MCQ
The standard deviation of first 10 natural numbers is:
- A5.5
- B3.87
- C2.97
- D2.97
Solution:
We know that the standard deviation of first n natural number is $\sqrt{\frac{\text{n}^2-1}{12}}.$
$\therefore$ Standard deviation of first 10 natural numbers
$=\sqrt{\frac{10^2-1}{12}}$
$=\sqrt{\frac{99}{12}}$
$=\sqrt{8.25}$
$=2.87$
Hence, the correct answer is option (d).
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