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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

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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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Debashis Roy
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..


Yes, the SD will be \(\sqrt{2} = \sqrt{\frac{(-2)^2 + (-1)^2 + 0 + 1^2 + 2^2}{5}}\)

In any case, the answer doesn't change. Each unique distribution will have a unique SD. For median, you will need the exact position where the distribution is placed on the number line so you need both statements to answer the question.

Answer (C)
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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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GMATinsight
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



How did you calculate the number of terms as 5? Can you please elaborate?

Posted from my mobile device
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i done it with the help of range and standard deviation formula!
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chetan2u
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 \((2)^½\)

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hi..


(1) The smallest number in set A is 4.
we just know the smallest number .
we require to know the number of items or the largest number to know the median
insuff

(2) The standard deviation of all the numbers in set A is \((2)^½\)
since theterms are consecutive, we can know the number of elements in set..
\(\sqrt{2}=\sqrt{1^2+0+1^2}\), so 3 elements
but where are these 3 elements...
insuff

combined
3 elements starting with 4..
4,5,6
median 5
suff

c

souvonik2k, in response to your query mentioned in post below

the consecutive numbers have a difference of 1 ...
so two cases ..
1) ODD numbers in the list..
MEDIAN is also the MEAN..
SD depends on how each element is away from median..
If 3 elements, say 5,6,7,.... middle is 0 away from mean, the smaller and bigger are 1 away from mean so SD = \(\sqrt{1^2+0+1^2}=\sqrt{2}\)
If 5 elements say 2,3,4,5,6... middle is 0 away, the biggest and smallest are 2 away from mean and other two 1 away, so SD =\(\sqrt{2^2+1^2+0+1^2+2^2}=\sqrt{10}\)

and so on, it will keep increasing with more elements added
2) EVEN numbers in the list
median=mean = average of centre two numbers
if 2 elements.. 3,4...SD = \(\sqrt{(1/2)^2+(1/2)^2}=\sqrt{1/2}\)


hope it helps
­Hi,

When we calculate standard deviation, shouldn't we also divide by the size of the population. So, based on the example you provided, std should be equal to (2/3)^0.5 instead of (2)^0.5?
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nijat00
chetan2u
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 \((2)^½\)

Please give kudos, if u liked my post!
hi..


(1) The smallest number in set A is 4.
we just know the smallest number .
we require to know the number of items or the largest number to know the median
insuff

(2) The standard deviation of all the numbers in set A is \((2)^½\)
since theterms are consecutive, we can know the number of elements in set..
\(\sqrt{2}=\sqrt{1^2+0+1^2}\), so 3 elements
but where are these 3 elements...
insuff

combined
3 elements starting with 4..
4,5,6
median 5
suff

c

souvonik2k, in response to your query mentioned in post below

the consecutive numbers have a difference of 1 ...
so two cases ..
1) ODD numbers in the list..
MEDIAN is also the MEAN..
SD depends on how each element is away from median..
If 3 elements, say 5,6,7,.... middle is 0 away from mean, the smaller and bigger are 1 away from mean so SD = \(\sqrt{1^2+0+1^2}=\sqrt{2}\)
If 5 elements say 2,3,4,5,6... middle is 0 away, the biggest and smallest are 2 away from mean and other two 1 away, so SD =\(\sqrt{2^2+1^2+0+1^2+2^2}=\sqrt{10}\)

and so on, it will keep increasing with more elements added
2) EVEN numbers in the list
median=mean = average of centre two numbers
if 2 elements.. 3,4...SD = \(\sqrt{(1/2)^2+(1/2)^2}=\sqrt{1/2}\)


hope it helps
­Hi,

When we calculate standard deviation, shouldn't we also divide by the size of the population. So, based on the example you provided, std should be equal to (2/3)^0.5 instead of (2)^0.5?
­Your doubt has laready been addressed here: 
https://gmatclub.com/forum/set-a-consis ... l#p2234518

Hope this helps.
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chetan2u
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 \((2)^1⁄2\)

Please give kudos, if u liked my post!

hi..


(1) The smallest number in set A is 4.
we just know the smallest number .
we require to know the number of items or the largest number to know the median
insuff

(2) The standard deviation of all the numbers in set A is \((2)^1⁄2\)
since theterms are consecutive, we can know the number of elements in set..
\(\sqrt{2}=\sqrt{1^2+0+1^2}\), so 3 elements
but where are these 3 elements...
insuff

combined
3 elements starting with 4..
4,5,6
median 5
suff

c

souvonik2k, in response to your query mentioned in post below

the consecutive numbers have a difference of 1 ...
so two cases ..
1) ODD numbers in the list..
MEDIAN is also the MEAN..
SD depends on how each element is away from median..
If 3 elements, say 5,6,7,.... middle is 0 away from mean, the smaller and bigger are 1 away from mean so SD = \(\sqrt{1^2+0+1^2}=\sqrt{2}\)
If 5 elements say 2,3,4,5,6... middle is 0 away, the biggest and smallest are 2 away from mean and other two 1 away, so SD =\(\sqrt{2^2+1^2+0+1^2+2^2}=\sqrt{10}\)

and so on, it will keep increasing with more elements added
2) EVEN numbers in the list
median=mean = average of centre two numbers
if 2 elements.. 3,4...SD = \(\sqrt{(1/2)^2+(1/2)^2}=\sqrt{1/2}\)


hope it helps
­chetan2u am I wrong here ?
we are calculating the std deviation right?
isnt it root of (sum of sq of difference of each number with mean / n)
shouldnt the numbers be 4,5,6,7,8 ?
sqrt (4+1+0+1+4/ 5) ?
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nisen20
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.
Great explanation!
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