Statistics – JEE Main MathematicsPractice Questions & PYQs

Finding the unknown frequency $x$ begins by identifying the mid-points of the class intervals 10-20, 20-30, and 30-40, which are 15, 25, and 35. Variance is independent of shifting the origin, so subtracting 25 from each mid-point simplifies the data set to -10, 0, and 10, with corresponding frequencies 2, $x$, and 2. The mean of these transformed values is calculated as $\frac{(-10 \times 2) + (0 \times x) + (10 \times 2)}{x + 4} = 0$. Applying the variance formula $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2$, we substitute the given variance of 50 to establish the following relationship:

$50 = \frac{2(-10)^2 + x(0)^2 + 2(10)^2}{x + 4} - 0^2$

This simplifies to $50 = \frac{400}{x + 4}$, which rearranges to $50(x + 4) = 400$. Dividing by 50, we find $x + 4 = 8$, which results in $x = 4$.

Variance measures the spread of data through the relationship between the mean of squares and the square of the mean. Correcting variance after finding an error in the data requires isolating the original sum of observations and the sum of their squares. Given 20 observations with a mean of 10, the total sum is 200, and the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ confirms that $4 = \frac{\sum x_i^2}{20} - 100$, meaning the original sum of squares is 2080.

Replacing the incorrect value of 9 with 11 shifts the total sum to $200 - 9 + 11 = 202$, resulting in a new mean of 10.1. Simultaneously, the sum of squares is updated by removing the squared incorrect value and adding the square of the correct value, yielding $2080 - 81 + 121 = 2120$. With these updated totals, the variance is $\frac{2120}{20} - 10.1^2$, which simplifies to $106 - 102.01$, resulting in a final variance of 3.99.

Determining a quadratic equation from its roots relies on finding their sum and product using the statistical properties of the dataset. Since the mean of the five observations 3, 5, 7, $a$, and $b$ is 5, their total sum is $3 + 5 + 7 + a + b = 25$, which simplifies to $a + b = 10$.

The standard deviation of 2 implies a variance of 4. The variance is calculated by averaging the squared deviations from the mean, expressed as the equation $\frac{(3-5)^2 + (5-5)^2 + (7-5)^2 + (a-5)^2 + (b-5)^2}{5} = 4$. Multiplying by 5 gives the total sum of squared deviations as 20. Expanding the terms leads to $4 + 0 + 4 + (a-5)^2 + (b-5)^2 = 20$, which reduces to $(a-5)^2 + (b-5)^2 = 12$. Expanding further gives $a^2 - 10a + 25 + b^2 - 10b + 25 = 12$, or $a^2 + b^2 - 10(a+b) + 50 = 12$. Substituting $a+b = 10$ into this expression yields $a^2 + b^2 - 100 + 50 = 12$, meaning $a^2 + b^2 = 62$. Using the algebraic identity $(a+b)^2 = a^2 + b^2 + 2ab$, we have $10^2 = 62 + 2ab$, which simplifies to $100 = 62 + 2ab$, resulting in $ab = 19$. Constructing the quadratic equation with sum 10 and product 19 results in:

$x^2 - 10x + 19 = 0$

The mean of a data set multiplied by the number of observations provides the total sum. Given 7 observations with a mean of 8, the total sum is $7 \times 8 = 56$. Subtracting the sum of the five provided values, $2 + 4 + 10 + 12 + 14 = 42$, results in the equation $x + y = 14$ for the two unknown numbers.

The variance measures the dispersion of the data, defined by the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. Applying the known values of variance 16 and mean 8, we get $16 = \frac{\sum x_i^2}{7} - 8^2$, which simplifies to $\sum x_i^2 = 560$. The sum of the squares of the five known observations is $2^2 + 4^2 + 10^2 + 12^2 + 14^2 = 460$, which implies that the sum of the squares of the remaining two values is $x^2 + y^2 = 560 - 460 = 100$.

Using the algebraic identity $(x - y)^2 = 2(x^2 + y^2) - (x + y)^2$, we substitute the values to find $(x - y)^2 = 2(100) - (14)^2 = 200 - 196 = 4$. Taking the square root of both sides results in $|x - y| = 2$.

The variance of a collection of values is independent of any uniform shift applied to the data, meaning we can evaluate the dispersion of $x_i$ using the given shifted values $x_i - p$. The formula for variance effectively calculates the mean of the squares of the deviations from the mean, and in this context, it is represented by:
$Var(X) = \frac{\sum_{i=1}^{n} (x_i - p)^2}{n} - \left( \frac{\sum_{i=1}^{n} (x_i - p)}{n} \right)^2$
Applying the provided totals of $9$ and $3$ for $n = 10$ observations, we calculate the variance as:
$Var(X) = \frac{9}{10} - \left( \frac{3}{10} \right)^2 = \frac{9}{10} - \frac{9}{100} = \frac{90 - 9}{100} = \frac{81}{100}$
Since the standard deviation is the square root of the variance, we find the final result by taking the root of this value:
$\sqrt{\frac{81}{100}} = \frac{9}{10}$

The variance of the first $n$ natural numbers is given by the formula
$\sigma^2 = \frac{n^2 - 1}{12}$
Given that this variance is $10$, we set up the equation $\frac{n^2 - 1}{12} = 10$. Multiplying both sides by $12$ results in $n^2 - 1 = 120$, so $n^2 = 121$, which identifies $n$ as $11$.

For the first $m$ even natural numbers, notice that this set corresponds to the first $m$ natural numbers each multiplied by $2$. Since multiplying every data point in a dataset by a constant $k$ increases the variance by a factor of $k^2$, the variance of the first $m$ even natural numbers is $2^2$ times the variance of the first $m$ natural numbers. Given this value is $16$, we have $4 \times \frac{m^2 - 1}{12} = 16$. Dividing by $4$, we obtain $\frac{m^2 - 1}{12} = 4$, which simplifies to $m^2 - 1 = 48$, or $m^2 = 49$, resulting in $m = 7$. Combining these values, $m + n$ is $7 + 11$, which totals $18$.

Mean and variance possess fundamental properties that describe how data distributions change when values are adjusted by a constant. Adding or subtracting a constant from every observation shifts the mean by that exact amount but leaves the variance unchanged, as the relative spread of the data points remains constant.

With the sum of $(x_i - 5)$ equal to $10$ for $10$ observations, the mean of $(x_i - 5)$ is calculated as $\frac{10}{10} = 1$. Since the new set of observations is $(x_i - 3)$, which is equivalent to adding $2$ to $(x_i - 5)$, the new mean $\mu$ is $1 + 2 = 3$. Regarding the variance, the dispersion of the initial values $(x_i - 5)$ is calculated using the formula $\frac{1}{10} \sum (x_i - 5)^2 - (\text{mean of } x_i - 5)^2$. Substituting the known values gives $\frac{40}{10} - 1^2$, which simplifies to $4 - 1 = 3$. Because additive shifts do not alter variance, the variance $\lambda$ for the set $(x_i - 3)$ is the same as the variance for $(x_i - 5)$, which is $3$. Consequently, the ordered pair $(\mu, \lambda)$ results in $(3, 3)$.

Variance measures the spread of data points around their mean, which for an arithmetic progression with an odd number of terms is simply the middle term. In a sequence of 11 terms, the mean is the 6th term, expressible as $a + 5d$, where $a$ is the first term and $d$ is the common difference. Because the progression is symmetrical, the deviations of the terms from this mean are $0, \pm d, \pm 2d, \pm 3d, \pm 4d,$ and $\pm 5d$.

Summing the squares of these deviations gives $2 \times (d^2 + 4d^2 + 9d^2 + 16d^2 + 25d^2) = 110d^2$. Since variance is the average of these squared deviations, dividing this sum by 11 leads to the relationship:
$\text{Variance} = \frac{110d^2}{11} = 10d^2$
Equating this to the provided variance of 90 gives the equation $10d^2 = 90$. Solving for $d$ yields $d^2 = 9$, and since the progression is defined as increasing, we take the positive root to find $d = 3$.

Variance is determined by calculating the mean of the squared values minus the square of the mean, represented by the formula $\frac{\sum x_i^2}{n} - \bar{x}^2$. For the initial five observations with a mean of 10 and a standard deviation of 3, the sum of these values is $5 \times 10 = 50$, and the sum of squares is found by solving $3^2 = \frac{\sum x_i^2}{5} - 10^2$, which indicates the mean of the squared values is 109, leading to an initial sum of squares of $5 \times 109 = 545$. Adding the new observation of -50 updates the total count to six observations, resulting in a new sum of $50 + (-50) = 0$ and a new mean of 0. The sum of squares for this expanded dataset is the original 545 plus the square of the new observation, $(-50)^2$, totaling $545 + 2500 = 3045$. Applying the variance formula to this new set yields $\frac{3045}{6} - 0^2 = 507.5$.

The arithmetic mean and the variance together provide a system of equations sufficient to uncover unknown variables in a dataset. Since the mean of five observations is 5, their total sum must be $5 \times 5 = 25$. Given three of the observations are 1, 3, and 8, their sum is 12, which implies the two unknown values, let us call them $x$ and $y$, must satisfy $x + y = 13$.

The variance of 9.20 describes how the squared observations relate to the square of the mean. Applying the standard formula for variance, which is the mean of the squared terms minus the square of the mean, we can write:
$9.20 = \frac{1^2 + 3^2 + 8^2 + x^2 + y^2}{5} - 5^2$
Adding 25 to both sides results in $34.20 = \frac{74 + x^2 + y^2}{5}$, which simplifies to $x^2 + y^2 = 171 - 74 = 97$. With the sum $x + y = 13$ and the sum of squares $x^2 + y^2 = 97$, we can determine the product $xy$ using the identity $(x+y)^2 = x^2 + y^2 + 2xy$. Substituting the known values, we get $169 = 97 + 2xy$, which means $2xy = 72$ or $xy = 36$. Two numbers that sum to 13 and multiply to 36 are 4 and 9, making their ratio 4:9.

Standard deviation quantifies how much individual scores within a data set differ from the mean. It is computed as the square root of the variance, which is the average of the squared differences between each score and the overall mean.

Begin by finding the missing sixth test score, knowing that the sum of all six marks must equal the mean multiplied by the number of tests, which is $6 \times 48 = 288$. Subtracting the five given scores of 45, 54, 41, 57, and 43 from 288 confirms the sixth score is 48. Now, find the deviation of each score from the mean of 48. These deviations are -3, 6, -7, 9, -5, and 0. Squaring each of these results in 9, 36, 49, 81, 25, and 0, with a sum of 200. The variance is therefore the average of these squared deviations, calculated as $\frac{200}{6} = \frac{100}{3}$. Finally, the standard deviation is the square root of the variance, yielding $\sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}$.

To determine the mean of the squared terms $(x_{i}-4)^{2}$, we rely on the fundamental relationship between variance and raw data points. The formula for variance is $\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - \bar{x}^{2}$, which allows us to isolate the average value of the squared observations as $\frac{\sum x_{i}^{2}}{n} = \sigma^{2} + \bar{x}^{2}$. Given that both the mean and standard deviation are 16 for 50 observations, the average of the squared values becomes $16^{2} + 16^{2}$, resulting in $256 + 256 = 512$.

Expanding the expression $(x_{i}-4)^{2}$ gives $x_{i}^{2} - 8x_{i} + 16$. When we calculate the mean of these expanded terms across the 50 observations, the expression simplifies to $\frac{\sum x_{i}^{2}}{50} - 8\left(\frac{\sum x_{i}}{50}\right) + 16$. Substituting the known value of 512 for $\frac{\sum x_{i}^{2}}{50}$ and the mean of 16 for $\frac{\sum x_{i}}{50}$, the calculation becomes $512 - 8(16) + 16$. Evaluating this results in $512 - 128 + 16$, which equals 400.

Since the variance of a dataset remains constant even if you subtract the same constant value from every observation, we can simplify this problem by shifting all data points by $\frac{1}{2}$. Subtracting $\frac{1}{2}$ from the original values leaves us with ten instances of $-d$, ten instances of $0$, and ten instances of $d$. Because the mean of this new set is zero, the variance is simply the average of the squared terms:

$\frac{10(-d)^2 + 10(0)^2 + 10(d)^2}{30} = \frac{4}{3}$

Simplifying the numerator results in $\frac{20d^2}{30}$, which reduces to $\frac{2d^2}{3}$. Equating this expression to the given variance of $\frac{4}{3}$ allows us to solve for $d^2 = 2$, leading to the final result $|d| = \sqrt{2}$.

The median of an even-numbered data set is defined by the average of the two middle terms. In this series of ten numbers, the middle positions are occupied by 34 and $x$, meaning their average is given by $\frac{34+x}{2}=35$, which simplifies to $x=36$. With the mean of these ten values established as 42, the total sum of the set must be $42 \times 10=420$. Adding the known values $10, 22, 26, 29, 34, 36, 42, 67,$ and $70$ results in a sum of 336, implying that $y=420-336=84$. Finally, computing the ratio $\frac{y}{x}$ provides $\frac{84}{36}$, which simplifies to $\frac{7}{3}$.

Standard deviation measures the dispersion of data points relative to their mean, defined by the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2$. For the set containing $-1, 0, 1, k$ with a standard deviation of $\sqrt{5}$, the variance $\sigma^2$ is 5 and the number of terms $n = 4$. Calculating the sum of squares results in $k^2 + 2$, while the mean of these values is $\frac{k}{4}$. Substituting these components into the variance formula yields:

$5 = \frac{k^2 + 2}{4} - \left(\frac{k}{4}\right)^2$

Multiplying the entire equation by 16 simplifies the expression to $80 = 4(k^2 + 2) - k^2$, which further resolves to $80 = 3k^2 + 8$. Subtracting 8 from both sides gives $3k^2 = 72$, and dividing by 3 results in $k^2 = 24$. Since the problem specifies $k > 0$, taking the square root gives $k = 2\sqrt{6}$.

The arithmetic mean of a dataset relates directly to the sum of its deviations from any chosen constant. When the sum of deviations of $n$ observations from an assumed value $A$ is known, the true mean $\bar{x}$ is calculated using the relationship $\bar{x} = A + \frac{\sum (x_i - A)}{n}$. Applying the provided values where $A$ is 30, $n$ is 50, and the sum of deviations is 50, we simply divide the total sum of deviations by the number of observations to find the required adjustment, which is $\frac{50}{50} = 1$. Adding this adjustment to the assumed value provides the true mean, calculated as $30 + 1 = 31$.

The total number of students is determined by the sum of the frequencies associated with each mark value, providing the constraint needed to solve for the unknown variable. Summing the given frequencies $(x+1)^{2} + (2x-5) + (x^{2}-3x) + x$ and setting the result equal to 20 leads to the simplified quadratic equation $x^{2} + x - 12 = 0$. Solving for $x$ yields a positive value of 3, which confirms that all frequency counts are valid. With $x=3$, the frequencies for marks 2, 3, 5, and 7 are 16, 1, 0, and 3, respectively. Calculating the mean by summing the products of each mark and its corresponding frequency divided by the total number of students results in the following:

$\text{Mean} = \frac{2(16) + 3(1) + 5(0) + 7(3)}{20} = \frac{32 + 3 + 0 + 21}{20} = 2.8$

The mean and variance provide a bridge between individual values and their aggregated properties. Because the mean of 5 observations is 4, the total sum of all values must be 20. Removing the three known observations of 3, 4, and 4 leaves a remainder of 9 for the sum of the final two values.

Variance describes the dispersion of these points, following the relationship $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. By substituting the given variance of 5.2 and mean of 4 into this expression, the sum of the squares of all five observations is calculated to be 106. After accounting for the squares of the known values, which are 9, 16, and 16, the sum of the squares of the two remaining unknown observations is determined to be 65.

To find the absolute difference between these two unknowns, we utilize the algebraic relationship $(x_4 - x_5)^2 = 2(x_4^2 + x_5^2) - (x_4 + x_5)^2$. Substituting the sum of the squares, 65, and the square of their sum, 81, results in the calculation $(x_4 - x_5)^2 = 2(65) - 81$, which simplifies to 49. Taking the square root indicates that the absolute difference between the other two observations is 7.

The standard deviation of a dataset is calculated using the formula $\sigma = \sqrt{\frac{\sum x_i^2}{n} - \bar{x}^2}$. Determining the necessary components begins by expanding the given summations: the first expression $\sum (x_i+1)^2 = 9n$ simplifies to $\sum x_i^2 + 2\sum x_i + n = 9n$, while the second expression $\sum (x_i-1)^2 = 5n$ expands to $\sum x_i^2 - 2\sum x_i + n = 5n$.

Combining these two equations yields the variables required for the final calculation. Adding the expressions together eliminates the linear term, resulting in $2\sum x_i^2 + 2n = 14n$, which simplifies to $\frac{\sum x_i^2}{n} = 6$. Subtracting the second equation from the first isolates the linear term, yielding $4\sum x_i = 4n$, which means the mean $\bar{x} = 1$. Consequently, the variance is found by subtracting the square of the mean from the mean of the squares, resulting in $6 - 1^2 = 5$, and the standard deviation is the square root of this value, $\sqrt{5}$.

The fundamental definitions of mean and variance allow us to set up algebraic relationships based on the sum and the sum of the squares of the observations. With a total of 7 observations, the mean of 8 implies that their total sum must be 56, and since the sum of the five provided numbers is 42, the two unknown values, $a$ and $b$, must satisfy $a + b = 14$. The variance of 16 follows the relation $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, which leads to $\frac{a^2 + b^2 + 460}{7} - 64 = 16$ and thus shows that $a^2 + b^2 = 100$. By applying the algebraic identity $(a+b)^2 = a^2 + b^2 + 2ab$ with our known values, we see that $196 = 100 + 2ab$, which means the product is 48.