Question

Standard deviation of first 50 natural numbers is
al 45.43 b) 14.43 (c) 20.43 d) 16.43​

Answer 1

Answer:

14.4308

Step-by-step explanation:

S.D. of the first n natural numbers is σ=  { (n^2  −1  ) / 12 }^(1/2)

Here n = 50 ,  

so , S.D. of the first 50 natural numbers =  { (50^2  −1  ) / 12 }^(1/2)

=   { (2500-1  ) / 12 }^(1/2)

=   { (2499  ) / 12 }^(1/2)

= 208.25 ^(1/2)

= 14.4308

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Answer 2

Answer:

The standard deviation of the first 50 natural numbers is

= 14.43

Step-by-step explanation:

The standard deviation of the first n natural numbers is given by

$S.D=\sqrt{\frac{n^{2}-1 }{12} }$

To find the standard deviation of the  first 50 natural numbers

put n = 50

$S.D= \sqrt{\frac{50^{2}-1 }{12} }$

      $= \sqrt{\frac{2500-1}{12} }$

      $=\sqrt{\frac{2499}{12} }$

      $=\sqrt{208.25}$

Standard deviation = 14.4308

Therefore standard deviation of the first 50 natural numbers is 14.43