1
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)
2
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
3
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
4
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
If the data x1, x2,......., x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :
A
$$\sqrt 2$$
B
2
C
2$$\sqrt 2$$
D
4
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