Statistics – JEE Main MathematicsPractice Questions & PYQs

Variance represents the dispersion of a data distribution, determined by subtracting the square of the mean from the average of the squared observations. The sum of the given observations is defined by the mean of 5 across five terms, leading to the equation $1+3+a+7+b=25$, which simplifies to $a+b=14$. Incorporating the given variance of 10 into the standard formula requires the sum of the squares, resulting in the equation $\frac{1+9+a^{2+}49+b^2}{5} - 25 = 10$. Rearranging this expression provides $a^{2+}b^2=116$, and solving this system alongside the sum equation with the constraint $a>b$ confirms the values $a=10$ and $b=4$.

The modified observations, created by adding the index $n$ to each term, yield the sequence 2, 5, 13, 11, 9. The mean of this new sequence is calculated as 8. To find the variance, compute the difference between the mean of the squares of these observations and the square of the mean:
$\frac{2^{2+}5^{2+}13^{2+}11^{2+}9^2}{5} - 8^2$
Evaluating this expression results in $\frac{4+25+169+121+81}{5} - 64$, which simplifies to $80-64 = 16$.

The core principle for problems involving mean and variance is that they are entirely determined by the sum of the data points and the sum of their squares. For a set of $n$ numbers with mean $\bar{x}$, the total sum is $n \cdot \bar{x}$, while the variance $\sigma^2$ relates to these values via the computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

With $8$ observations, calculating the mean as $\frac{6+4+a+8+b+12+10+13}{8} = 9$ leads directly to $53+a+b = 72$, which simplifies to $a+b = 19$. Substituting the known values into the variance equation gives $\frac{36+16+a^{2+}64+b^{2+}144+100+169}{8} - 9^2 = 9.25$, or $\frac{529+a^{2+}b^2}{8} = 90.25$, which results in $a^{2+}b^2 = 193$. By applying the algebraic identity $(a+b)^2 = a^{2+}b^{2+}2ab$, we can write $19^2 = 193 + 2ab$, resulting in $361 = 193 + 2ab$ and therefore $2ab = 168$, or $ab = 84$. Combining these findings, the value of $a+b+ab$ is $19 + 84$, which equals $103$.

The relationship between a dataset and its linear transformation is governed by fundamental properties of mean and variance. For a set of values $x$ with mean $\bar{x}$ and variance $\sigma_x^2$, any linear transformation of the form $ax + b$ results in a new mean of $a\bar{x} + b$ and a new variance of $a^2 \sigma_x^2$.

To determine the initial parameters, we analyze the sum of the observations. Since $\sum_{i=1}^{10} (x_i - 2) = 30$, we have $\sum x_i - 20 = 30$, implying $\sum x_i = 50$, which yields a mean $\bar{x} = 5$. Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, we substitute the known variance of $\frac{4}{5}$ to get $\frac{4}{5} = \frac{\sum x_i^2}{10} - 25$, leading to $\sum x_i^2 = 258$. Expanding the equation $\sum_{i=1}^{10} (x_i - \beta)^2 = 98$ gives $\sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 98$. Substituting the calculated sums results in $258 - 2\beta(50) + 10\beta^2 = 98$, which simplifies to the quadratic equation $\beta^2 - 10\beta + 16 = 0$. Factoring this yields $(\beta - 8)(\beta - 2) = 0$, and given the condition $\beta > 2$, we find $\beta = 8$.

The new observations are defined as $y_i = 2(x_i - 1) + 4\beta$. Substituting $\beta = 8$, this expression simplifies to $y_i = 2x_i - 2 + 32$, or $y_i = 2x_i + 30$. Applying the transformation properties, the new mean $\mu = 2(5) + 30 = 40$, and the new variance $\sigma^2 = 2^2 \times (\frac{4}{5}) = \frac{16}{5}$. Computing the final required ratio gives $\frac{\beta \mu}{\sigma^2} = \frac{8 \times 40}{16/5} = \frac{320}{16/5} = \frac{320 \times 5}{16} = 100$.

Correcting statistics involves reversing the error in the total sum and the sum of squared values before recalculating the mean and variance. Since the mean is the total sum divided by the number of observations, the original sum is $100 \times 40 = 4000$. Subtracting the erroneous value $50$ and adding the correct value $40$ yields a new total sum of $3990$, making the corrected mean $\mu = \frac{3990}{100} = 39.9$.

The updated standard deviation relies on the relationship $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2$. The incorrect variance of $5.1^2$ corresponds to an initial sum of squares of $100 \times (5.1^2 + 40^2) = 162601$. Removing the contribution of the incorrect observation $50^2 = 2500$ and adding the square of the correct one $40^2 = 1600$ results in an adjusted sum of squares of $161701$. Dividing this by $100$ and subtracting the square of the new mean $39.9^2$ gives a variance of $25$, which means $\sigma = 5$. Combining these values, $10(\mu + \sigma) = 10(39.9 + 5) = 449$.

Variance describes the dispersion of a dataset around its mean, determined for a frequency distribution by evaluating $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$, where $N$ is the total frequency count. The total frequency $N$ is found by summing the individual frequencies to reach 22, and the mean $\bar{x}$ is calculated by dividing the sum of the products of data and frequency, $\sum f_i x_i = 176$, by this total, resulting in 8. Squaring each observation, multiplying by its frequency, and summing these values produces $\sum f_i x_i^2 = 2048$. Substituting these figures into the variance formula gives $\frac{2048}{22} - 8^2$, which simplifies to approximately $93.09 - 64$, yielding a value of 29.09.

The variance of a dataset represents the spread of values relative to the mean, mathematically expressed by the formula $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$. With a mean of 56 and a variance of 66.2 for 10 data points, we can determine the aggregate sum of squares for the entire set. By substituting the given parameters into the equation, we get $\frac{\sum x_i^2}{10} - (56)^2 = 66.2$. This rearranges to $\frac{\sum x_i^2}{10} = 3136 + 66.2$, which identifies the sum of all squared terms as 32022. Summing the squares of the eight known observations-65, 68, 58, 44, 48, 45, 60, and 60-results in 25678. Subtracting this known total from the aggregate sum yields $\alpha^2 + \beta^2 = 32022 - 25678 = 6344$.

Finding the median of a grouped frequency distribution involves identifying the specific class interval that contains the middle observation. First, we compute the cumulative frequencies, which are 3, 12, 22, 30, and 36, indicating a total frequency $N$ of 36. Because the median position corresponds to $\frac{N}{2}$, which is 18, we locate the median class as 8-12, as this is the first interval where the cumulative frequency reaches 18.

Applying the median formula $M = L + \left( \frac{\frac{N}{2} - C}{f} \right) h$ with the lower limit $L=8$, cumulative frequency of the preceding class $C=12$, median class frequency $f=10$, and class width $h=4$, we arrive at $M = 8 + \frac{18-12}{10} × 4$. Calculating this expression yields $M = 8 + 2.4$, which equals 10.4. Multiplying this value by 20 results in 208.

The mean deviation about the median provides a measure of dispersion by calculating the average absolute difference between each data point and the median value. Since the total number of students in the class is 40, the sum of all individual frequencies $5+8+5+12+x+y$ must equal 40, which simplifies to the relationship $x+y=10$.

To determine the unknown values, we apply the mean deviation formula $\frac{\sum f_i |x_i - M|}{\sum f_i} = 1.25$ where $M=18$ is the given median. Summing the absolute deviations for each age group, we compute $5|15-18| + 8|16-18| + 5|17-18| + 12|18-18| + x|19-18| + y|20-18|$, which simplifies to $15 + 16 + 5 + 0 + x + 2y$, or $36+x+2y$. Setting the total deviation over the total frequency equal to 1.25, we find $36+x+2y = 50$, leading to the second equation $x+2y=14$. Solving the system of equations $x+y=10$ and $x+2y=14$ reveals $y=4$ and $x=6$. Substituting these values into the final expression $4x+5y$ gives $4(6) + 5(4)$, which equals 44.

Since the sum of probabilities in any valid distribution must equal unity, the equation $\frac{1}{3} + k + \frac{1}{6} + \frac{1}{4} = 1$ leads directly to $k = \frac{1}{4}$. The mean $\mu$ is defined by the sum of each value multiplied by its probability, which simplifies to $\mu = \frac{\alpha}{3} - \frac{1}{2}$. The variance $\sigma^2$ is then determined by the expectation of the squared variable minus the squared mean, represented by \sigma^2 = E\[X^2]$ - \mu^2$. Substituting the provided distribution values gives:
$\sigma^2 = \frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4}$

Given the relationship $\sigma - \mu = 2$, we can square both sides to get $\sigma^2 = (\mu + 2)^2 = \mu^2 + 4\mu + 4$. Expanding this using the previously derived expressions for $\mu$ leads to the following equality:
$\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4} = \frac{\alpha^2}{9} + \alpha + \frac{9}{4}$
Simplifying this equation results in $\frac{\alpha^2}{9} - \frac{2\alpha}{3} = 0$, which yields $\alpha = 0$ or $\alpha = 6$. Since the value $0$ is already present in the distribution, we must select $\alpha = 6$. Substituting this back into the expression for the mean, we find $\mu = \frac{6}{3} - \frac{1}{2} = 1.5$. Using $\sigma = \mu + 2$, the required value $\sigma + \mu$ is $2\mu + 2$, which evaluates to $2(1.5) + 2 = 5$.

The variance of a dataset is fundamentally related to the difference between the mean of the squares of the observations and the square of their mean, expressed by the identity $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$. For the initial set of 15 observations, the incorrect mean of 12 and standard deviation of 3 allow us to calculate the incorrect sum $\sum x_i = 15 \times 12 = 180$ and the incorrect sum of squares $\sum x_i^2 = 15 \times (3^2 + 12^2) = 2295$. Correcting the misread data point by removing 10 and including 12 updates the total sum to $\sum x_i = 180 - 10 + 12 = 182$ and the sum of squares to $\sum x_i^2 = 2295 - 100 + 144 = 2339$. Substituting these corrected values into the required expression $15 (\mu + \mu^2 + \sigma^2)$, which algebraically simplifies to $15 (\mu + \frac{\sum x_i^2}{n}) = 15\mu + \sum x_i^2$, leads to $15 (\frac{182}{15}) + 2339$, which results in a final value of 2521.

The mean and variance of a dataset describe its central tendency and dispersion, allowing for the determination of unknown parameters when aggregate sums of observations are provided. By leveraging the definition of the mean and the properties of variance relative to constant shifts, we can isolate the required values for $\alpha$ and $\beta$ through algebraic manipulation.

Starting with the sum $\sum_{i=1}^{10} (x_i - \alpha) = 2$, we can expand this to $\sum_{i=1}^{10} x_i - 10\alpha = 2$. Given the mean of the ten observations is $\frac{6}{5}$, the total sum of the observations is $\sum_{i=1}^{10} x_i = 10 \times \frac{6}{5} = 12$. Substituting this total into the expanded equation yields $12 - 10\alpha = 2$, which simplifies to $10 = 10\alpha$, confirming that $\alpha = 1$.

To find $\beta$, we apply the variance formula involving deviations from a constant, which is expressed as $\sigma^2 = \frac{1}{n} \sum (x_i - \beta)^2 - \left(\frac{\sum (x_i - \beta)}{n}\right)^2$. Inserting the known values $n=10$, $\sigma^2 = \frac{84}{25}$, and $\sum_{i=1}^{10} (x_i - \beta)^2 = 40$, the expression becomes $\frac{84}{25} = \frac{40}{10} - \left(\frac{12 - 10\beta}{10}\right)^2$. This simplifies to $\frac{84}{25} = 4 - \left(\frac{6 - 5\beta}{5}\right)^2$. Rearranging to isolate the squared term gives $\left(\frac{6 - 5\beta}{5}\right)^2 = 4 - \frac{84}{25} = \frac{16}{25}$, which implies $6 - 5\beta = \pm 4$. Solving this yields $5\beta = 6 \mp 4$, resulting in $\beta = \frac{2}{5}$ or $\beta = 2$. Since $\beta$ must be a positive integer, we identify $\beta = 2$, making the final ratio $\frac{\beta}{\alpha} = 2$.

The calculation of standard deviation relies on the relationship between the variance, the mean, and the sum of squares, defined by the formula $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$. Identifying the corrected standard deviation requires first deriving the initial parameters from the given mean of 10 and standard deviation of 2 across 20 observations. Since the mean is 10, the initial sum of observations is $\Sigma x_i = 20 \times 10 = 200$. Substituting the variance of $2^2 = 4$ into the formula, we find the initial sum of squares as $\frac{\Sigma x_i^2}{20} - 10^2 = 4$, which yields $\Sigma x_i^2 = 20(104) = 2080$.

Adjusting for the incorrect value of 8 by replacing it with 12 alters both the total sum and the sum of squares. The new sum of observations becomes $200 - 8 + 12 = 204$, producing an updated mean of $\frac{204}{20} = 10.2$. Similarly, the new sum of squares is calculated by subtracting the square of the incorrect value and adding the square of the correct value, resulting in $2080 - 64 + 144 = 2160$. Applying these updated figures to the variance formula gives $\frac{2160}{20} - (10.2)^2$, which simplifies to $108 - 104.04 = 3.96$. Consequently, the corrected standard deviation is $\sqrt{3.96}$.

The variance of a frequency distribution quantifies how far individual data points deviate from the mean, calculated by taking the difference between the mean of the squared values and the square of the mean. This is expressed using the formula $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$, where $N$ represents the total frequency of the distribution.

To find the unknown constant, start by determining the total frequency $N$ by summing the given frequencies, which is 7. Next, calculate the mean $\bar{x}$ using the values provided in the distribution as $\bar{x} = \frac{2c + 2c + 3c + 4c + 5c + 6c}{7} = \frac{22c}{7}$. The sum of the squares of the observations weighted by their respective frequencies is $\sum f_i x_i^2 = 2c^2 + 1(4c^2) + 1(9c^2) + 1(16c^2) + 1(25c^2) + 1(36c^2)$, which simplifies to $92c^2$.

Substitute these values into the variance formula to obtain $\frac{92c^2}{7} - (\frac{22c}{7})^2 = 160$. Simplifying the square of the mean results in $\frac{92c^2}{7} - \frac{484c^2}{49} = 160$. By finding a common denominator of 49, the expression becomes $\frac{644c^2 - 484c^2}{49} = 160$, which simplifies to $\frac{160c^2}{49} = 160$. Dividing both sides by 160 and multiplying by 49 reveals that $c^2 = 49$, which confirms that the value of the constant is 7.

Mean deviation about a central measure such as the median or the mean is calculated by finding the average of the absolute differences between each data point and that central value. For the given set of seven observations $125, a, b, 170, 190, 210, 230$, the median is given as $170$. By calculating the mean deviation about the median, which is the average of $|125-170|$, $|a-170|$, $|b-170|$, $|170-170|$, $|190-170|$, $|210-170|$, and $|230-170|$, we obtain $\frac{45 + (170-a) + (170-b) + 0 + 20 + 40 + 60}{7} = \frac{205}{7}$. Simplifying this expression yields $505 - (a+b) = 205$, which confirms that the sum of the unknown variables is $a+b = 300$.

With the sum $a+b$ established, the arithmetic mean is determined by dividing the sum of all seven observations by $7$. This gives $\frac{170 + 125 + 230 + 190 + 210 + (a+b)}{7} = \frac{925 + 300}{7} = \frac{1225}{7} = 175$. The mean deviation about the mean is then calculated as the average of the absolute differences between each observation and $175$, given by $\frac{|125-175| + |a-175| + |b-175| + |170-175| + |190-175| + |210-175| + |230-175|}{7}$. Substituting the values, this expression becomes $\frac{50 + (175-a) + (175-b) + 5 + 15 + 35 + 55}{7}$. Since $a+b=300$, the terms $(175-a) + (175-b)$ simplify to $350 - 300 = 50$, leading to a total deviation of $\frac{50 + 50 + 5 + 15 + 35 + 55}{7} = \frac{210}{7} = 30$.

Calculating the standard deviation requires identifying the relationship between the sum of the squares of the observations and the square of their sum. The standard deviation $\sigma$ is defined by the formula $\sigma = \sqrt{\frac{\sum a_i^2}{n} - (\frac{\sum a_i}{n})^2}$. Given 10 observations with a sum of 50, the mean is $\frac{50}{10} = 5$, making the square of the mean 25.

We determine the sum of the squares by expanding the square of the sum of the observations according to the algebraic identity:
$(\sum_{i=1}^{10} a_i)^2 = \sum_{i=1}^{10} a_i^2 + 2 \sum_{k < j} a_k a_j$
Substituting the given totals yields $50^2 = \sum a_i^2 + 2(1100)$, which simplifies to $2500 = \sum a_i^2 + 2200$. Isolating the sum of squares results in $\sum a_i^2 = 300$. Substituting these values into the variance formula, we find the variance is $\frac{300}{10} - 25 = 5$. Consequently, the standard deviation is $\sqrt{5}$.

Statistics problems involving unknown variables rely on the fundamental definitions of mean and variance to create a system of equations that allows us to isolate the missing values. Since the arithmetic mean is the sum of all observations divided by the total count, we start by calculating the sum of our six observations. Given that the mean is 55 for these six values, the total sum must satisfy the equation $a + b + 68 + 44 + 48 + 60 = 55 \times 6 = 330$, which simplifies to $a + b = 110$.

Next, we utilize the formula for variance, which measures the average of the squared differences from the mean, expressed as $\frac{\sum (x_i - \bar{x})^2}{n} = 194$. Substituting our known values into the equation yields $(a-55)^2 + (b-55)^2 + (68-55)^2 + (44-55)^2 + (48-55)^2 + (60-55)^2 = 194 \times 6$. Simplifying the known squared differences results \in $169 + 121 + 49 + 25 = 364$, which leads to the equation $(a-55)^2 + (b-55)^2 = 1164 - 364 = 800$. Expanding these squares gives $a^2 + b^2 - 110(a+b) + 6050 = 800$, and by substituting our earlier finding of $a+b=110$, we determine that $a^2 + b^2 = 6850$.

With both the sum and the sum of squares identified, we can determine the individual values for $a$ and $b$. Using the system of equations where $a+b=110$ and $a^{2+}b^2=6850$, we can deduce that the values must be 75 and 35. Since the problem specifies that $a > b$, we assign $a=75$ and $b=35$. Finally, calculating the expression $a+3b$ results \in $75 + 3(35) = 75 + 105 = 180$.

Mean deviation is defined as the average of the absolute differences between each observation and the distribution's mean. Given the mean is 2, the sum of all 6 observations is 12, which implies $a + b = 10$ because the known values total 2. Applying the variance formula
$\frac{\sum x_i^2}{6} - 2^2 = 23$
shows that the sum of the squares of the observations is 162; subtracting the sum of the known squares, 110, gives $a^2 + b^2 = 52$, identifying $a$ and $b$ as 4 and 6. Finally, computing the mean deviation using the completed dataset $\{-3, 4, 7, -6, 4, 6\}$ and the mean 2, we sum the absolute distances:
$\frac{|-3-2| + |4-2| + |7-2| + |-6-2| + |4-2| + |6-2|}{6} = \frac{26}{6}$
This simplifies to $\frac{13}{3}$.

The properties of central tendency and dispersion allow us to define algebraic constraints on unknown observations within a dataset. Given the mean of 18 for 5 observations 9, 25, a, b, and c, the sum of the variables is $a + b + c = 56$. The mean deviation of 4 implies that the absolute deviations sum to $|a-18| + |b-18| + |c-18| = 4$, and the variance of 136/5 indicates that the squared deviations sum to $(a-18)^2 + (b-18)^2 + (c-18)^2 = 6$. Identifying three integers whose absolute differences from 18 sum to 4 and whose squares sum to 6 reveals that the differences from 18 must be -1, 1, and 2. Arranging these such that $a < b < c$, we find $a = 17$, $b = 19$, and $c = 20$. This set satisfies the sum and the ordering condition, leading to the result calculated as
$2a + b - c = 2(17) + 19 - 20 = 33$

The variance of a dataset represents the average squared deviation from the mean, calculated as the difference between the mean of the squares of the observations and the square of their mean. Given that the mean of five observations is $\frac{24}{5}$, their total sum is $24$. Since the mean of the first four observations is $\frac{7}{2}$, their total sum is $14$, which implies that the fifth observation is $24 - 14 = 10$.

To determine the variance of the first four observations, we must first find the sum of the squares of all five observations using the provided variance:
$\frac{\sum_{i=1}^{5} x_i^2}{5} - \left(\frac{24}{5}\right)^2 = \frac{194}{25}$
Solving this equation, we find that $\frac{\sum_{i=1}^{5} x_i^2}{5} = \frac{194}{25} + \frac{576}{25} = \frac{770}{25}$, meaning the sum of the squares of all five observations is 154. Subtracting the square of the fifth observation, $10^2 = 100$, leaves 54 as the sum of the squares for the first four observations. Calculating the variance of this subset, we take the mean of these squares and subtract the square of their mean: $\frac{54}{4} - (\frac{7}{2})^2$, which simplifies to $\frac{54}{4} - \frac{49}{4}$, resulting in $\frac{5}{4}$.

The mean of a data distribution is the weighted average of the observations, calculated by dividing the sum of the products of each value and its corresponding frequency by the total number of observations. Since the mean is 5, we determine the unknown frequency $\alpha$ by setting up the following equation:
$5 = \frac{1(4) + 3(24) + 5(28) + 7(\alpha) + 9(8)}{4 + 24 + 28 + \alpha + 8}$
Simplifying this expression leads to $288 + 7\alpha = 5(64 + \alpha)$, which simplifies further to $288 + 7\alpha = 320 + 5\alpha$, resulting in $\alpha = 16$. The total frequency of the distribution is therefore 80.

Once the data set is complete, we calculate the mean deviation about the mean, $m$, which represents the average absolute difference between the observations and the mean, and the variance, $\sigma^2$, which captures the average of the squared differences. The absolute differences of the data points from the mean of 5 are 4, 2, 0, 2, and 4, respectively. Thus, the mean deviation is:
$m = \frac{4(4) + 24(2) + 28(0) + 16(2) + 8(4)}{80} = \frac{128}{80}$
Similarly, the variance is found by weighting the squared differences:
$\sigma^2 = \frac{4(4)^2 + 24(2)^2 + 28(0)^2 + 16(2)^2 + 8(4)^2}{80} = \frac{352}{80}$

With these two parameters calculated over the same total frequency, their sum is $m + \sigma^2 = \frac{128 + 352}{80} = \frac{480}{80} = 6$. Substituting $\alpha = 16$ and the calculated sum into the required expression yields:
$\frac{3\alpha}{m+\sigma^2} = \frac{3(16)}{6} = \frac{48}{6} = 8$