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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Mathematics
Chapter
Statistics

JEE Advanced 2025 · Paper 1 · Mathematics · Statistics

Consider the following frequency distribution : Value 4 5 8 9 6 12...

Consider the following frequency distribution :
Value 4 5 8 9 6 12 11
Frequency 5 f1f_1 f2f_2 2 1 1 3

Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For the given frequency distribution, let α\alpha denote the mean deviation about the mean, β\beta denote the mean deviation about the median, and σ2\sigma^2 denote the variance. Match each entry in List-I to the correct entry in List-II and choose the correct option. (
List-IList-II
(P)7 f_1+9 f_2 is equal to(1)146
(Q)19 \alpha is equal to(2)47
(R)19 \beta is equal to(3)48
(S)19 \sigma^2 is equal to(4)145
(5)55
)

Options

  1. A(P)→(5),(Q)→(3),(R)→(2),(S)→(4)(\mathrm{P}) \rightarrow(5),(\mathrm{Q}) \rightarrow(3),(\mathrm{R}) \rightarrow(2),(\mathrm{S}) \rightarrow(4)
  2. B(P)→(5),(Q)→(2),(R)→(3),(S)→(1)(\mathrm{P}) \rightarrow(5),(\mathrm{Q}) \rightarrow(2),(\mathrm{R}) \rightarrow(3),(\mathrm{S}) \rightarrow(1)
  3. C(P)→(5),(Q)→(3),(R)→(2),(S)→(1)(\mathrm{P}) \rightarrow(5),(\mathrm{Q}) \rightarrow(3),(\mathrm{R}) \rightarrow(2),(\mathrm{S}) \rightarrow(1)
  4. D(P)→(3),(Q)→(2),(R)→(5),(S)→(4)(\mathrm{P}) \rightarrow(3),(\mathrm{Q}) \rightarrow(2),(\mathrm{R}) \rightarrow(5),(\mathrm{S}) \rightarrow(4)

Correct answer

C. (P)→(5),(Q)→(3),(R)→(2),(S)→(1)(\mathrm{P}) \rightarrow(5),(\mathrm{Q}) \rightarrow(3),(\mathrm{R}) \rightarrow(2),(\mathrm{S}) \rightarrow(1)

Step-by-step solution

(\begin{array}{|c|c|c|c|c|c|c|} \hline X_i & f_{\mathrm{i}}& \begin{array}{l} \bar{x}=7 \\ d_i = \left|x_i-\bar{x}\right| \end{array} & M=6ei=∣xi−M∣\begin{array}{l} \mathrm{M}=6 \\ \mathrm{ei} =|\mathrm{xi}-\mathrm{M}| \end{array} & ∑fid\sum f_{\mathrm{i}} \mathrm{d}_i{\mathrm{i}} & \sum f_{\mathrm{i}}\mathrm{e}_{\mathrm{i}} & \Sigma f_{\mathrm{i}}\mathrm{d}_{\mathrm{i}^2\\ \hline 4 & 5 & 3 & 2 & 15 & 10 & 45 \\ \hline 5 & 4 & 2 & 1 & 8 & 4 & 16 \\ \hline 6 & 1 & 1 & 0 & 1 & 0 & 1 \\ \hline 8 & 3 & 1 & 2 & 3 & 6 & 3 \\ \hline 9 & 2 & 2 & 3 & 4 & 6 & 8 \\ \hline 11 & 3 & 4 & 5 & 12 & 15 & 48 \\ \hline 12 & 1 & 5 & 6 & 5 & 6 & 25 \\ \hline & & & & 48 & 47 & 146 \\ \hline \end{array}f1=4f2=3α=4819,β=4719,σ2=14619\begin{aligned} & f_1=4 \\ & f_2=3 \\ & \alpha=\frac{48}{19}, \beta=\frac{47}{19},\sigma^2=\frac{146}{19}\end{aligned}$
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