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QJEEAdv23Ma93ac: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Ma93ac Solution: Consider the given data with frequency distribution Match each…

Consider the given data with frequency distribution

xiamp;3amp;8amp;11amp;10amp;5amp;4fiamp;5amp;2amp;3amp;2amp;4amp;4\begin{array}{ccccccc}x_i & 3 & 8 & 11 & 10 & 5 & 4 \\f_i & 5 & 2 & 3 & 2 & 4 & 4\end{array}

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:

Multiple Choice Options

  • A (P)→(3)  (Q)→(2)  (R)→(4)  (S)→(5)(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(5)
  • B (P)→(3)  (Q)→(2)  (R)→(1)  (S)→(5)(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~ (R) \rightarrow(1) ~~ (S) \rightarrow(5)
  • C (P)→(2)  (Q)→(3)  (R)→(4)  (S)→(1)(P) \rightarrow(2) ~~ (Q) \rightarrow(3) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(1)
  • D (P)→(3)  (Q)→(3)  (R)→(5)  (S)→(5)(P) \rightarrow(3) ~~ (Q) \rightarrow(3) ~~ (R) \rightarrow(5) ~~ (S) \rightarrow(5)

Step-by-Step Verified Solution

x1amp;fiamp;fixiamp;fi∣xi−xˉ∣amp;fi∣xi−N∣3amp;5amp;15amp;15amp;104amp;4amp;16amp;8amp;45amp;4amp;20amp;4amp;08amp;2amp;16amp;4amp;610amp;2amp;20amp;8amp;1011amp;3amp;33amp;15amp;18amp;Σfi=20amp;Σfixi=120amp; sum =54amp; sum =48\begin{array}{|c|c|c|c|c|}\hline x_1 & f_i & f_i x_i & f_i\left|x_i-\bar{x}\right| & f_i\left|x_i-N\right| \\\hline 3 & 5 & 15 & 15 & 10 \\\hline 4 & 4 & 16 & 8 & 4 \\\hline 5 & 4 & 20 & 4 & 0 \\\hline 8 & 2 & 16 & 4 & 6 \\\hline 10 & 2 & 20 & 8 & 10 \\\hline 11 & 3 & 33 & 15 & 18 \\\hline & \Sigma f_i=20 & \Sigma f_i x_i=120 & \text { sum }=54 & \text { sum }=48 \\\hline\end{array}

(P) \quad Mean =ΣfixiΣfi=12020=6=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{120}{20}=6

(Q)

 Median amp;=(10th +11th ) observation 2amp;=5+52=5\begin{aligned}\text { Median } & =\frac{\left(10^{\text {th }}+11^{\text {th }}\right) \text { observation }}{2} \\\\& =\frac{5+5}{2}=5\end{aligned}

(both observation are same)

(R) Mean deviation

amp;=Σfi∣xi−xˉ∣Σfi=5420amp;=2.7\begin{aligned}& =\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{54}{20} \\\\& =2.7\end{aligned}

(S) Mean deviation about median

amp;=Σfi∣xi−M∣Σfi=4820amp;=2.4\begin{aligned}& =\frac{\Sigma f_i\left|x_i-\mathrm{M}\right|}{\Sigma f_i}=\frac{48}{20} \\\\& =2.4\end{aligned}

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