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JEE Advanced 2023 · Paper 1, Mathematics question 15

Consider the given data with frequency distribution

xix_i 3 8 11 10 5 4
fif_i 5 2 3 2 4 4

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

The correct option is:

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
    2.52.5
  • 2
    55
  • 3
    66
  • 4
    2.72.7
  • 5
    2.42.4
Choose one option
Answer and worked solution

Official answer: A

Here N=∑fi=20N=\sum f_i=20.

Mean. ∑fixi=15+16+33+20+20+16=120\sum f_ix_i=15+16+33+20+20+16=120, so the mean is 66 → (3).

Median. Sorted cumulative frequencies:

xx 3 4 5 8 10 11
cum. ff 5 9 13 15 17 20

The 10th and 11th values are both 5, so the median is 55 → (2).

Mean deviation about the mean 6.
3⋅5+2⋅4+1⋅4+2⋅2+4⋅2+5⋅320=5420=2.7 → (4).\frac{3\cdot5+2\cdot4+1\cdot4+2\cdot2+4\cdot2+5\cdot3}{20}=\frac{54}{20}=2.7\ \to\ \textbf{(4)}.

Mean deviation about the median 5.
2⋅5+1⋅4+0+3⋅2+5⋅2+6⋅320=4820=2.4 → (5).\frac{2\cdot5+1\cdot4+0+3\cdot2+5\cdot2+6\cdot3}{20}=\frac{48}{20}=2.4\ \to\ \textbf{(5)}.

P→3, Q→2, R→4, S→5.

Topics:Measures of central tendencyVariance and standard deviation
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