JEE Advance - Mathematics (2023 - Paper 1 Online - No. 15)
Consider the given data with frequency distribution
$$ \begin{array}{ccccccc} x_i & 3 & 8 & 11 & 10 & 5 & 4 \\ f_i & 5 & 2 & 3 & 2 & 4 & 4 \end{array} $$
Match each entry in List-I to the correct entries in List-II.
The correct option is:
$$ \begin{array}{ccccccc} x_i & 3 & 8 & 11 & 10 & 5 & 4 \\ f_i & 5 & 2 & 3 & 2 & 4 & 4 \end{array} $$
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:
$(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(5)$
$(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~ (R) \rightarrow(1) ~~ (S) \rightarrow(5)$
$(P) \rightarrow(2) ~~ (Q) \rightarrow(3) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(1) $
$(P) \rightarrow(3) ~~ (Q) \rightarrow(3) ~~ (R) \rightarrow(5) ~~ (S) \rightarrow(5)$
Explanation
$$
\begin{array}{|c|c|c|c|c|}
\hline x_1 & f_i & f_i x_i & f_i\left|x_i-\bar{x}\right| & f_i\left|x_i-N\right| \\
\hline 3 & 5 & 15 & 15 & 10 \\
\hline 4 & 4 & 16 & 8 & 4 \\
\hline 5 & 4 & 20 & 4 & 0 \\
\hline 8 & 2 & 16 & 4 & 6 \\
\hline 10 & 2 & 20 & 8 & 10 \\
\hline 11 & 3 & 33 & 15 & 18 \\
\hline & \Sigma f_i=20 & \Sigma f_i x_i=120 & \text { sum }=54 & \text { sum }=48 \\
\hline
\end{array}
$$
(P) $\quad$ Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{120}{20}=6$
(Q)
$$ \begin{aligned} \text { Median } & =\frac{\left(10^{\text {th }}+11^{\text {th }}\right) \text { observation }}{2} \\\\ & =\frac{5+5}{2}=5 \end{aligned} $$
(both observation are same)
(R) Mean deviation
$$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{54}{20} \\\\ & =2.7 \end{aligned} $$
(S) Mean deviation about median
$$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-\mathrm{M}\right|}{\Sigma f_i}=\frac{48}{20} \\\\ & =2.4 \end{aligned} $$
(P) $\quad$ Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{120}{20}=6$
(Q)
$$ \begin{aligned} \text { Median } & =\frac{\left(10^{\text {th }}+11^{\text {th }}\right) \text { observation }}{2} \\\\ & =\frac{5+5}{2}=5 \end{aligned} $$
(both observation are same)
(R) Mean deviation
$$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{54}{20} \\\\ & =2.7 \end{aligned} $$
(S) Mean deviation about median
$$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-\mathrm{M}\right|}{\Sigma f_i}=\frac{48}{20} \\\\ & =2.4 \end{aligned} $$
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