1
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :
A
1
B
7
C
3
D
5
2
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :
A
31
B
50
C
51
D
30
3
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
The outcome of each of 30 items was observed; 10 items gave an outcome $${1 \over 2}$$ – d each, 10 items gave outcome $${1 \over 2}$$ each and the remaining 10 items gave outcome $${1 \over 2}$$+ d each. If the variance of this outcome data is $${4 \over 3}$$ then |d| equals :
A
$${2 \over 3}$$
B
$${{\sqrt 5 } \over 2}$$
C
$${\sqrt 2 }$$
D
2
4
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to
A
582.5
B
507.5
C
586.5
D
509.5
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