calculate the mean and standard deviation of n natural numbers
Answers
Answered by
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Step-by-step explanation:
The first n natural numbers are 1,2,3,.....,n.
Their mean,
x
ˉ
=
n
∑x
=
n
1+2+3+.....+n
=
2n
n(n+1)
=
2
n+1
Sum of the square of the first n natural numbers is ∑x
2
=
6
n(n+1)(2n+1)
Thus, the standard deviation σ=
n
∑x
2
−(
n
∑x
)
2
=
6n
n(n+1)(2n+1)
−(
2
n+1
)
2
=
6
(n+1)(2n+1)
−(
2
n+1
)
2
=
(
2
n+1
)[
3
(2n+1
−
2
(n+1)
]
=
(
2
n+1
)[
6
2(2n+1)−3(n+1)
]
=
(
2
n+1
)(
6
4n+2−3n−3
)
=
(
2
n+1
)(
6
n−1
)
=
12
n
2
−1
.
Hence, the S.D. of the first n natural numbers is σ=
12
n
2
−1
.
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