1
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
2
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
3
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
If the mean of the data : 7, 8, 9, 7, 8, 7, $$\lambda$$, 8 is 8, then the variance of this data is :
A
$${7 \over 8}$$
B
1
C
$${9 \over 8}$$
D
2
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number $$\lambda$$ and then each of them is decreased by 25, their mean remains the same. Then $$\lambda$$ is equal to :
A
$${1 \over 3}$$
B
$${2 \over 3}$$
C
$${4 \over 3}$$
D
$${10 \over 3}$$
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