Question

The Standard Deviation of the numbers 31, 32, 33……. 46, 47 is

• A. \(\sqrt{\frac{17}{12}}\)
• B. \(\sqrt{\frac{47^2-1}{12}}\)
• C. \(2\sqrt6\)
• D. \(4\sqrt3\)

Answer 1

The Correct Option is C

The numbers are in an arithmetic sequence: 31, 32, 33, ..., 46, 47. This sequence has: - The first term \( a = 31 \), - The common difference \( d = 1 \), - The last term \( l = 47 \). The number of terms \( n \) is given by: \[ n = \frac{l - a}{d} + 1 = \frac{47 - 31}{1} + 1 = 17 \] For an arithmetic sequence, the standard deviation is given by: \[ \sigma = \sqrt{\frac{n^2 - 1}{12}} \] Substituting \( n = 17 \): \[ \sigma = \sqrt{\frac{17^2 - 1}{12}} = \sqrt{\frac{289 - 1}{12}} = \sqrt{\frac{288}{12}} = \sqrt{24} = 2 \sqrt{6} \] Therefore, the standard deviation is \( 2 \sqrt{6} \). 

Answer 2

The given numbers are 31, 32, 33, ..., 46, 47. This is an arithmetic progression (AP) with first term a = 31, last term l = 47, and number of terms n = 47 - 31 + 1 = 17.

We know that the standard deviation of first n natural numbers is given by \(\sqrt{\frac{n^2-1}{12}}\).

Let's consider the series 1, 2, 3, ..., 17. The standard deviation of this series is \(\sqrt{\frac{17^2-1}{12}}\).

Now, the given series is 31, 32, ..., 47. We can obtain this series by adding 30 to each term of the series 1, 2, ..., 17.

Adding a constant to each term of a series does not change the standard deviation. Therefore, the standard deviation of the given series is the same as the standard deviation of the series 1, 2, ..., 17.

So, the standard deviation of the given series is \(\sqrt{\frac{17^2-1}{12}} = \sqrt{\frac{289-1}{12}} = \sqrt{\frac{288}{12}} = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\).

Therefore, the standard deviation of the numbers 31, 32, 33, ..., 46, 47 is \(2\sqrt{6}\).