Statistics – JEE Main MathematicsPractice Questions & PYQs

Standard deviation is a measure of dispersion that relies on the difference between the mean of the squared observations and the square of the mean of the observations. To determine this value for a set of ten observations, we must first isolate the sum of the variables, $\sum_{i=1}^{10} x_i$, and the sum of their squares, $\sum_{i=1}^{10} x_i^2$, by expanding the algebraic expressions provided.

Expanding the first expression, $\sum_{i=1}^{10} (x_i+2)^2 = 180$, gives $\sum_{i=1}^{10} x_i^2 + 4 \sum_{i=1}^{10} x_i + \sum_{i=1}^{10} 4 = 180$, which simplifies to $\sum_{i=1}^{10} x_i^2 + 4 \sum_{i=1}^{10} x_i = 140$ after subtracting 40 from both sides. Expanding the second expression, $\sum_{i=1}^{10} (x_i-1)^2 = 90$, results in $\sum_{i=1}^{10} x_i^2 - 2 \sum_{i=1}^{10} x_i + \sum_{i=1}^{10} 1 = 90$, which simplifies to $\sum_{i=1}^{10} x_i^2 - 2 \sum_{i=1}^{10} x_i = 80$. By subtracting the second simplified equation from the first, we find $6 \sum_{i=1}^{10} x_i = 60$, indicating that the sum of the observations is 10. Substituting this result back into $\sum_{i=1}^{10} x_i^2 - 2(10) = 80$ reveals that the sum of squares is 100.

Given $N=10$, $\sum_{i=1}^{10} x_i = 10$, and $\sum_{i=1}^{10} x_i^2 = 100$, the variance is calculated using the formula
$\sigma^2 = \frac{\sum_{i=1}^{10} x_i^2}{N} - \left( \frac{\sum_{i=1}^{10} x_i}{N} \right)^2$
Substituting the known values yields $\sigma^2 = \frac{100}{10} - \left( \frac{10}{10} \right)^2$, which simplifies to $10 - 1 = 9$. Taking the square root of the variance provides a standard deviation of 3.

The mean deviation about the mean is determined by finding the weighted average of the absolute differences between each data point and the arithmetic mean. We first calculate the mean $\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{234}{26} = 9$ by totaling the product of each value and its frequency. With the mean known, we identify the absolute distance of each data value from 9, resulting in $|5-9|=4$, $|7-9|=2$, $|9-9|=0$, $|10-9|=1$, $|12-9|=3$, and $|15-9|=6$. Multiplying these distances by their corresponding frequencies gives the set of values 32, 12, 0, 2, 6, 36, the sum of which is 88. Dividing this sum by the total frequency of 26 results in the mean deviation, $\frac{88}{26}$, which simplifies to $\frac{44}{13}$.

The mean of a data set defined by binomial frequencies is derived by dividing the sum of the products of each value and its frequency by the total frequency, which for ${}^n C_r$ is $2^n$. The numerator involves the sum $\sum_{r=0}^n r(r-1) {}^n C_r$, which simplifies using the property $r(r-1) {}^n C_r = n(n-1) {}^{n-2} C_{r-2}$ to $n(n-1) 2^{n-2}$. Dividing this by the total frequency $2^n$ gives the simplified formula for the mean:

$\frac{n(n-1) 2^{n-2}}{2^n} = \frac{n(n-1)}{4}$

Setting this expression equal to 60 gives the equation $n(n-1) = 240$, and solving for the positive integer $n$ yields $n = 16$. Since the frequencies follow the binomial distribution, the median of the data corresponds to the value at the central term of the expansion. With $n = 16$, the center of the distribution occurs at index $r = 8$. Plugging this value into the expression for the variable $X$, we calculate $8(8-1)$, resulting in 56.

The fundamental relationship between the mean, variance, and the sum of squared observations is key to solving this problem. Expanding the summations $\sum_{i=1}^{20} (x_i+5)^2 = 2500$ and $\sum_{i=1}^{20} (x_i-5)^2 = 100$ leads to the expressions $\sum x_i^2 + 10\sum x_i + 500 = 2500$ and $\sum x_i^2 - 10\sum x_i + 500 = 100$. Subtracting the second from the first reveals $20\sum x_i = 2400$, which implies $\sum x_i = 120$ and a mean of $\overline{x} = \frac{120}{20} = 6$.

Adding the two expanded equations gives $2\sum x_i^2 + 1000 = 2600$, simplifying to $\sum x_i^2 = 800$. Applying the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2$ results in $\sigma^2 = \frac{800}{20} - 6^2 = 40 - 36 = 4$, making the standard deviation $\sigma = \sqrt{4} = 2$. Consequently, the ratio of the mean to the standard deviation is $6 : 2$, which simplifies to $3 : 1$.

Determining the two missing values is the first step, as the mean of 10 observations being 9 implies a total sum of 90. Removing the sum of the known 8 observations, which is 80, leaves 10 for the missing pair; using 4 and 6 allows the data set to be ordered as $2, 3, 4, 5, 6, 10, 11, 13, 15, 21$.

The median of an even-numbered set is found by averaging the two central terms. With the fifth and sixth values being 6 and 10, the median is $M = \frac{6+10}{2} = 8$.

Mean deviation about the median is defined as the average of the absolute deviations of all data points from this median. Summing these differences results in the expression
$\sum |x_i - 8| = |2-8| + |3-8| + |4-8| + |5-8| + |6-8| + |10-8| + |11-8| + |13-8| + |15-8| + |21-8|$
which equates to $6 + 5 + 4 + 3 + 2 + 2 + 3 + 5 + 7 + 13 = 50$. Dividing this total sum by the count of 10 observations yields the final value of 5.

The mean deviation about the median $M$ of a series of $n$ numbers is defined by $MD = \frac{1}{n} \sum_{i=1}^{n} |x_i - M|$. For the sequence $k, 2k, \dots, 1000k$, the median is the average of the 500th and 501st terms, which yields $M = \frac{500k + 501k}{2} = 500.5k$. Substituting these values into the formula, we have $\frac{1}{1000} \sum_{i=1}^{1000} |ik - 500.5k| = \frac{|k|}{1000} \sum_{i=1}^{1000} |i - 500.5| = 500$. Evaluating the sum $\sum_{i=1}^{1000} |i - 500.5|$ results in $250000$. This leads to the simplified equation $250|k| = 500$, implying $|k| = 2$, so $k^2 = 4$.

Calculating the mean of a grouped frequency distribution requires the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$, where $x_i$ denotes the midpoint of each class interval and $f_i$ is the corresponding frequency. By identifying the midpoints for the class intervals from $5-10$ through $30-35$ and applying the provided frequencies, one can construct an equation where the weighted sum divided by the total frequency equals the given mean of 21. Solving this equation yields the value $k=10$. Substituting $x=10$ into the quadratic equation $2x^{2}-19x-10=0$ results in $2(10)^{2}-19(10)-10$, which simplifies to $200-190-10=0$, verifying that $k$ is a root of this quadratic equation.

When merging two distinct data groups, the overall variance is not simply a weighted average of the individual variances; you must also incorporate a correction factor that accounts for the distance between the two group means. This calculation is handled by a combined variance formula that weighs each group's variance by its size relative to the total population, and then adds the squared difference between the means scaled by the product of the sample sizes. For this specific scenario, the variables are $n_1 = 4$, $\sigma_1^2 = 13$, $\overline{x_1} = 1$ and $n_2 = 6$, $\sigma_2^2 = 1$, $\overline{x_2} = 2$, which fit into the formula:

$\sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{x_1} - \overline{x_2})^2$

Plugging in these values results in $\frac{4 \times 13 + 6 \times 1}{10} + \frac{4 \times 6}{10^2} (1 - 2)^2$. Computing the first term gives $\frac{58}{10}$, which is $5.8$, while the second term becomes $\frac{24}{100} \times 1$, or $0.24$. Summing these components yields a final combined variance of $6.04$.

The mean of a data set is defined by $\bar{x} = \frac{\sum x_i}{n}$ and the variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. Given the mean of the eight numbers $-10,-7,-1, x, y, 9,2,16$ is $\frac{7}{2}$, the sum is $\sum x_i = x+y+9$, leading to $\frac{x+y+9}{8} = \frac{7}{2}$, which simplifies to $x+y=19$. Substituting the variance $\frac{293}{4}$ and mean $\frac{7}{2}$ into the variance formula results in $\frac{293}{4} = \frac{100+49+1+x^{2+}y^{2+}81+4+256}{8} - \frac{49}{4}$, which implies $\frac{342}{4} = \frac{x^{2+}y^{2+}491}{8}$, solving to $x^{2+}y^2=193$. Utilizing the identity $(x+y)^2 = x^{2+}y^{2+}2xy$, we find $361 = 193 + 2xy$, thus $xy=84$. The numbers $x$ and $y$ are roots of the quadratic equation $t^2 - 19t + 84 = 0$, giving $x, y$ values of $7$ and $12$, which implies $|x-y|=5$. The four numbers are $7, 12, 19+1=20$, and $5$, and their mean is $\frac{7+12+20+5}{4} = \frac{44}{4} = 11$.

The mean of a transformed data set $Y = aX + b$ is given by $\bar{Y} = a\bar{X} + b$ and the variance by $\sigma_Y^2 = a^2 \sigma_X^2$, where $\bar{X}$ and $\sigma_X^2$ represent the mean and variance of the original set $X$. For $X = \{1, 2, \dots, 19\}$, the mean is $\bar{X} = \frac{1+19}{2} = 10$ and the variance is calculated using the formula for the first $n$ natural numbers:
$\sigma_X^2 = \frac{n^2 - 1}{12} = \frac{19^2 - 1}{12} = \frac{360}{12} = 30$
Given the variance of $Y$ is 750, we solve for $a$:
$750 = a^2 \cdot 30 \implies a^2 = 25 \implies a = \pm 5$
Using the relation for the mean $\bar{Y} = 30$, we substitute $\bar{X} = 10$:
$30 = 10a + b \implies b = 30 - 10a$
Substituting $a = 5$ gives $b = 30 - 50 = -20$, and substituting $a = -5$ gives $b = 30 - (-50) = 80$. The sum of all possible values of $b$ is $-20 + 80 = 60$.

The mean $\mu$ and variance $\sigma^2$ for a grouped frequency distribution are defined by $\mu = \frac{\sum f_i x_i}{N}$ and $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$, where $x_i$ are the class midpoints, $f_i$ are the frequencies, and $N = \sum f_i$. For the given intervals $4-8, 8-12, 12-16, 16-20$, the midpoints are $6, 10, 14, 18$ and the total frequency is $N = 3 + \lambda + 4 + 7 = 14 + \lambda$. Substituting these values into the mean formula gives $\mu = \frac{6(3) + 10(\lambda) + 14(4) + 18(7)}{14 + \lambda} = \frac{200 + 10\lambda}{14 + \lambda}$.

The weighted sum of the squares of the midpoints is $\sum f_i x_i^2 = 6^2(3) + 10^2(\lambda) + 14^2(4) + 18^2(7) = 108 + 100\lambda + 784 + 2268 = 3160 + 100\lambda$. Substituting the expression for variance yields the equation $\frac{3160 + 100\lambda}{14 + \lambda} - \left(\frac{200 + 10\lambda}{14 + \lambda}\right)^2 = 19$. Multiplying by $(14 + \lambda)^2$ and simplifying results in the quadratic equation $19\lambda^2 - 28\lambda - 516 = 0$.

Solving for the positive root gives $\lambda = \frac{28 + \sqrt{784 + 39216}}{38} = \frac{28 + 200}{38} = 6$. With $\lambda = 6$, the mean is $\mu = \frac{200 + 10(6)}{14 + 6} = \frac{260}{20} = 13$. Therefore, the value of $\lambda + \mu = 6 + 13 = 19$.

The sum of observations and the sum of their squares are related to the mean $\bar{x}$ and variance $\sigma^2$ by the formulas $\sum x_i = n\bar{x}$ and $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$. For the initial data with $n=10$, the sum of observations is $\sum x_i = 10 \times 10 = 100$ and the sum of squares is $\sum x_i^2 = 10(2 + 10^2) = 1020$. Replacing $\alpha$ with $\beta$ yields a new mean of $10.1$, so the new sum of observations is $\sum x_i' = 10 \times 10.1 = 101$, implying $\beta - \alpha = 1$. The new variance of $1.99$ results in a new sum of squares of $\sum x_i'^2 = 10(1.99 + 10.1^2) = 1040$. Since the change in the sum of squares is $\beta^2 - \alpha^2 = 1040 - 1020 = 20$, and this factors as $(\beta - \alpha)(\beta + \alpha) = 20$, substituting $\beta - \alpha = 1$ results in $\alpha + \beta = 20$.

The sum of probabilities for a random variable must equal one, which implies $\frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$. Simplifying this expression results in $\frac{14a+1}{30} + b = 1$, which simplifies further to $14a + 30b = 29$.

The given condition $\sigma^2 + \mu^2 = 2$ represents the second moment of the random variable, E\[X^2]$ = \sum x_i^2 P(x_i) = 2$. Substituting the values$0, 1, 2, 3$and their respective probabilities into this formula gives$0^2 \cdot \frac{2a+1}{30} + 1^2 \cdot \frac{8a-1}{30} + 2^2 \cdot \frac{4a+1}{30} + 3^2 \cdot b = 2$. This expands to$\frac{8a-1}{30} + \frac{16a+4}{30} + 9b = 2$, which simplifies to$\frac{24a+3}{30} + 9b = 2$, or$\frac{8a+1}{10} + 9b = 2$, leading to the linear equation$8a + 90b = 19$.

Solving the system of equations $14a + 30b = 29$ and $8a + 90b = 19$ involves multiplying the first equation by 3 to obtain $42a + 90b = 87$. Subtracting the second equation from this result yields $34a = 68$, resulting in $a = 2$. Substituting $a = 2$ into $14a + 30b = 29$ gives $28 + 30b = 29$, so $30b = 1$, or $b = \frac{1}{30}$. Consequently, the ratio $\frac{a}{b}$ is $\frac{2}{1/30} = 60$.

Variance is fundamentally linked to the mean through the relationship $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$, which allows us to relate the total sum of observations and the sum of their squares to the number of terms $n$. Given that the mean is 8 and the sum of the first $(n - 1)$ observations is 48, we define the $n$-th observation as $x_n = 8n - 48$. Substituting the known sum of squares 496 and this expression for $x_n$ into the variance formula results in $\frac{496 + (8n - 48)^2}{n} - 64 = 16$. Rearranging this equation leads to $496 + 64(n - 6)^2 = 80n$, which simplifies to the quadratic equation $4n^2 - 53n + 175 = 0$. Solving this expression yields $n = 7$ as the only valid integer solution.

The total sum of the nine given numbers must be $9 \times 40 = 360$, providing the initial equation $21 + 8 + 17 + a + 51 + 103 + b + 13 + 67 = 360$, which simplifies to $280 + a + b = 360$, or $a + b = 80$. Since the median of these nine numbers is 21, the data points must be arranged in non-decreasing order such that 21 occupies the fifth position, implying $b < 21$ and $a > 21$. Specifically, the sorted order of the numbers is $8, 13, b, 17, 21, 51, a, 67, 103$.

The mean deviation about the median is defined as the average of the absolute differences between each observation and the median, expressed as $\frac{\sum |x_i - 21|}{9} = 26$. Calculating the absolute differences for the known values gives $|8-21| + |13-21| + |b-21| + |17-21| + |21-21| + |51-21| + |a-21| + |67-21| + |103-21| = 234$. Given the conditions $b < 21 < a$, this simplifies to $13 + 8 + (21-b) + 4 + 0 + 30 + (a-21) + 46 + 82 = 234$. Combining the constants results in $183 + a - b = 234$, which simplifies to $a - b = 51$. Solving the system $a + b = 80$ and $a - b = 51$ by adding the two equations yields $2a = 131$.

The mean and variance of a dataset are tied to the sum and the sum of squares of the observations, which allows us to determine unknowns in the set. The sum of the seven values $2, 4, \alpha, 8, \beta, 12, 14$ is $7 \times 8 = 56$, which simplifies to $\alpha + \beta = 16$. Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$, we substitute the given values $16 = \frac{\sum x_i^2}{7} - 64$, implying $\sum x_i^2 = 560$. By summing the squares of the known numbers and subtracting from $560$, we find $\alpha^2 + \beta^2 = 136$. Solving the system $\alpha + \beta = 16$ and $\alpha^2 + \beta^2 = 136$ leads to $\alpha=6$ and $\beta=10$ given the constraint $\alpha < \beta$. Finally, the roots of the required quadratic equation are $3(6)+2=20$ and $2(10)+1=21$, and constructing the equation from these roots gives $x^2 - (20+21)x + (20 \times 21) = 0$, simplifying to $x^{2-}41x+420=0$.

The numbers form an arithmetic progression starting at $8$ with a common difference of $13$, allowing us to determine the total number of terms by solving $320 = 8 + (n - 1)13$, which gives $n = 25$. To find the variance, we first identify the mean, which for an arithmetic progression is simply the average of the first and last terms, calculated as $\frac{8 + 320}{2} = 164$.

The variance measures the dispersion of these values around the mean and can be computed efficiently for an arithmetic progression using the formula $\sigma^2 = d^2 \frac{n^2 - 1}{12}$. Substituting the known values $d = 13$ and $n = 25$ yields the following:
$\sigma^2 = 13^2 \times \frac{25^2 - 1}{12}$
$\sigma^2 = 169 \times \frac{624}{12} = 169 \times 52 = 8788$

This calculation demonstrates that the variance for the given sequence is 8788.

To find the variance of a dataset after correcting specific values, one must first update the sum of the observations and the sum of their squares, as these are the foundational components of the variance formula. Given the initial mean of 5.5 for 10 values, the total sum is $10 \times 5.5 = 55$, while the original sum of squares is 371. By removing the erroneous entries 4 and 5 and including the correct values 6 and 8, the updated sum becomes $55 - 4 - 5 + 6 + 8 = 60$, which results in a corrected mean of 6.

Adjust the sum of squares by subtracting the squares of the original errors and adding the squares of the correct entries, which calculation is $371 - 4^2 - 5^2 + 6^2 + 8^2 = 371 - 16 - 25 + 36 + 64 = 430$. Finally, the variance is determined by subtracting the square of the new mean from the average of the squared values, leading to $\frac{430}{10} - 6^2 = 43 - 36 = 7$.

The values of $a$ and $b$ are determined using the standard definitions of arithmetic mean and variance. Summing the observations gives $2+3+3+4+5+7+a+b = 8 \times 4 = 32$, which simplifies to $a+b=8$. Applying the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ results in $2 = \frac{112 + a^2 + b^2}{8} - 16$, simplifying to $a^2 + b^2 = 32$. Since $(a+b)^2 - (a^{2+}b^2) = 2ab = 32$, we have $ab=16$, which implies $a=4$ and $b=4$ by solving the quadratic equation $x^2 - 8x + 16 = 0$.

The full set of observations is $\{2, 3, 3, 4, 4, 4, 5, 7\}$, where the mode is 4. The mean deviation about the mode is calculated as $\frac{1}{n} \sum |x_i - \text{mode}| = \frac{|2-4| + 2|3-4| + 3|4-4| + |5-4| + |7-4|}{8}$, which evaluates to $\frac{2 + 2 + 0 + 1 + 3}{8} = 1$.

The median of a grouped frequency distribution is determined by the formula $M = L + \left( \frac{\frac{n}{2} - Cf}{f} \right) \times h$, where $L$ is the lower boundary, $n$ is the total frequency, $Cf$ is the cumulative frequency of the preceding class, $f$ is the frequency of the median class, and $h$ is the class width. Given a median of 14, a median class interval of 12-18, a class frequency of 12, a cumulative frequency of 18, and a class width of 6, applying these values to the formula results in $14 = 12 + \left( \frac{\frac{n}{2} - 18}{12} \right) \times 6$. Subtracting 12 from both sides gives $2 = \frac{\frac{n}{2} - 18}{2}$, and multiplying by 2 simplifies this to $4 = \frac{n}{2} - 18$. Adding 18 to both sides leads to $\frac{n}{2} = 22$, confirming that the total number of students is 44.