Q32 / 51 · JEE Advanced 2023 (Paper 1) · Mathematics

Consider the given data with frequency distribution

xi38111054fi523244

Match each entry in List-I to the correct entries in List-II.

  List-I   List-II
P The mean of the above data is 1 2.5
Q The median of the above data is 2 5
R The mean deviation about the
mean of the above data is
3 6
S The mean deviation about the
median of the above data is
4 2.7
    5 2.4

The correct option is

  • Correct answer
    A
    P→3 Q→2 R→4 S→5
  • B
    P→3 Q→2 R→1 S→5
  • C
    P→2 Q→3 R→4 S→1
  • D
    P→3 Q→3 R→5 S→5

Solution

Arranging the given data in ascending order

we get,

xi34581011fi544223

Now finding the mean of the above data we get,

Mean =3×5+8×2+11×3+10×2+5×4+4×45+2+3+2+4+4

=15+16+33+20+20+1620=12020=6

Now, median =1210th+11th observation

=125+5=5

Hence, Mean deviation about mean will be,

=3×5+2×4+1×4+2×2+4×2+5×320

=5420=2.7

And Mean deviation about median

=2×5+1×4+0+3×2+5×2+6×320

=4.820=2.4

P→3; Q→2; R→4; S→5

∴ Option A is correct.

Hence, the correct answer is option: 'A'.

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