Statistics – JEE Main MathematicsPractice Questions & PYQs

The mean of the constructed data is determined by expanding the squared term in the expression $(x_i - i)^2$ into $x_i^2 - 2ix_i + i^2$ and calculating the average of these three components over the twenty observations. This requires finding the individual sums $\sum_{i=1}^{20} x_i^2$, $\sum_{i=1}^{20} i x_i$, and $\sum_{i=1}^{20} i^2$ separately.

For the first component, the squared geometric progression terms $x_i^2$ form a new geometric progression with a first term of $3^2 = 9$ and a common ratio of $(1/2)^2 = 1/4$, which sums to $12(1 - \frac{1}{2^{40}})$. The second component involves twice the sum of an arithmetico-geometric series, simplifying to $6(4 - \frac{11}{2^{18}})$. The third component, $\sum_{i=1}^{20} i^2$, is the sum of the first twenty squares, which is calculated as $\frac{20(21)(41)}{6} = 2870$.

Combining these results leads to the expression for the mean:
$\overline{x} = \frac{1}{20} \left[ 12\left(1 - \frac{1}{2^{40}}\right) - 6\left(4 - \frac{11}{2^{18}}\right) + 2870 \right]$
Given that the terms containing powers of two in the denominator are extremely small and approach zero, the mean effectively evaluates to $\frac{12 - 24 + 2870}{20} = \frac{2858}{20} = 142.9$. Calculating the greatest integer less than or equal to this value gives 142.

For the sequence 1, 2, 3, ..., $n$, the arithmetic mean is $\bar{x} = \frac{n+1}{2}$. Since $n$ is odd, the numbers are distributed symmetrically around this mean, and the mean deviation is the average of the absolute differences between each term and the mean. These absolute differences occur in pairs of integers from 0 to $\frac{n-1}{2}$, where each non-zero difference appears twice.

Summing these differences is equivalent to calculating $2 \sum_{k=1}^{(n-1)/2} k$, which simplifies to $\frac{(n-1)(n+1)}{4}$. Dividing this total sum by $n$ provides the expression for the mean deviation as $\frac{(n-1)(n+1)}{4n}$. When this expression is equated to $\frac{5(n+1)}{n}$, the common factor $\frac{n+1}{n}$ can be canceled from both sides, resulting in $\frac{n-1}{4} = 5$. Solving this linear equation confirms that $n = 21$.

Variance quantifies how far data points deviate from the mean, where the sum of squared deviations is calculated as the product of the number of students and the variance. Given 7 students, a mean of 62, and a variance of 20, the sum of squared differences $(x_i - 62)^2$ across the class is:
$\sum_{i=1}^{7} (x_i - 62)^2 = 7 \times 20 = 140$
If a student were to fail by scoring less than 50, their squared difference from the mean would be $(50 - 62)^2 = 144$. Because the total sum of squared deviations is strictly limited to 140, it is mathematically impossible for even one student to have a squared deviation of 144, ensuring that zero students fail.

To determine the correct standard deviation, we must recover the true sum of observations and the true sum of their squares by reversing the impact of the incorrect data point. The mean and variance of a dataset are derived directly from the summation of the values and the summation of their squares, respectively, so an error in a single observation propagates into both these metrics. By calculating the total sum of 10 observations as 150 and using the variance formula $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$, we can extract the exact values of the initial sums needed for correction.

Since the mean of 10 observations was 15, the wrong sum is $\sum x_i = 150$. Replacing the erroneous value of 25 with the correct value of 15 adjusts this to $150 - 25 + 15 = 140$, resulting in a corrected mean of 14. For the variance, the wrong sum of squares is found by rearranging $\sigma^2 + (\bar{x})^2 = \frac{\sum x_i^2}{N}$, which gives $15 + 225 = 240$, so $\sum x_i^2 = 2400$. We update this to the correct sum of squares using $2400 - 25^2 + 15^2 = 2000$. Applying the variance formula with the corrected values yields $\frac{2000}{10} - 14^2$, which simplifies to $200 - 196 = 4$, making the correct standard deviation $\sqrt{4} = 2$.

The mean and variance are inherently linked to the sum of observations and the sum of their squares, which allows us to identify the impact of a single incorrect data point. With 50 observations and an initial mean of 15, the total sum is 750, while the variance of $2^2 = 4$ indicates that the sum of the squares of the observations divided by 50 is $4 + 15^2 = 229$, resulting in an initial sum of squares equal to $50 \times 229 = 11450$.

Replacing the incorrect observation $b$ with the correct one $a$ adjusts the total sum to $750 - b + a = 16 \times 50 = 800$, which leads to the equation $a - b = 50$. Since the problem states that the sum of the correct and incorrect observations is $a + b = 70$, solving this system yields $a = 60$ and $b = 10$.

Updating the sum of squares requires subtracting the square of the incorrect value and adding the square of the correct value, resulting in $11450 - 10^2 + 60^2 = 14950$. The new variance is then found by dividing this updated sum by 50 and subtracting the square of the new mean, calculated as $\frac{14950}{50} - 16^2 = 299 - 256$, which equals 43.

The variance of a dataset is determined by the relationship between the sum of the squares of the observations and the square of their mean, defined by the formula $\sigma^2 = \frac{\sum x_i^2}{N} - \bar{x}^2$. When modifying a dataset by removing specific observations, one must update the sum of the observations and the sum of their squares before calculating the new variance for the reduced sample size.

For 40 observations with a mean of 30 and a standard deviation of 5, the total sum is 1200. Using the variance formula $25 = \frac{\sum x_i^2}{40} - 30^2$, the original sum of squares is 37000. Upon omitting the observations 12 and 10, the new sum of the observations becomes $1200 - 22 = 1178$, and the new sum of squares becomes $37000 - (144 + 100) = 36756$. With the new observation count of 38, the variance of this modified set is given by $\sigma^2 = \frac{\sum y_i^2}{38} - (\frac{\sum y_i}{38})^2$. Multiplying this equation by 38 results in $38\sigma^2 = \sum y_i^2 - 38 \times (\frac{\sum y_i}{38})^2$, which simplifies to $36756 - \frac{1178^2}{38}$. Given that $1178 / 38 = 31$, this further reduces to $36756 - 38 \times 31^2$, or $36756 - 36518$, resulting in 238.

The mean of a data set establishes the total sum of its values, while the variance measures the average squared dispersion around that mean, both of which serve as mathematical constraints to identify missing variables. Given the eight data points 4, 5, 6, 6, 7, 8, $x$, and $y$, the condition for the mean can be expressed as
$\frac{36+x+y}{8}=6$
which simplifies to $x+y=12$. The variance, defined as the average of the squared deviations from the mean of 6, is given by
$\frac{(4-6)^{2+}(5-6)^{2+}(6-6)^{2+}(6-6)^{2+}(7-6)^{2+}(8-6)^{2+}(x-6)^{2+}(y-6)^2}{8}=\frac{9}{4}$
Summing the known squared deviations results in 10, transforming the equation into $10+(x-6)^{2+}(y-6)^2=18$, or $(x-6)^{2+}(y-6)^2=8$. Substituting $y=12-x$ into this expression leads to $2(x-6)^2=8$, which means $(x-6)^2=4$. Solving for $x$ under the constraint $x < y$ yields $x=4$ and $y=8$. Evaluating the final expression gives $4^{4+}8^2=256+64=320$.

In a data set of eight numbers arranged in ascending order, the median is the average of the two middle terms. These terms are the fourth and fifth values, $2k$ and 12, which means the median $M$ is defined by the formula $M = \frac{2k+12}{2} = k+6$.

The mean deviation is calculated by taking the average of the absolute differences between each data point and the median. Using the given values 3, 5, 7, $2k$, 12, 16, 21, and 24, the sum of these absolute deviations is $|3-M| + |5-M| + |7-M| + |2k-M| + |12-M| + |16-M| + |21-M| + |24-M|$. Substituting $M=k+6$ and simplifying the terms according to the ascending order yields the expression $(k+3) + (k+1) + (k-1) + (6-k) + (6-k) + (10-k) + (15-k) + (18-k)$, which results in $58-2k$.

Equating this total divided by the count of 8 to the provided mean deviation of 6 gives the equation:
$\frac{58-2k}{8} = 6$
Multiplying by 8 results in $58-2k = 48$, which implies $2k = 10$ or $k = 5$. Substituting this value back into the expression for the median $k+6$ results in 11.

Variance measures the dispersion of data based on the mean and the sum of the squares of individual observations. Correcting a calculation involving a misread value requires determining the original sum of observations and the initial sum of squares using the given statistics. With 15 observations and an incorrect mean of 8, the original sum of values is 120, and the initial sum of squares is derived from the following relationship:
$\frac{\sum x^2}{15} - 8^2 = 9$
Solving this confirms the initial sum of squares is 1095.

Replacing the incorrect entry of 5 with the correct entry of 20 changes the sum of observations to 135, resulting in a corrected mean of 9. Correspondingly, update the sum of squares by removing the square of the incorrect value and adding the square of the correct value:
$\sum x_c^2 = 1095 - 5^2 + 20^2 = 1470$
Calculating the corrected variance with these updated values using $\frac{1470}{15} - 9^2$ simplifies to $98 - 81$, yielding a final result of 17.

The fundamental relationship between the mean and variance allows us to connect the spread of a data set to the sum of the squares of its values. Since variance is defined as the mean of the squares minus the square of the mean, we can determine the sum of the squares for a given set of observations using the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2$

For the 20 observations provided, the mean $\bar{x} = 15$ implies that the sum of the observations $\sum x_i$ is $20 \times 15 = 300$. Using the variance of 9, we substitute the known values into the equation to find the sum of the squares: $9 = \frac{\sum x_i^2}{20} - (15)^2$. This simplifies to $9 = \frac{\sum x_i^2}{20} - 225$, which means $\frac{\sum x_i^2}{20} = 234$, or $\sum x_i^2 = 4680$.

When each observation $x_i$ is transformed into $(x_i + \alpha)$, the mean of these new squared observations is given by $\frac{1}{20} \sum (x_i + \alpha)^2 = 178$. Expanding the term inside the summation leads to $\frac{1}{20} \sum (x_i^2 + 2\alpha x_i + \alpha^2) = 178$, which can be distributed as $\frac{\sum x_i^2}{20} + \frac{2\alpha \sum x_i}{20} + \alpha^2 = 178$. Substituting the previously derived values gives $234 + 2\alpha(15) + \alpha^2 = 178$, resulting in the quadratic equation $\alpha^2 + 30\alpha + 56 = 0$. Factoring this equation as $(\alpha + 28)(\alpha + 2) = 0$ provides two possible solutions for $\alpha$, which are $-28$ and $-2$. The maximum of these two values is $-2$, and squaring this value results in 4.

The relationship between mean and variance relies on two fundamental identities involving the sum of the observations and the sum of their squares. By determining these aggregate values for both the full set of five observations and the subset of the first four, we can isolate the individual fifth observation and calculate the variance of the smaller set through algebraic substitution.

For the total set of five observations, the sum of all values is the product of the count and the mean, which is $5 \times \frac{24}{5} = 24$. Similarly, the sum of the first four observations is calculated as $4 \times \frac{7}{2} = 14$. Subtracting this sum from the total reveals that the fifth observation, $x_5$, is equal to $24 - 14 = 10$.

To find the variance $a$ of the first four observations, we must first determine the sum of the squares of all five values. Using the variance formula, $\frac{\sum_{i=1}^5 x_i^2}{5} - (\frac{24}{5})^2 = \frac{194}{25}$, we can rearrange the equation to find that $\sum_{i=1}^5 x_i^2 = 5 \times \left(\frac{194}{25} + \frac{576}{25}\right) = \frac{770}{5} = 154$. Since $x_5$ is $10$, the sum of the squares of the first four observations is $154 - 10^2 = 54$. Applying the variance definition to the first four observations gives $a = \frac{54}{4} - (\frac{7}{2})^2 = \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$. Consequently, the final expression evaluates as $4 \times \frac{5}{4} + 10 = 5 + 10 = 15$.

To determine the unknown numbers, we start with the fundamental properties of the mean and variance. Given the mean of the five numbers $a, b, 8, 5, 10$ is $6$, their total sum must be $30$, implying that $a + b = 7$. We then apply the variance formula $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$, which tells us that the sum of squared deviations from the mean is $5 \times 6.8 = 34$. Substituting the known values leads to $(a-6)^2 + (b-6)^2 + (8-6)^2 + (5-6)^2 + (10-6)^2 = 34$, which simplifies to $(a-6)^2 + (b-6)^2 + 21 = 34$, leaving $(a-6)^2 + (b-6)^2 = 13$.

By substituting $b = 7 - a$, we solve the equation $(a-6)^2 + (1-a)^2 = 13$ to find $a^2 - 7a + 12 = 0$, which gives $a = 3$ or $a = 4$. This confirms the set of numbers is $3, 4, 8, 5, 10$. The mean deviation $M$ is calculated by taking the average of the absolute differences from the mean, yielding $M = \frac{|3-6| + |4-6| + |8-6| + |5-6| + |10-6|}{5} = \frac{3 + 2 + 2 + 1 + 4}{5} = \frac{12}{5}$. Finally, multiplying this result by $25$ gives $25 \times 2.4 = 60$.

The mean of the data set $3, 7, 12, a, 43-a$ is calculated to be $13$. To find the number of natural values of $a$ such that the variance is a natural number, one must express the variance using the formula $\frac{\sum x_i^2}{n} - \bar{x}^2$ and equate it to an arbitrary natural number. This process leads to a quadratic equation in $a$ where the discriminant $D$ must be a perfect square. The discriminant takes the form $40n-7$, which consistently results in a value with a units digit of $3$. Since no perfect square can end in the digit $3$, it is impossible for the variance to be a natural number, confirming that there are no such values for $a$.

The mean and variance of a dataset provide a system of equations that allow for the determination of unknown values. Calculating the average of 3, 7, $x,$ and $y$ yields $3+7+x+y=20$, which establishes that $x+y=10$. Applying the variance formula, $\frac{3^{2+}7^{2+}x^{2+}y^2}{4} - 5^2=10$, simplifies to $58+x^{2+}y^2=140$, or $x^{2+}y^2=82$.

Substituting $y=10-x$ into the equation $x^{2+}y^2=82$ produces $x^{2+}(10-x)^2=82$, which expands to $2x^{2-}20x+18=0$. Factoring the quadratic $x^{2-}10x+9=0$ yields the roots 1 and 9; since the condition $x>y$ must hold, we assign $x=9$ and $y=1$. The mean of the new numbers $3+2x, 7+2y, x+y,$ and $x-y$ is found by summing the terms to get $10+4x+2y$ and dividing by 4. Substituting $x=9$ and $y=1$ into this expression results in $\frac{10+4(9)+2(1)}{4}$, which simplifies to $\frac{48}{4}=12$.

The combined variance of two data sets is derived from the aggregate of their squared deviations, necessitating the sum of squares for each group, which can be determined from the relation $\sigma^2 = \frac{\sum x^2}{N} - \bar{x}^2$ rearranged as $\sum x^2 = N(\sigma^2 + \bar{x}^2)$. For the first group of 10 observations with a mean of 2 and variance of 2, the sum of squares is $10(2 + 2^2) = 60$, whereas for the second group of $n$ observations with a mean of 3 and variance of 1, the sum of squares is $n(1 + 3^2) = 10n$.

The variance of the combined set is given by the formula $\sigma_{combined}^2 = \frac{\sum x_I^2 + \sum x_{II}^2}{N_{total}} - (\bar{x}_{combined})^2$. Substituting the values leads to the equation $\frac{17}{9} = \frac{60 + 10n}{n + 10} - (\frac{20 + 3n}{n + 10})^2$. Simplifying by multiplying the entire equation by $9(n+10)^2$ yields the quadratic expression $8n^2 - 20n - 100 = 0$, which reduces to $2n^2 - 5n - 25 = 0$. Factoring this quadratic as $(2n + 5)(n - 5) = 0$ confirms that the positive number of observations is $n = 5$.

Statistical parameters like the mean and standard deviation are entirely dependent on the total sum of the data points and the sum of their squares. To find the accurate metrics after discovering that a data value was incorrectly recorded as 25 instead of 35, we simply adjust these two fundamental totals. Starting with a mean of 10 for 20 observations, the initial total sum is 200. Adjusting this sum by subtracting the incorrect value of 25 and adding the correct value of 35 provides a new total of 210, which leads to a corrected mean of $\alpha = 10.5$.

Calculating the new variance requires the updated sum of the squares of the observations. Given the standard deviation is 2.5, the original variance is 6.25. Using the standard variance formula $\sigma^2 = \frac{\Sigma x_i^2}{n} - \bar{x}^2$, we can solve for the initial sum of squares: $6.25 = \frac{\Sigma x_i^2}{20} - 100$. This confirms the initial sum of squares was 2125. We then update this value by subtracting the square of the incorrect entry and adding the square of the correct entry, resulting in a new sum of squares equation: $2125 - 625 + 1225 = 2725$.

With the corrected sum of squares now available, we find the new variance $\beta$ by substituting the updated values into the variance formula: $\beta = \frac{2725}{20} - (10.5)^2$. This simplifies to $136.25 - 110.25$, which equals 26. Therefore, the resulting pair $(\alpha, \beta)$ is determined to be $(10.5, 26)$.

Variance quantifies how data points deviate from their mean, calculated by subtracting the square of the mean from the mean of the squared values. With a data set consisting of nine 1s and one unknown natural number $k$, the mean is $\frac{9+k}{10}$ and the mean of the squared observations is $\frac{9+k^2}{10}$. Substituting these into the variance inequality $\sigma^2 < 10$ provides the relationship:

$\frac{9+k^2}{10} - \left(\frac{9+k}{10}\right)^2 < 10$

Multiplying through by 100 eliminates the denominators, resulting in $10(9+k^2) - (81 + 18k + k^2) < 1000$. Simplifying the expression leads to $9k^2 - 18k + 9 < 1000$, which factors neatly into $9(k-1)^2 < 1000$. Dividing by 9 gives $(k-1)^2 < 111.11$, so $k-1$ must be less than approximately 10.54. Since $k$ must be a natural number, the largest integer $k$ satisfying $k < 11.54$ is 11.

The arithmetic mean and the computational formula for variance provide the necessary tools to determine the relationship between the variables in the data set. Calculate the total sum of the observations by multiplying the mean by the number of entries, which gives $8 \times 9 = 72$. Since the sum of the known terms is $6+10+7+13+12+12 = 60$, the sum of the two unknown variables is $a+b = 72 - 60 = 12$.

The variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ allows for the calculation of the squares of $a$ and $b$. Summing the squares of the known terms gives $36+100+49+169+144+144 = 642$, making the total sum of squares $a^2 + b^2 + 642$. Incorporating the known variance $\frac{37}{4}$ into the equation $\frac{a^2 + b^2 + 642}{8} - 81 = \frac{37}{4}$ reveals that $\frac{a^2 + b^2 + 642}{8} = 90.25$. Multiplying by 8 results in $a^2 + b^2 + 642 = 722$, confirming that $a^2 + b^2 = 80$.

Relating the sum and difference of the two variables through algebraic identities provides the final result. Expand the identity $(a+b)^2 = 12^2 = 144$ into $a^2 + b^2 + 2ab = 144$. Substituting the value $a^2 + b^2 = 80$ into this equation shows that $80 + 2ab = 144$, which means $2ab = 64$. Consequently, applying the identity $(a-b)^2 = a^2 + b^2 - 2ab$ results in $80 - 64 = 16$.

The mean of a frequency distribution is determined by dividing the sum of the products of each value and its frequency by the total number of observations. For the initial data provided, this relationship is expressed as $32+8\alpha+9\beta=6(8+\alpha+\beta)$, which simplifies to $2\alpha+3\beta=16$. Factoring in the variance of 6.8 allows one to solve for the missing frequencies, yielding $\alpha=5$ and $\beta=2$.

With the frequencies now established, the total number of observations becomes $4+4+5+2=15$. When the value $x_3$ is adjusted from 8 to 7, the new sum of the products is calculated as $2(4)+6(4)+7(5)+9(2)$, which equals 85. Dividing this new sum by the total frequency results in the mean calculated as $\frac{85}{15} = \frac{17}{3}$.

Combining statistical groups requires accounting for both the internal distribution of each subgroup and the dispersion created by the differences between their individual means and the total average. To determine the average marks for girls, denoted as $\mu$, we balance the weighted sum of marks against the total population average. With 50 candidates averaging 15 marks, the total sum is 750. Subtracting the contribution of 20 boys who averaged 12 leaves 510 marks accounted for by the 30 girls, resulting in $\mu = \frac{510}{30} = 17$.

When calculating the variance for the combined group, we combine the internal variances of each subgroup with the effect of the squared difference between the subgroup means and the total mean. The first part is the weighted average of the group variances, which simplifies to $\frac{20 \times 2 + 30 \times 2}{50} = 2$. The second part, which captures the dispersion between the groups, is determined by the formula:

$\sigma_{diff}^{2} = \frac{n_B n_G}{(n_B + n_G)^2}(\bar{X}_B - \mu)^2$

Substituting the known values, we calculate $\frac{20 \times 30}{50^2} \times (12 - 17)^2 = \frac{600}{2500} \times 25 = 6$. Adding these components gives a total variance $\sigma^2 = 2 + 6 = 8$. Finally, adding the average marks of the girls to the total variance of the candidates yields $17 + 8 = 25$.