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Consider $5$ observations $0 \leq a \leq b \leq c \leq d \leq e \leq 100$. If $\sigma$ is the standard deviation then it is always less than or equal to

(a) $20$ $\qquad$ (b) $30$ $\qquad$ (c) $40$ $\qquad$ (d) $50$

Now I know that the formula for standard deviation(as taught to us, I do not know if there are any other formulae) is

= √(∑(ₖ)²-²)

where is the mean of given data

On simplifying using given data,

25²=(∑(-)²)

I do not know how to proceed after this. My intuition says that if the last two numbers be 100 and the others be 0 standard deviation should be maximum(as it's 'spaced out' more, if that makes any sense). Then

² = 2400

but the answer is given as (d)50

Is the answer given wrong? Is there any inequality that would help in solving this question that I am not aware of? I tried using root mean square ≥ mean but that gives a lower bound not an upper bound.

Any help/hint would be appreciated

Rodrigo de Azevedo
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Drayon Coolagon
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    I MathJax'ed a bit. Please do the rest – Rodrigo de Azevedo May 15 '23 at 06:51
  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User May 15 '23 at 06:58
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    Look at the post Maximum of the Variance Function for Given Set of Bounded Numbers, particularly the answer of Dilip. Your answer is perfectly correct, the option of $50$ is correct for this question but not optimal (which is exactly your point of confusion) because you have $5$ observations, which is an odd number. Using $N=5,c=100$ in Dilip's working gives $2400$ as the maximum variance, so you actually identified the best value. – Sarvesh Ravichandran Iyer May 15 '23 at 07:00
  • @SarveshRavichandranIyer Thank you!! – Drayon Coolagon May 15 '23 at 07:09
  • All your own good work, so it's nice to see it ratified. As a principle, remember that in multiple choice exams (I've set some of them, so I know this!) people purposely set non-optimal values to confuse people, as is the case here. When the variance is less than $\sqrt{2400}$, it is less than $50$ as well (and that's the only option!) but as it turns out students will be scratching their heads if they expect to obtain exactly $50$ as an upper bound somehow. – Sarvesh Ravichandran Iyer May 15 '23 at 07:22
  • Note that is you used $\frac{1}{n-1}$ in the variance calculation then the maximum (sample) standard deviation would be $\sqrt{3000} \approx 54.8$ – Henry May 15 '23 at 07:51

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