^2$. Substituting the known values into this relationship results in:
$\frac{20}{3} = \frac{49+100+121+225+a^{2+}b^2}{6} - 100$
Multiplying by 6 gives $40 = 495 + a^2 + b^2 - 600$, which simplifies to $a^{2+}b^2=145$. The absolute difference $|a-b|$ is determined by the identity $(a-b)^2 = (a+b)^2 - 4ab$. Since $2ab = (a+b)^2 - (a^{2+}b^2) = 289 - 145 = 144$, it follows that $4ab = 288$. Substituting these values into the identity yields $(a-b)^2 = 289 - 288 = 1$, and thus $|a-b|=1$.

Variance depends on both the sum of the observations and the sum of their squares, allowing us to isolate subsets of data by adjusting these totals. For 7 observations with a mean of 8 and a variance of 16, the total sum is $7 \times 8 = 56$, and the sum of their squares is $7 \times (16 + 8^2) = 560$. Removing the values 6 and 8 leaves a remaining sum of $56 - 14 = 42$ and a remaining sum of squares of $560 - (36 + 64) = 460$.

The mean for these 5 observations becomes $\frac{42}{5}$. Applying the variance formula, we calculate $\frac{460}{5} - (\frac{42}{5})^2$. This simplifies to $92 - \frac{1764}{25}$, which resolves to $\frac{2300 - 1764}{25}$, ultimately resulting in $\frac{536}{25}$.

To determine the unknown frequencies, we utilize the standard definitions for mean and median in grouped frequency distributions. First, the sum of all frequencies is $N = a + b + 12 + 9 + 5 = a + b + 26$. By multiplying each class midpoint-3, 9, 15, 21, and 27-by its corresponding frequency, the mean is expressed as the sum of these products divided by $N$. Equating this to the given mean yields the equation:

$\frac{3a + 9b + 504}{a + b + 26} = \frac{309}{22}$

Cross-multiplying and simplifying this expression leads to $66a + 198b + 11088 = 309a + 309b + 8034$, which further reduces to the linear equation $81a + 37b = 1018$.

Next, we apply the median formula $L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$. Since the median is 14, it falls within the 12-18 interval, where the lower limit $L$ is 12, the class frequency $f$ is 12, the class size $h$ is 6, and the cumulative frequency of the classes preceding the median class is $a + b$. Substituting these values gives:

$14 = 12 + \left( \frac{\frac{a + b + 26}{2} - (a + b)}{12} \right) \times 6$

Simplifying this equation, we find $2 = \frac{\frac{a + b + 26}{2} - (a + b)}{2}$, which rearranges to $4 = \frac{a + b + 26 - 2(a + b)}{2}$, eventually resulting in $a + b = 18$. Solving the system formed by $81a + 37b = 1018$ and $a + b = 18$ provides the values $a = 8$ and $b = 10$. Finally, computing $(a - b)^2$ gives $(8 - 10)^2$, which is 4.

The defining characteristics of a data set-its mean and variance-are intrinsically linked to the sum of its values and the sum of their squares. By leveraging these two statistical relationships, we can isolate the combined contribution of unknown observations.

Given that the mean of 8 observations is 10, the total sum must be $8 \times 10 = 80$, so subtracting the sum of the 6 known values ($63$) shows the remaining two observations must satisfy $a+b = 17$. Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$, we determine the total \sum of squares is $8(13.5 + 100) = 908$. Subtracting the \sum of the squares of the known values ($739$) leaves $a^{2+}b^2 = 169$. Finally, using the algebraic identity $(a-b)^2 = 2(a^{2+}b^2) - (a+b)^2$, we calculate $(a-b)^2 = 2(169) - 17^2 = 49$, which gives an absolute difference of 7.

The mean of a frequency distribution is determined by the ratio of the total sum of products of each value and its frequency to the total frequency. Given the values 0, 2, 4, 8, ..., $2^n$ and frequencies ${}^{n}C_{0}, {}^{n}C_{1}, \dots, {}^{n}C_{n}$, the weighted sum is $\sum_{r=0}^{n} 2^r \cdot {}^{n}C_{r}$. Since the first value is 0, its product is 0, making the effective sum equal to $\sum_{r=1}^{n} 2^r \cdot {}^{n}C_{r}$, which simplifies to $3^n - 1$. The total sum of frequencies is given by the sum of binomial coefficients, which is $2^n$. Equating the mean $\frac{3^n - 1}{2^n}$ to the provided $\frac{728}{2^n}$ leads to the equation $3^n - 1 = 728$, so $3^n = 729$, which confirms $n = 6$.

Linear transformations affect the mean and standard deviation of a dataset in predictable ways: the mean changes by the same linear function applied to the individual observations, whereas the standard deviation is only scaled by the absolute value of the multiplier and remains unaffected by any constant shift. Specifically, if each observation is transformed into $px - q$, the new mean becomes $p\bar{x} - q$ and the new standard deviation becomes $|p|\sigma_x$.

Given the original mean is 20 and the original standard deviation is 2, the new mean $20p - q = 10$ and the new standard deviation $2|p| = 1$ must hold true. Solving the latter for $p$ results in:
$|p| = 0.5$
This implies $p$ is either $0.5$ or $-0.5$. Substituting $p = 0.5$ into the mean equation results in $10 - q = 10$, which gives $q = 0$, violating the condition that $q \neq 0$. Consequently, $p$ must be $-0.5$, leading to the equation $20(-0.5) - q = 10$. This simplifies to $-10 - q = 10$, which ultimately yields $q = -20$.

Standard deviation measures the dispersion of a data set and remains unchanged when every value is shifted by a constant, allowing us to simplify the calculations by working directly with the deviations $d_i = x_i - a$. The variance is computed by subtracting the square of the mean of these deviations from the mean of their squares, expressed by the formula $\sigma^2 = \frac{\sum_{i=1}^{n} d_i^2}{n} - \left(\frac{\sum_{i=1}^{n} d_i}{n}\right)^2$. Given that the sum of the deviations is $n$, their mean is $\frac{n}{n} = 1$, while the sum of the squared deviations is $na$, making the mean of the squares $\frac{na}{n} = a$. Substituting these values into the variance formula yields $\sigma^2 = a - 1^2$, which simplifies to $a - 1$, resulting in a standard deviation of $\sqrt{a - 1}$.

The mean and variance of a dataset under a linear transformation $Y = aX + b$ are governed by the rules $\bar{Y} = a\bar{X} + b$ and $\sigma_Y^2 = a^2 \sigma_X^2$. For the set $X$ containing integers from $1$ to $17$, the mean $\bar{X}$ is $9$ and the variance $\sigma_X^2$ is $24$. Given the variance of the transformed set $Y$ is $216$, we substitute these into the variance rule to get $216 = a^2 \cdot 24$, which simplifies to $a^2 = 9$. Since $a$ must be positive, $a = 3$. Using the mean rule with the given mean of $Y$ as $17$, we solve $17 = 3(9) + b$ to find $b = -10$. Adding these values together yields $a + b = 3 - 10 = -7$.

The mean and variance of a dataset are inherently tied to the sum of the observations and the sum of their squares. Given eight numbers with a mean of 10, their total sum is 80, and subtracting the known values of 3, 7, 9, 12, 13, and 20 leaves the equation $x + y = 16$.

The variance formula relates these values as follows:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\text{mean})^2$
Substituting the known values into the expression $25 = \frac{3^2 + 7^2 + 9^2 + 12^2 + 13^2 + 20^2 + x^2 + y^2}{8} - 100$ leads to $\frac{852 + x^2 + y^2}{8} = 125$, which simplifies to $x^2 + y^2 = 148$.

Finally, the algebraic identity $(x + y)^2 = x^2 + y^2 + 2xy$ allows us to isolate the product. Substituting the values determined above into $256 = 148 + 2xy$ leaves $2xy = 108$, confirming that $xy = 54$.

In statistics, the standard deviation of a dataset is inherently limited by its range, specifically governed by the inequality $\sigma \le \frac{M - m}{2}$ where $M$ and $m$ are the maximum and minimum values respectively. For a distribution where every observation $x_i$ is contained within the interval between 0 and 10, we substitute these bounds to find $\sigma \le \frac{10 - 0}{2}$, which simplifies to $\sigma \le 5$. Since the calculation dictates that the standard deviation must be less than 5, it is impossible for the value to be 6.