JEE Advanced
Mathematics
Statistics
2025
JEE Advanced 2025 (Paper 1)
JEE Advanced Mathematics Question (2025) — Solution
Question
Consider the following frequency distribution : Value 4 5 8 9 6 12 11 Frequency 5 f_1 f_2 2 1 1 3 Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For the given frequency distribution, let denote the mean deviation about the mean, denote the mean deviation about the median, and ^2 denote the variance. Match each entry in List-I to the correct entry in List-II and choose the correct option. \( array |l|l|l|l| & LIST - I & & LIST - II \\ (P) & 7 f_1+9 f_2 is equal to & (1) & 146 \\ (Q) & 19 is equal to & (2) & 47 \\ (R) & 19 is equal to & (3) & 48 \\ (S) & 19 ^2 is equal to & (4) & 145 \\ & & (5) & 55 \\ array \)
Options
- A. ( P ) (5),( Q ) (3),( R ) (2),( S ) (4)
- B. ( P ) (5),( Q ) (2),( R ) (3),( S ) (1)
- C. ( P ) (5),( Q ) (3),( R ) (2),( S ) (1)
- D. ( P ) (3),( Q ) (2),( R ) (5),( S ) (4)
Answer
C. ( P ) (5),( Q ) (3),( R ) (2),( S ) (1)
Step-by-step solution
\( array |c|c|c|c|c|c|c| X_i & f_ i & array l x =7 \\ d_i = |x_i- x | array & array l M =6 \\ ei = | xi - M | array & f_ i d _ i & f_ i e _ i & f_ i d _ i ^2 \\ 4 & 5 & 3 & 2 & 15 & 10 & 45 \\ 5 & 4 & 2 & 1 & 8 & 4 & 16 \\ 6 & 1 & 1 & 0 & 1 & 0 & 1 \\ 8 & 3 & 1 & 2 & 3 & 6 & 3 \\ 9 & 2 & 2 & 3 & 4 & 6 & 8 \\ 11 & 3 & 4 & 5 & 12 & 15 & 48 \\ 12 & 1 & 5 & 6 & 5 & 6 & 25 \\ & & & & 48 & 47 & 146 \\ array \) aligned & f_1=4 \\ & f_2=3 \\ & = 48 19 , = 47 19 , ^2= 146 19 aligned
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