Statistics – JEE Main MathematicsPractice Questions & PYQs

Variance is defined as the mean of the squares of the data points minus the square of the mean of those points. The initial set of five students has an average height of 150 cm and a variance of 18 square centimeters, making the total sum of their heights 750. According to the formula $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$, the original sum of their squared heights is calculated as $5(18 + 150^2) = 112590$.

Adding the new student with a height of 156 cm brings the total sum to 906, which results in a new mean of $906 / 6 = 151$. The new sum of squared heights becomes $112590 + 156^2 = 136926$. Dividing this sum by the new total of 6 students and subtracting the square of the new mean yields the final variance as $136926 / 6 - 151^2 = 22821 - 22801 = 20$.

Standard deviation measures the spread of a dataset by finding the square root of the variance, calculated as the difference between the mean of the squares and the square of the mean. First, determine the total sum of the ten values by adding 44, derived from the first four terms, and 96, derived from the remaining six terms, resulting in 140. Dividing this total by 10 yields a mean of 14. To find the variance, subtract the square of this mean from the average of the squared values:

$\sigma = \sqrt{\frac{\sum x_i^2}{n} - \bar{x}^2} = \sqrt{\frac{2000}{10} - 14^2}$

This calculation, $\sqrt{200 - 196}$, simplifies to $\sqrt{4}$, which results in a standard deviation of 2.

The arithmetic mean follows the linear transformation property, which states that if every observation in a dataset with mean $\bar{x}$ is multiplied by a constant $\lambda$ and then a constant $k$ is subtracted, the new mean is given by $\lambda \bar{x} - k$. Applying this to the given values, where the original mean is $75$, the multiplier is $\lambda$, and the constant is $25$, the transformed mean is expressed as $\lambda(75) - 25$.

Setting the new mean equal to the original mean of $75$ results in the equation $\lambda(75) - 25 = 75$. Adding $25$ to both sides yields $\lambda(75) = 100$, and dividing by $75$ simplifies to $\lambda = \frac{100}{75} = \frac{4}{3}$.

The standard deviation of a dataset is zero if and only if all observations are identical, which means the initial five observations were each 9. Let $x$ be the changed observation. Given the new mean is 10, the sum of the new set of five observations is $4 \times 9 + x = 5 \times 10$, which leads to $36 + x = 50$, identifying the new observation as $x = 14$.

The updated set of observations is $\{9, 9, 9, 9, 14\}$. Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, the sum of the squares of the observations is $4(9^2) + 14^2 = 324 + 196 = 520$. The variance is thus $\frac{520}{5} - 10^2 = 104 - 100 = 4$. Consequently, the standard deviation is $\sqrt{4} = 2$.

The arithmetic mean is defined by $\bar{x} = \frac{\sum x_i}{n}$ and the variance is defined as $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$. With the mean given as 8 for the data set $\{7, 8, 9, 7, 8, 7, \lambda, 8\}$, the equation $\frac{7+8+9+7+8+7+\lambda+8}{8} = 8$ simplifies to $\frac{54+\lambda}{8} = 8$, which leads to $54+\lambda = 64$ and $\lambda = 10$. Using the updated set $\{7, 8, 9, 7, 8, 7, 10, 8\}$, the squared differences $(x_i - 8)^2$ result in values of $1, 0, 1, 1, 0, 1, 4, 0$ respectively. The sum of these squared deviations is 8, resulting in a variance of $\frac{8}{8} = 1$.

Standard deviation measures the spread of a dataset and remains invariant under a shift of origin, meaning the standard deviation of the original set is identical to that of the set defined by the differences $x_i - 5$. The variance is determined by subtracting the square of the mean of these differences from the mean of their squares.

The mean of the squared terms is $\frac{45}{9}$, which equals 5, while the mean of the differences is $\frac{9}{9}$, which equals 1. Applying the variance formula $\frac{45}{9} - (\frac{9}{9})^2 = 5 - 1 = 4$, the standard deviation is the square root of 4, which is 2.

The average of a group is defined as the total sum of its members divided by the count, implying that the total sum equals the product of the mean and the count. Starting with 25 teachers and a mean age of 40, the original combined age is $1000$ years. After a 60-year-old teacher is replaced by a newcomer, the count remains 25 while the mean shifts to 39, resulting in a new total age of $975$ years. The age of the new teacher accounts for this net change in the total sum, expressed as:

$1000 - 60 + \text{New Age} = 975$

This calculation identifies the newly appointed teacher as 35 years old.

The variance of a dataset is calculated using the formula $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$. After omitting the incorrect observations 3, 4, and 5, the new total number of observations becomes $n = 100 - 3 = 97$.

The new sum of observations is $\sum x_i = 400 - (3 + 4 + 5) = 388$, and the new sum of squares is $\sum x_i^2 = 2475 - (3^2 + 4^2 + 5^2) = 2475 - 50 = 2425$. Substituting these updated values into the variance formula yields $\sigma^2 = \frac{2425}{97} - \left(\frac{388}{97}\right)^2$. Dividing $2425$ by $97$ gives $25$, and dividing $388$ by $97$ gives $4$, leading to $\sigma^2 = 25 - 4^2 = 25 - 16 = 9$.

The mean deviation of a data set is defined as the average of the absolute differences between every observation and the arithmetic mean. For the given sequence of 101 terms ranging from 1 to $1+100d$, the arithmetic progression property ensures that the mean is the average of the first and last terms, which is $1+50d$. To find the mean deviation, we calculate the absolute difference of each term from this mean, resulting in a series of values $|-50d|, |-49d|, \dots, 0, \dots, |50d|$.

Summing these deviations requires adding the first 50 multiples of $d$ twice, which yields $2 \times d \times \frac{50 \times 51}{2} = 2550d$. Dividing this total sum by the 101 terms gives the mean deviation $\frac{2550d}{101} = 255$. Solving the equation for $d$ results in $d = \frac{255 \times 101}{2550}$, which simplifies to 10.1.

The standard deviation measures data dispersion as the square root of the variance, defined as the mean of the squares minus the square of the mean. For the numbers 2, 3, a, and 11, the mean is $\frac{a+16}{4}$, and the variance is $3.5^2$, or $\frac{49}{4}$.

Equating the variance to $\frac{2^2 + 3^2 + a^2 + 11^2}{4} - \left(\frac{a+16}{4}\right)^2$ simplifies to $\frac{49}{4} = \frac{4a^2 + 536 - a^2 - 32a - 256}{16}$. Multiplying both sides by 16 removes the denominators to yield $196 = 3a^2 - 32a + 280$, which rearranges to the following equation:
$3a^2 - 32a + 84 = 0$

The relationship between the mean, variance, and individual observations is defined by the equations $\bar{x} = \frac{\sum x_i}{n}$ and $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. For a set of 5 observations with a mean of 5, the total sum is 25. Given the known observations 1, 2, and 6, the two remaining unknowns must sum to 16.

Applying the variance formula, $124 = \frac{\sum x_i^2}{5} - 25$ implies that the sum of the squares of the five observations must be 745. Subtracting the sum of the squares of the known values, $1^2 + 2^2 + 6^2 = 41$, leaves 704 for the sum of the squares of the two unknown values.

Solving the resulting system of equations, $a+b=16$ and $a^{2+}b^2=704$, leads to values for the unknowns that are irrational. These irrational values prevent the mean deviation from matching any of the numerical options provided in the question. Consequently, the problem constraints are mathematically inconsistent, rendering none of the available options valid.

The mean of a data set is defined as the total sum of the observations divided by the number of observations. Given 16 observations with a mean of 16, the original total sum was $16 \times 16 = 256$. Removing the value 16 from this sum and adding the new observations 3, 4, and 5 results in a new total sum of $256 - 16 + 3 + 4 + 5 = 252$. Since the total number of observations has shifted from 16 to $16 - 1 + 3 = 18$, the updated mean is calculated by dividing the new sum by the new count, which results in $\frac{252}{18} = 14$.

Variance quantifies the dispersion of a data set by calculating the average of the squared differences from the mean. The first 50 even numbers form the sequence 2, 4, 6, ..., 100. Dividing the sum of these 50 terms by 50 results in a mean of 51.

Applying the standard variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ involves replacing each observation with $2r$, where $r$ ranges from 1 to 50. This transforms the sum of squared values into $\sum_{r=1}^{50} (2r)^2$, which simplifies to $4 \sum_{r=1}^{50} r^2$. Substituting these figures into the variance formula yields $\frac{\sum_{r=1}^{50} 4r^2}{50} - (51)^2$, which simplifies to 833.

Statistical measures are classified into those that represent central tendencies and those that define the spread of a dataset. When a constant value of 10 is added to every student's score, the mean, median, and mode all shift by that same amount because the entire distribution is translated along the number line. However, variance measures the dispersion of scores relative to the mean, and because the definition of variance is $\sigma^2 = \frac{ \sum (x_i - \bar{x})^2 }{ n }$, the constant added to each observation $x_i$ and the resulting constant added to the mean $\bar{x}$ cancel each other out, leaving the spread of the data unchanged.

Multiplying every observation in a dataset of size $n$ by a scaling factor $c$ scales the arithmetic mean by $c$ and the variance by $c^2$. With a factor $c = 2$, the new variance becomes $2^2 \sigma^2 = 4\sigma^2$, confirming that the first statement is true. Regarding the arithmetic mean, applying the factor $c = 2$ to each term results in the following calculation:
$\frac{\sum_{i=1}^{n} 2x_i}{n} = 2 \left( \frac{\sum_{i=1}^{n} x_i}{n} \right) = 2\overline{x}$
Because the resulting mean is $2\overline{x}$ rather than the $4\overline{x}$ suggested in the statement, the second assertion is false.

The mean deviation about the median is defined as the average of the absolute differences between each data point and the median of the set. For the sequence of $50$ terms ranging from $a$ to $50a$, the median is calculated by taking the average of the middle two terms, the $25$th and $26$th terms, which is $\frac{25a + 26a}{2} = 25.5a$.

Calculating the mean deviation involves finding the absolute difference of each term $ia$ from this median, summing them, and dividing by the count of terms. This sum is represented by $|a| \sum_{i=1}^{50} |i - 25.5|$. The sequence of differences $|i - 25.5|$ consists of values starting from $0.5$ and ending at $24.5$, with each value appearing twice. Using the sum formula for an arithmetic progression, the sum of these $50$ differences is $2 \times (0.5 + 1.5 + \dots + 24.5) = 2 \times \frac{25}{2} \times (0.5 + 24.5) = 625$. Therefore, the total sum of absolute deviations is $625|a|$.

Equating the mean deviation to the given value of $50$ produces the equation $\frac{625|a|}{50} = 50$. Simplifying this results in $625|a| = 2500$, which leads to $|a| = 4$.

The variance of a combined data set is determined by pooling the individual sums of squares and adjusting for the deviation of the combined mean from the separate means. Since the variance formula is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2$, the sum of squares for any data set can be calculated as $\sum x_i^2 = n(\sigma^2 + \overline{x}^2)$. For the first data set with $n=5$, variance 4, and mean 2, the sum of squares is $5(4 + 2^2) = 40$. For the second data set with $n=5$, variance 5, and mean 4, the sum of squares is $5(5 + 4^2) = 105$. The total sum of squares for the combined set of $10$ observations is $40 + 105 = 145$. Because the two sets are of equal size, the combined mean is the average of the two individual means, resulting in $\overline{X} = \frac{2 + 4}{2} = 3$. The combined variance is then calculated as $\frac{145}{10} - 3^2$, which simplifies to $14.5 - 9 = 5.5$, or $\frac{11}{2}$.

The variance of the first $n$ even natural numbers is derived by first finding the mean, which is $\frac{2(1+2+\dots+n)}{n} = n+1$, and then applying the formula $\frac{1}{n} \sum (2k)^2 - (n+1)^2$. Substituting the standard sum of squares formula, $4 \frac{n(n+1)(2n+1)}{6}$, into this expression simplifies the variance to $\frac{n^{2-}1}{3}$, demonstrating that the value given in the first statement is false. Conversely, the second statement correctly identifies the standard formulas for the sum of the first $n$ natural numbers, $\frac{n(n+1)}{2}$, and the sum of their squares, $\frac{n(n+1)(2n+1)}{6}$, making that statement true.

The mean deviation of a sequence is calculated by finding the arithmetic mean and then averaging the absolute differences of each term from that value. Since the sequence $1, 1+d, 1+2d, \dots, 1+100d$ is an arithmetic progression with 101 terms, the mean is the middle value, $1+50d$. The absolute deviations are expressed as $|(1+kd) - (1+50d)|$, which simplifies to $|k-50|d$ for $k$ ranging from 0 to 100. Summing these absolute differences gives $d(50+49+\dots+1+0+1+\dots+50)$, which simplifies to $2d(1+2+\dots+50)$ or $2550d$. Dividing this sum by the total number of terms provides the mean deviation:
$\frac{2550d}{101} = 255$
Multiplying by 101 and dividing by 2550 determines that $d = 10.1$.

The mean and variance of a dataset provide the foundation for describing its central tendency and spread, which allows us to set up a system of algebraic equations to determine unknown values. Given that the mean of the five numbers is 6, their sum is $5 \times 6 = 30$, leading to the relation $a + b + 8 + 5 + 10 = 30$, or $a + b = 7$. By applying the definition of variance, $\frac{\sum(x_i - 6)^2}{n} = 6.8$, we calculate the sum of the squared deviations from the mean as $5 \times 6.8 = 34$. This produces the equation $(a - 6)^2 + (b - 6)^2 + (8 - 6)^2 + (5 - 6)^2 + (10 - 6)^2 = 34$, which simplifies to $(a - 6)^2 + (b - 6)^2 + 21 = 34$, or $(a - 6)^2 + (b - 6)^2 = 13$. Substituting $b = 7 - a$ into this equation results in $(a - 6)^2 + (1 - a)^2 = 13$, which expands and reduces to the quadratic form $a^2 - 7a + 12 = 0$. Solving this quadratic equation provides the possible values $a = 3$ and $a = 4$, which correspond to the pair of numbers 3 and 4.