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CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Descriptive Statistics
Quiz 12
If the variance of the data 2,4,5,6,8,17 is 23,33, then variance of 5,11,14,17,23,50 is ________________.
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0%
46,66
0%
96.32
0%
209,97
0%
213,97
Ina a class of 20 students, each student can score 10 or 0 marks in a certain examination. The maximum possible variance in the marks of the students in the class is
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0%
24
0%
22
0%
20
0%
25
In a series of
100
100
observations, half of them is equal to
p
p
and remaining half equal to
−
p
−
p
if the standard deviation of these observations is
10
10
then
|
p
|
=
|
p
|
=
__
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0%
10
10
0%
1
10
1
10
0%
1
√
10
1
10
0%
√
10
10
10
∑
i
=
1
(
X
i
−
10
)
2
=
2000
10
∑
i
=
1
X
i
−
10
=
100
∑
i
=
1
10
(
X
i
−
10
)
2
=
2000
∑
i
=
1
10
X
i
−
10
=
100
then the standard deviation of data set
X
i
X
i
.
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0%
1
1
0%
5
5
0%
10
10
0%
2
2
If the standard deviation of
x
1
,
x
2
,
.
.
.
,
x
n
x
1
,
x
2
,
.
.
.
,
x
n
is 3.5 then the standard deviation of
2
x
1
−
3
,
2
x
2
−
3
,
.
.
.
,
2
x
n
−
3
2
x
1
−
3
,
2
x
2
−
3
,
.
.
.
,
2
x
n
−
3
is
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0%
-7
0%
10.5
0%
7
0%
-10
Find the standard deviation of 10 observation 111, 211, 311,.......,1011.
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0%
100
√
3
3
0%
250
0%
300
0%
50
√
33
33
The variance of the first
n
n
natural number is
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0%
1
12
(
n
2
−
1
)
1
12
(
n
2
−
1
)
0%
1
6
(
n
2
−
1
)
1
6
(
n
2
−
1
)
0%
1
6
(
n
2
+
1
)
1
6
(
n
2
+
1
)
0%
1
12
(
n
2
+
1
)
1
12
(
n
2
+
1
)
Explanation
Variance
(
σ
2
)
=
1
n
n
∑
i
=
1
(
x
i
−
¯
x
)
2
(
σ
2
)
=
1
n
∑
i
=
1
n
(
x
i
−
x
¯
)
2
=
1
n
n
∑
i
=
1
[
x
i
−
(
n
+
1
2
)
]
2
=
1
n
∑
i
=
1
n
[
x
i
−
(
n
+
1
2
)
]
2
=
1
n
n
∑
i
=
1
x
2
i
−
1
n
n
∑
i
=
1
2
(
n
+
1
2
)
x
i
+
1
n
n
∑
i
=
1
(
n
+
1
2
)
2
=
1
n
∑
i
=
1
n
x
i
2
−
1
n
∑
i
=
1
n
2
(
n
+
1
2
)
x
i
+
1
n
∑
i
=
1
n
(
n
+
1
2
)
2
=
1
n
n
(
n
+
1
)
(
2
n
+
1
)
6
−
(
n
+
1
n
)
[
n
(
n
+
1
)
2
]
+
(
n
+
1
)
2
4
n
×
n
=
1
n
n
(
n
+
1
)
(
2
n
+
1
)
6
−
(
n
+
1
n
)
[
n
(
n
+
1
)
2
]
+
(
n
+
1
)
2
4
n
×
n
=
(
n
+
1
)
(
2
n
+
1
)
6
−
(
n
+
1
)
2
2
+
(
n
+
1
)
2
4
=
(
n
+
1
)
(
2
n
+
1
)
6
−
(
n
+
1
)
2
2
+
(
n
+
1
)
2
4
=
(
n
+
1
)
(
2
n
+
1
)
6
−
(
n
+
1
)
2
4
=
(
n
+
1
)
(
2
n
+
1
)
6
−
(
n
+
1
)
2
4
=
(
n
+
1
)
[
4
n
+
2
−
3
n
−
3
12
]
=
(
n
+
1
)
[
4
n
+
2
−
3
n
−
3
12
]
=
(
n
+
1
)
(
n
−
1
)
12
=
(
n
+
1
)
(
n
−
1
)
12
=
n
2
−
1
12
=
n
2
−
1
12
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is
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0%
√
52
7
52
7
0%
√
6
6
0%
52
7
52
7
0%
6
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