Hi AIMGMAT770,
Let's consider a series of n consecutive numbers: \(a, a+1, a+2......a+n-1\)
To find the standard deviation, we first calculate the mean, which in this case is same as the average of first and last number.
Mean = \((a+a+n-1)/2 = a + (n-1)/2\)
Next we need to find the squared distance of each number of the series from the mean:
Case 1: n is odd
For the 1st number, a -> \([a - (a + (n-1)/2)]^2 = [(n-1)/2]^2\)
For the 2nd number, a +1 -> \([a + 1 - (a + (n-1)/2)]^2 = [1-(n-1)/2]^2\)
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For the number preceding the middle number, a + (n-1)/2 -1 -> \([a + (n-1)/2 -1- (a + (n-1)/2)]^2 = 1\)
For the middle number, a + (n-1)/2 (Mean is also the (n-1)/2th term of the series) -> \([a + (n-1)/2 - (a + (n-1)/2)]^2 = 0\)
For the number succeeding the middle number, a + (n-1)/2 + 1 ->\([a + (n-1)/2 +1- (a + (n-1)/2)]^2 = 1\)
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For the second lat number, a +n -2 -> \([a + n -2 - (a + (n-1)/2)]^2 = [(n-1)/2-1]^2\)
For the last number, a +n -1 -> \([a + n -1 - (a + (n-1)/2)]^2 = [(n-1)/2]^2\)
Sum of the squared distance = \(2[1^2 + 2^2 + 3^2......[(n-1)/2]^2]\)
Variance = \(2[1^2 + 2^2 + 3^2......[(n-1)/2]^2]/n\)
Standard deviation = \(\sqrt{[2[1^2 + 2^2 + 3^2......[(n-1)/2]^2]/n]}\)
As sum of the squares of consecutive n numbers is n(n+1)(2n+1)/6
Hence, the Standard deviation simplifies to \(\sqrt{(n^2 -1)/12}\)
Case 2: n is even
For the 1st number, a -> \([a - (a + (n-1)/2)]^2 = [(n-1)/2]^2\)
For the 2nd number, a +1 -> \([a + 1 - (a + (n-1)/2)]^2 = [1-(n-1)/2]^2\)
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For the number preceding the mean, a + (n-1)/2 - 0.5 -> \([a + (n-1)/2 -0.5 - (a + (n-1)/2)]^2 = 0.5^2\)
{a + (n-1)/2 (Mean is the average of the two middle terms and hence at a distance of 0.5 from them)}
For the number succeeding the mean, a + (n-1)/2 + 0.5 ->\([a + (n-1)/2 + 0.5 - (a + (n-1)/2)]^2 = 0.5^2\)
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For the second last number, a +n -2 -> \([a + n -2 - (a + (n-1)/2)]^2 = [(n-1)/2-1]^2\)
For the last number, a +n -1 -> \([a + n -1 - (a + (n-1)/2)]^2 = [(n-1)/2]^2\)
Sum of the squared distance = \(2[0.5^2 + 1.5^2 + 2.5^2......[(n-1)/2]^2]\)
Variance = \(2[0.5^2 + 1.5^2 + 2.5^2......[(n-1)/2]^2]/n\)
Standard deviation = \(\sqrt{2[0.5^2 + 1.5^2 + 2.5^2......[(n-1)/2]^2]/n}\)
As you can see, in both the cases you only need to know the value of n to find the Standard deviation.
It is very unlikely that you will be asked to find the standard deviation but you can definitely get a DS question where the question stem asks you to find the standard deviation of a series of n consecutive numbers and one of the statements gives you the value of n.
All you need to understand here that if you know the number of terms, you can find the standard deviation.