Question

he variance of first 49 natural numbers is-​

Answer 1

Given:

no of terms n = 49

To find:

variance σ =?

Step-by-step explanation:

• The term variance is outlined as a statistical mensuration of the spread between the numbers in an information set.
• Additionally, variance measures, however, way every variety in the set is from the mean and therefore from each alternative variety within the set.
• Variance is also denoted by this image :  $\sigma^2$.
• Mathematically, it is expressed as :

           $\displaystyle \sigma=\sqrt{\frac{1}{n}\sum(x- \bar_x )}^2 }$

where :

        σ = variance

        n= number of terms

        x= total  sum of terms

        $\bar_x$ = average mean of terms

• Here n = 49

• The summation of the first 49 natural numbers is 1225.

i.e. x = 1225

• The average mean of the first 49 natural numbers is

     $\displaystyle \bar_x}= \frac{1225}{49} =25$

• Put given values in the above equation

    $\displaystyle \sigma=\sqrt{\frac{1}{49}(1225-25 )^2 }$

    σ = 171.42

Hence the variance of the first 49 natural numbers is 171.42.

Answer 2

Given :  first 49 natural numbers

To Find : Variance of   first 49 natural numbers

Solution:

Variance of   first n natural numbers   =  (n + 1)(n - 1) /12

n = 49

Variance  = (49 + 1) (49 - 1) /12

= 50 * 48 / 12

= 50 * 4

= 200

Variance = 200

Variance of   first n natural numbers   =  (n + 1)(n - 1) /12 can be found using

Variance = ( ∑n² ) / n  -  ((∑n) / n)²  

= n(n + 1)(2n +1)/6n   -   (n (n + 1) /2n)²

 (n + 1)(n - 1) /12

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