1
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
2
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
3
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -
A
6 : 7
B
10 : 3
C
4 : 9
D
5 : 8
4
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
$$\sqrt 7$$
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