Statistics – JEE Main MathematicsPractice Questions & PYQs

The average marks of the class function as a central point balancing the scores of the two groups. Since the boys' average of 52 is 2 units above the overall average of 50 and the girls' average of 42 is 8 units below it, the number of boys and girls must be in a ratio that compensates for these differences, which is 8 to 2, or 4 to 1. Since the total class consists of 4 parts boys and 1 part girls, the portion of boys is $\frac{4}{4+1} \times 100 = 80$. This results in the boys making up 80% of the class.

Variance measures the dispersion of a dataset, which is determined by the spread of values around their mean, not by the absolute positions of those values on the number line. A fundamental property of variance is that it is invariant to a change of origin, meaning that adding or subtracting a constant value from every observation in a set does not alter the variance. In this case, each value in population B is exactly 50 greater than the corresponding value in population A, representing a simple shift of the origin for the entire dataset. Since this shift does not change the distance between data points, the variance remains the same for both populations, such that $V_A = V_B$ and the ratio $\frac{V_A}{V_B} = 1$.

For moderately skewed frequency distributions, the three measures of central tendency are connected by the empirical relationship $Mode = 3 \times \text{Median} - 2 \times \text{Mean}$. When the mean is 21 and the median is 22, the mode is determined as follows:

$Mode = 3(22) - 2(21)$
$Mode = 66 - 42$
$Mode = 24$

The variance of a dataset is non-negative, as it represents the average squared deviation from the mean, which implies $\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2 \geq 0$. Substituting the values $\sum x_i^2 = 400$ and $\sum x_i = 80$ into this expression yields $\frac{400}{n} - (\frac{80}{n})^2 \geq 0$, which simplifies to $\frac{400}{n} - \frac{6400}{n^2} \geq 0$. Multiplying by $n^2$ leads to $400n \geq 6400$, meaning the number of observations $n$ must be at least $16$. Among the provided values, 18 is the only option that meets this requirement.

Identifying the mode through a histogram is a standard statistical practice, as the highest frequency peak visually indicates where the modal concentration lies. Because the median relies on the specific magnitude of observations, applying a change in scale to the data set directly alters the median value, confirming it is not independent of scale changes. Conversely, variance measures the spread of data points; while shifting the data by a constant value does not affect this dispersion, scaling the data by a factor $a$ changes the variance to $a^2$ times the original value, meaning it is not independent of scale.

Standard deviation measures the spread of data around the mean, which is zero in this case because the observations $a$ and $-a$ are distributed equally. The variance is therefore the mean of the squared observations, calculated as the sum of $n$ instances of $a^2$ and $n$ instances of $(-a)^2$ divided by the total count $2n$. This yields the expression $\frac{n a^2 + n a^2}{2n} = a^2$ Since the standard deviation is the square root of the variance, we have $\sqrt{a^2} = |a|$. With the standard deviation provided as 2, it follows that $|a| = 2$.

Variance is determined by the relationship between the mean of the squared values and the square of the mean, expressed as:

$\sigma^2 = \frac{\sum x^2}{N} - \left(\frac{\sum x}{N}\right)^2$

Because the initial dataset of 15 observations contained an incorrect value of 20 that should have been 30, both the sum of the variables and the sum of their squares must be updated. Calculate the corrected sum as $\sum x = 170 - 20 + 30 = 180$. Similarly, update the sum of squares by removing the squared value of the error and adding the square of the correct value: $\sum x^2 = 2830 - 20^2 + 30^2 = 3330$.

Find the corrected mean squared value by dividing the new sum of squares by the number of observations, resulting in $\frac{3330}{15} = 222$. Calculate the square of the mean as $\left(\frac{180}{15}\right)^2 = 12^2 = 144$. Subtracting the squared mean from the mean squared value, expressed as $222 - 144 = 78$, yields the final corrected variance of 78.

The median of an odd number of observations is the middle value, which in a set of 9 distinct numbers is the 5th observation when arranged in ascending order. Since the process only increases the four largest observations, the first 5 terms remain entirely undisturbed, meaning the 5th term-represented as $x_5$-remains unchanged. Consequently, the median of the new set stays the same as the original median of 20.5.

The fundamental principle here is that the total sum of all student marks equals the sum of marks obtained by the boys combined with the sum of marks obtained by the girls. With 100 students averaging 72, the collective marks for the entire class amount to $100 \times 72 = 7200$. Since 70 boys have an average of 75, their total contribution is $70 \times 75 = 5250$. Subtracting this sum from the class total leaves the girls' combined marks, which are $7200 - 5250 = 1950$. Because there are $100 - 70 = 30$ girls in the class, dividing their total marks by the number of girls yields an average of $\frac{1950}{30} = 65$.

The variance of a combined dataset composed of two subsets is calculated by aggregating the variances of each subset, adjusted for the squared deviations of their respective means from the global mean. Since both sets $X$ and $Y$ consist of 31 terms in an arithmetic progression with a common difference of 1, the variance of each set is determined by the formula for the variance of the first $n$ consecutive integers, which is $\frac{n^{2-}1}{12}$. With $n=31$, the individual variance for both sets is $\frac{31^{2-}1}{12} = 80$. The means of these sets are $\overline{x} = \frac{11 + 41}{2} = 26$ and $\overline{y} = \frac{61 + 91}{2} = 76$, which results in a combined mean of $\mu = \frac{26 + 76}{2} = 51$.

The overall variance $\sigma^2$ is found by calculating the average of the individual variances plus the average of the squared deviations of the subset means from the combined mean. Because the two sets have an equal number of elements, this simplifies to the formula $\sigma^2 = \frac{\sigma_x^2 + (\overline{x}-\mu)^2 + \sigma_y^2 + (\overline{y}-\mu)^2}{2}$. Substituting the values, we obtain $\sigma^2 = \frac{80 + (26-51)^2 + 80 + (76-51)^2}{2}$, which simplifies to $\frac{80 + 625 + 80 + 625}{2} = 705$. Completing the final arithmetic, the expression becomes $|26 + 76 - 705| = |102 - 705| = 603$.