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souvonik2k
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 kunalsinghNS

to 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. Insufficient

Statement 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. Insufficient

Combining 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 C

kunalsinghNS, 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
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This seems to be a pretty long calculation !!
:sad:
approximation is the best option i guess !
but thank you !!
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kunalsinghNS
This seems to be a pretty long calculation !!
:sad:
approximation is the best option i guess !
but thank you !!
Yup. But you only need to visualise the process and need to note that if st. Deviation of consecutive nos is given we can calculate the number of terms in the set. Mathematical derivation is just for academics purposes

Posted from my mobile device
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we also need to divide the summation of the squares by the total no of elements. Can someone point out where are we dividing the summation before taking the square root?

Thanks in advance
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Debashis Roy
we also need to divide the summation of the squares by the total no of elements. Can someone point out where are we dividing the summation before taking the square root?

Thanks in advance


Hello ,

In this step niks18 divided by n ( after approximation )

Step 3: take the average of Step 2 =kn2n=kn
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chetan2u VeritasKarishma
Hi, In your explanation : √=1^2+0+1^2−−−−−−−−−√2=1^2+0+1^2, so 3 elements...
shouldnt we divide the sum of the squares of the terms by the no of terms also...
In gthat case for 3 terms 4,5,6...
SD= √[(1^2+0+1^2)/3]...gives √(2/3)...

Please explain..
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souvonik2k
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 VeritasKarishma avigutman Bunuel

Can you please provide easy solution for this problem? I am not able to use information given in statement 2.
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AkhilAggarwal


Can you please provide easy solution for this problem? I am not able to use information given in statement 2.

AkhilAggarwal this problem doesn't really require a 'solution', as it's not a math question.
I like to think of standard deviation as a measurement of the sizes of the gaps among a set of numbers on the number line. A small standard deviation means the data points are close together, and a large standard deviation means the data points are spread out with large gaps among them.
In this case we know from the free info that we have a set of consecutive integers, so the gaps among them on the number line are essentially given. The only missing piece there is the number of data points. Think about it: the more consecutive integers we have, the larger the gaps among them, on average. If the set had only a single data point, there would be no gaps at all, and the standard deviation would be zero (and a range of zero). If there are 100 data points, you'll have some very large gaps (the largest of which is the range = 99).
With all of that in mind, what is statement (2) really telling us? It's enabling us to find the number of data points in the set.
There's a really important takeaway here, applicable to MANY DS problems:
If a statement gives information that leads to a single possibility, there's no need to compute that possibility. Another, much simpler example of this principle:
If x is a positive number, what is the value of x?
(1) x^2 = 3,492
This statement is sufficient, and I needn't (I mustn't) actually compute the value of x... I know that x will be located exactly root(3,492) away from zero, and from the free info I know which side of zero it will be on (the right side). Statement (1) in my example, together with the free info, leads to a single possible value of x, so I don't need to compute that value.
In conclusion, in the original problem, statement (2) provides us with a means to figure out the number of terms in set A (and we shouldn't worry about actually figuring out what that number is).
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souvonik2k
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 \(\sqrt{2}\)

Question: Median value = ?

Given Information: Gaps between terms is 1 as the terms are consecutive integers

Statement 1: The smallest number in set A is 4.
Number of terms is unknown which is detrimental to find the median
NOT SUFFICIENT

Statement 2: The standard deviation of all the numbers in set A is \(\sqrt{2}\)
i.e. there are 5 consecutive terms in the set but to find median we need one reference value which we don't have in second statement hence
NOT SUFFICIENT

Combining the statement
1)We know that first term is 4
2) We know that there is a fixed number of terms (5 terms) in teh set for given SD=√2 of consecutive numbers hence
SUFFICIENT

ANswer: Option C
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For any two sets consisting of the same numbers of consecutive integers, their standard deviations are equal.
Each set with different numbers of consecutive integers has a unique standard deviation.

e.g.
SD{0, 1, 2} = SD{1878, 1879, 1880} = 0.816
SD{4, 5, 6, 7, 8} = SD{55, 56, 57, 58, 59} = 1.414

Now, we know the standard deviation(statement 2) of this set. Theoretically, the length of this set is fixed.
And we also know the first number(statement 1) in this set. So, the median is given.
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What about the total number of terms here is it 3 or 5 ?


after searching I found the formula as std_dev=\sqrt{ (n^2 -1)/12 } for consecutive terms


so the concept is for equally spaced set {Arithmetic Progression} if we know the standard deviation then we can find out the total number of terms there.
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Nsp10
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 \(\sqrt{2}\)

What about the total number of terms here is it 3 or 5 ?

Five. The set of consecutive integers with the smallest term 4 and a standard deviation of \(\sqrt{2}\) is {4, 5, 6, 7, 8}. However, as mentioned above, any set of five consecutive integers will have the same standard deviation. For example, {-11, -10, -9, -8, -7} also has a standard deviation of \(\sqrt{2}\).
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