JEE Advanced 2025 Paper 1Question 1
Consider the following frequency distribution:\\ $\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Value} & 4 & 5 & 8 & 9 & 6 & 12 & 11\\ \hline \text{Frequency} & 5 & f_1 & f_2 & 2 & 1 & 1 & 3\\ \hline\end{array}$\\ Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.\\ For the given frequency distribution, let $\alpha$ denote the mean deviation about the mean, $\beta$ denote the mean deviation about the median, and $\sigma^2$ denote the variance.\\ Match each entry in List-I to the correct entry in List-II and choose the correct option.\\ $\begin{array}{ll}\textbf{List-I} & \textbf{List-II}\\ \text{(P) } 7f_1+9f_2 \text{ is equal to} & \text{(1) } 146\\ \text{(Q) } 19\alpha \text{ is equal to} & \text{(2) } 47\\ \text{(R) } 19\beta \text{ is equal to} & \text{(3) } 48\\ \text{(S) } 19\sigma^2 \text{ is equal to} & \text{(4) } 145\\ & \text{(5) } 55\end{array}$
Statistics
Solution
$f_1+f_2=7$; median $6$ as the 10th value forces $f_1=4$, $f_2=3$, so $7f_1+9f_2=55$. Mean $=\frac{133}{19}=7$, $19\alpha=48$, $19\beta=47$ (about $6$), $19\sigma^2=1077-19\cdot49=146$. So P-5, Q-3, R-2, S-1.