Math, asked by chakshusharma40, 4 months ago

calculate the mean and standard deviation of n natural numbers​

Answers

Answered by Anonymous
17

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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