1
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let xi (1 $$ \le $$ i $$ \le $$ 10) be ten observations of a random variable X. If
$$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$$ and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$$
where 0 $$ \ne $$ p $$ \in $$ R, then the standard deviation of these observations is :
A
$${7 \over {10}}$$
B
$${9 \over {10}}$$
C
$${4 \over 5}$$
D
$$\sqrt {{3 \over 5}} $$
2
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For the frequency distribution :
Variate (x) :      x1   x2   x3 ....  x15
Frequency (f) : f1    f2   f3 ...... f15
where 0 < x1 < x2 < x3 < ... < x15 = 10 and
$$\sum\limits_{i = 1}^{15} {{f_i}} $$ > 0, the standard deviation cannot be :
A
6
B
1
C
4
D
2
3
JEE Main 2020 (Online) 2nd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let X = {x $$ \in $$ N : 1 $$ \le $$ x $$ \le $$ 17} and
Y = {ax + b: x $$ \in $$ X and a, b $$ \in $$ R, a > 0}. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :
A
7
B
9
C
-7
D
-27
4
JEE Main 2020 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let the observations xi (1 $$ \le $$ i $$ \le $$ 10) satisfy the
equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40.
If $$\mu $$ and $$\lambda $$ are the mean and the variance of the
observations, x1 – 3, x2 – 3, ...., x10 – 3, then
the ordered pair ($$\mu $$, $$\lambda $$) is equal to :
A
(6, 6)
B
(3, 3)
C
(3, 6)
D
(6, 3)
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