1
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A variable $X$ takes values $0,0,2,6,12,20, \ldots, n(n-1)$ with frequencies ${ }^n C_0,{ }^n C_1,{ }^n C_2,{ }^n C_3,{ }^n C_4,{ }^n C_5, \ldots,{ }^n C_n$, respectively. If the mean of this data is 60 , then its median is :

A

56

B

42

C

72

D

90

2
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The mean deviation about the mean for the data

$$ \begin{array}{|c|c|c|c|c|c|c|} \hline x_i & 5 & 7 & 9 & 10 & 12 & 15 \\ \hline f_i & 8 & 6 & 2 & 2 & 2 & 6 \\ \hline \end{array} $$

$$ \text { is equal to: } $$

A

$$ \frac{40}{13} $$

B

$$ \frac{42}{13} $$

C

$$ \frac{44}{13} $$

D

$$ \frac{46}{13} $$

3
JEE Main 2026 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For 10 observations $x_1, x_2, \ldots, x_{10}$, if $\sum\limits_{i=1}^{10}\left(x_i+2\right)^2=180$ and $\sum\limits_{i=1}^{10}\left(x_i-1\right)^2=90$, then their standard deviation is :

A

2

B

$$ \sqrt{3} $$

C

$$ 2\sqrt{2} $$

D

3

4
JEE Main 2026 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Suppose that the mean and median of the non-negative numbers $21,8,17, a, 51,103, b, 13,67,(a>b)$, are 40 and 21 , respectively. If the mean deviation about the median is 26 , then $2 a$ is equal to :

A

109

B

117

C

161

D

131

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