1
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
Let X = {x $$\in$$ N : 1 $$\le$$ x $$\le$$ 17} and
Y = {ax + b: x $$\in$$ X and a, b $$\in$$ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :
A
7
B
9
C
-7
D
-27
2
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
Let the observations xi (1 $$\le$$ i $$\le$$ 10) satisfy the
equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)}$$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}}$$ = 40.
If $$\mu$$ and $$\lambda$$ are the mean and the variance of the
observations, x1 – 3, x2 – 3, ...., x10 – 3, then
the ordered pair ($$\mu$$, $$\lambda$$) is equal to :
A
(6, 6)
B
(3, 3)
C
(3, 6)
D
(6, 3)
3
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is
A
3.98
B
3.99
C
4.01
D
4.02
4
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 resepectively. Each of these 10 observations is multiplied by p and then reduced by q, where p $$\ne$$ 0 and q $$\ne$$ 0. If the new mean and new s.d. become half of their original values, then q is equal to
A
10
B
-20
C
-10
D
-5
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