souvonik2k wrote:
Set A consists of consecutive integers. What is the median of all the numbers in set A?
(1) The smallest number in set A is 4.
(2) The standard deviation of all the numbers in set A is \(\sqrt{2}\)
Hi
kunalsinghNSto find the median we need to know the number of elements in the set and as the set has consecutive elements, we need any one number from the set.
Statement 1: provides one number from the set but we don't know the number of elements in the set.
InsufficientStatement 2: let the number of elements in the set we \(n\) and \(a\) be the first number so our set will be {\(a, a+1, a+2.....a+(n-1)\)}
Use simple mathematical process to calculate st. deviation of this set.
So average of the set will be \(= \frac{{a+a+1......a+(n-1)}}{n}\) \(=\frac{{an+1+2...n-1}}{n}\) \(= [an+n(n-1)/2]/n\) ---------[{1+2...{n-1} this is a simple AP series with 1st term 1, last term n-1, common difference 1 and number of terms n-1. so you can easily calculate the sum in terms of n]
so average of the set \(= a+\frac{(n-1)}{2}\)
to calculate standard deviation we need to reduce each element in the set by the average, then square it, then take the average of the resultant no and finally take the square root
Step 1: \(a-a-\frac{(n-1)}{2}\), \(a+1-a-\frac{(n-1)}{2}\),.................., \(a+(n-1)-a-\frac{(n-1)}{2}\) \(= \frac{-(n-1)}{2}\), \(1-\frac{(n-1)}{2}\)........., \((n-1)-\frac{(n-1)}{2}\)
Step 2: now square each of the elements and add. On squaring & adding each of the elements, you will get something in terms of \(n^2\). let's call it \(kn^2\), where \(k\) is any constant resulting from the summation of the AP series
Step 3: take the average of Step 2 \(= \frac{kn^2}{n} = kn\)
Step 4: standard deviation \(= \sqrt{kn}\)
so we have \(\sqrt{kn}=\sqrt{2}\)
hence \(n=\frac{2}{k}\). So we get the value of \(n\) but we don't know any of the elements in the set.
InsufficientCombining 1 & 2: we get all the required parameters. Statement 1 provides the 1st element in the set and Statement 2 provides the number of elements in the set.
Hence
CkunalsinghNS, instead of going for the mathematical derivation, you can test by using simple nos, for e.g 1,2,3 etc. to know that if you are given st. deviation/variance, then you can calculate the number of terms in the set because each element is consecutive.
I have skipped some calculation because working with variables was becoming cumbersome
. but the point is statement 2 will give you the number of elements in the set as the set is consecutive. Hope this helps