Statistics
PYP_JEE_ADV_2023_P1
Grade None
Question:
Consider the given data with frequency distribution:
$x_i$: 3 8 11 10 5 4
$f_i$: 5 2 3 2 4 4
Match each entry in List-I to the correct entries in List-II.
**List-I**
(P) The mean of the above data is
(Q) The median of the above data is
(R) The mean deviation about the mean of the above data is
(S) The mean deviation about the median of the above data is
**List-II**
(1) 2.5
(2) 5
(3) 6
(4) 2.7
(5) 2.4
(P)→(3) (Q)→(2) (R)→(4) (S)→(5)
(P)→(3) (Q)→(2) (R)→(1) (S)→(5)
(P)→(2) (Q)→(3) (R)→(4) (S)→(1)
(P)→(3) (Q)→(3) (R)→(5) (S)→(5)
Step-by-Step Solution
Key Concept: Sort data to find median; compute mean deviations about mean and median separately
$n=\sum f_i=5+2+3+2+4+4=20$.
(P) Mean $=\dfrac{3(5)+8(2)+11(3)+10(2)+5(4)+4(4)}{20}=\dfrac{15+16+33+20+20+16}{20}=\dfrac{120}{20}=6$→(3).
(Q) Ordered values: 3(×5),4(×4),5(×4),8(×2),10(×2),11(×3). Cumulative: 5,9,13,... 10th value falls in '5' group. Median $=5$→(2).
(R) Mean deviation about mean (=6):
$\dfrac{|3-6|(5)+|8-6|(2)+|11-6|(3)+|10-6|(2)+|5-6|(4)+|4-6|(4)}{20}=\dfrac{15+4+15+8+4+8}{20}=\dfrac{54}{20}=2.7$→(4).
(S) Mean deviation about median (=5):
$\dfrac{|3-5|(5)+|8-5|(2)+|11-5|(3)+|10-5|(2)+|5-5|(4)+|4-5|(4)}{20}=\dfrac{10+6+18+10+0+4}{20}=\dfrac{48}{20}=2.4$→(5).
Answer: (P)→(3),(Q)→(2),(R)→(4),(S)→(5) → A.
Correct Answer: A