1
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
A data consists of n observations : x1, x2, . . . . . . ., xn.

If     $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$$    and

$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$$

then the standard deviation of this data is :
A
2
B
$$\sqrt 5$$
C
5
D
$$\sqrt 7$$
2
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :
A
16
B
22
C
20
D
18
3
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :
A
0
B
1
C
2
D
4
4
JEE Main 2018 (Offline)
+4
-1
If $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9$$ and

$$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$$, then the standard deviation of the 9 items
$${x_1},{x_2},.......,{x_9}$$ is
A
3
B
9
C
4
D
2
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