1
JEE Main 2020 (Online) 6th September Morning Slot
+4
-1
If $$\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$$ and $$\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$$
(n, a > 1) then the standard deviation of n
observations x1 , x2 , ..., xn is :
A
$$a$$ – 1
B
$$n\sqrt {a - 1}$$
C
$$\sqrt {n\left( {a - 1} \right)}$$
D
$$\sqrt {a - 1}$$
2
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
If the mean and the standard deviation of the
data 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation :
A
x2 – 20x + 18 = 0
B
2x2 – 20x + 19 = 0
C
x2 – 10x + 18 = 0
D
x2 – 10x + 19 = 0
3
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :
A
2
B
3
C
1
D
4
4
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :
A
5
B
3
C
7
D
9
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