1
JEE Main 2025 (Online) 3rd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=\mathrm{b}, a>\mathrm{b}$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n, n=1,2, \ldots, 5$ is
A
17
B
16
C
16.4
D
17.4
2
JEE Main 2025 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If the mean and the variance of $6,4, a, 8, b, 12,10,13$ are 9 and 9.25 respectively, then $a+b+a b$ is equal to :
A
103
B
106
C
100
D
105
3
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $x_1, x_2, ..., x_{10}$ be ten observations such that $\sum\limits_{i=1}^{10} (x_i - 2) = 30$, $\sum\limits_{i=1}^{10} (x_i - \beta)^2 = 98$, $\beta > 2$, and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2(x_{10} - 1) + 4\beta$, then $\frac{\beta\mu}{\sigma^2}$ is equal to :

A

100

B

90

C

120

D

110

4
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For a statistical data $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i=1}^{10} x_i^2=371$. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

A
5
B
7
C
9
D
4
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